means test null hypothesis

9.1 Null and Alternative Hypotheses

The actual test begins by considering two hypotheses . They are called the null hypothesis and the alternative hypothesis . These hypotheses contain opposing viewpoints.

H 0 , the — null hypothesis: a statement of no difference between sample means or proportions or no difference between a sample mean or proportion and a population mean or proportion. In other words, the difference equals 0.

H a —, the alternative hypothesis: a claim about the population that is contradictory to H 0 and what we conclude when we reject H 0 .

Since the null and alternative hypotheses are contradictory, you must examine evidence to decide if you have enough evidence to reject the null hypothesis or not. The evidence is in the form of sample data.

After you have determined which hypothesis the sample supports, you make a decision. There are two options for a decision. They are reject H 0 if the sample information favors the alternative hypothesis or do not reject H 0 or decline to reject H 0 if the sample information is insufficient to reject the null hypothesis.

Mathematical Symbols Used in H 0 and H a :

H 0 always has a symbol with an equal in it. H a never has a symbol with an equal in it. The choice of symbol depends on the wording of the hypothesis test. However, be aware that many researchers use = in the null hypothesis, even with > or < as the symbol in the alternative hypothesis. This practice is acceptable because we only make the decision to reject or not reject the null hypothesis.

Example 9.1

H 0 : No more than 30 percent of the registered voters in Santa Clara County voted in the primary election. p ≤ 30 H a : More than 30 percent of the registered voters in Santa Clara County voted in the primary election. p > 30

A medical trial is conducted to test whether or not a new medicine reduces cholesterol by 25 percent. State the null and alternative hypotheses.

Example 9.2

We want to test whether the mean GPA of students in American colleges is different from 2.0 (out of 4.0). The null and alternative hypotheses are the following: H 0 : μ = 2.0 H a : μ ≠ 2.0

We want to test whether the mean height of eighth graders is 66 inches. State the null and alternative hypotheses. Fill in the correct symbol (=, ≠, ≥, <, ≤, >) for the null and alternative hypotheses.

H 0 : μ __ 66
H a : μ __ 66

Example 9.3

We want to test if college students take fewer than five years to graduate from college, on the average. The null and alternative hypotheses are the following: H 0 : μ ≥ 5 H a : μ < 5

We want to test if it takes fewer than 45 minutes to teach a lesson plan. State the null and alternative hypotheses. Fill in the correct symbol ( =, ≠, ≥, <, ≤, >) for the null and alternative hypotheses.

H 0 : μ __ 45
H a : μ __ 45

Example 9.4

An article on school standards stated that about half of all students in France, Germany, and Israel take advanced placement exams and a third of the students pass. The same article stated that 6.6 percent of U.S. students take advanced placement exams and 4.4 percent pass. Test if the percentage of U.S. students who take advanced placement exams is more than 6.6 percent. State the null and alternative hypotheses. H 0 : p ≤ 0.066 H a : p > 0.066

On a state driver’s test, about 40 percent pass the test on the first try. We want to test if more than 40 percent pass on the first try. Fill in the correct symbol (=, ≠, ≥, <, ≤, >) for the null and alternative hypotheses.

H 0 : p __ 0.40
H a : p __ 0.40

Collaborative Exercise

Bring to class a newspaper, some news magazines, and some internet articles. In groups, find articles from which your group can write null and alternative hypotheses. Discuss your hypotheses with the rest of the class.

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Teach yourself statistics

Hypothesis Test for a Mean

This lesson explains how to conduct a hypothesis test of a mean, when the following conditions are met:

The sampling method is simple random sampling .
The sampling distribution is normal or nearly normal.

Generally, the sampling distribution will be approximately normally distributed if any of the following conditions apply.

The population distribution is normal.
The population distribution is symmetric , unimodal , without outliers , and the sample size is 15 or less.
The population distribution is moderately skewed , unimodal, without outliers, and the sample size is between 16 and 40.
The sample size is greater than 40, without outliers.

This approach consists of four steps: (1) state the hypotheses, (2) formulate an analysis plan, (3) analyze sample data, and (4) interpret results.

State the Hypotheses

Every hypothesis test requires the analyst to state a null hypothesis and an alternative hypothesis . The hypotheses are stated in such a way that they are mutually exclusive. That is, if one is true, the other must be false; and vice versa.

The table below shows three sets of hypotheses. Each makes a statement about how the population mean μ is related to a specified value M . (In the table, the symbol ≠ means " not equal to ".)

The first set of hypotheses (Set 1) is an example of a two-tailed test , since an extreme value on either side of the sampling distribution would cause a researcher to reject the null hypothesis. The other two sets of hypotheses (Sets 2 and 3) are one-tailed tests , since an extreme value on only one side of the sampling distribution would cause a researcher to reject the null hypothesis.

Formulate an Analysis Plan

The analysis plan describes how to use sample data to accept or reject the null hypothesis. It should specify the following elements.

Significance level. Often, researchers choose significance levels equal to 0.01, 0.05, or 0.10; but any value between 0 and 1 can be used.
Test method. Use the one-sample t-test to determine whether the hypothesized mean differs significantly from the observed sample mean.

Analyze Sample Data

Using sample data, conduct a one-sample t-test. This involves finding the standard error, degrees of freedom, test statistic, and the P-value associated with the test statistic.

SE = s * sqrt{ ( 1/n ) * [ ( N - n ) / ( N - 1 ) ] }

SE = s / sqrt( n )

Degrees of freedom. The degrees of freedom (DF) is equal to the sample size (n) minus one. Thus, DF = n - 1.

t = ( x - μ) / SE

P-value. The P-value is the probability of observing a sample statistic as extreme as the test statistic. Since the test statistic is a t statistic, use the t Distribution Calculator to assess the probability associated with the t statistic, given the degrees of freedom computed above. (See sample problems at the end of this lesson for examples of how this is done.)

Sample Size Calculator

As you probably noticed, the process of hypothesis testing can be complex. When you need to test a hypothesis about a mean score, consider using the Sample Size Calculator. The calculator is fairly easy to use, and it is free. You can find the Sample Size Calculator in Stat Trek's main menu under the Stat Tools tab. Or you can tap the button below.

Interpret Results

If the sample findings are unlikely, given the null hypothesis, the researcher rejects the null hypothesis. Typically, this involves comparing the P-value to the significance level , and rejecting the null hypothesis when the P-value is less than the significance level.

Test Your Understanding

In this section, two sample problems illustrate how to conduct a hypothesis test of a mean score. The first problem involves a two-tailed test; the second problem, a one-tailed test.

Problem 1: Two-Tailed Test

An inventor has developed a new, energy-efficient lawn mower engine. He claims that the engine will run continuously for 5 hours (300 minutes) on a single gallon of regular gasoline. From his stock of 2000 engines, the inventor selects a simple random sample of 50 engines for testing. The engines run for an average of 295 minutes, with a standard deviation of 20 minutes. Test the null hypothesis that the mean run time is 300 minutes against the alternative hypothesis that the mean run time is not 300 minutes. Use a 0.05 level of significance. (Assume that run times for the population of engines are normally distributed.)

Solution: The solution to this problem takes four steps: (1) state the hypotheses, (2) formulate an analysis plan, (3) analyze sample data, and (4) interpret results. We work through those steps below:

Null hypothesis: μ = 300

Alternative hypothesis: μ ≠ 300

Formulate an analysis plan . For this analysis, the significance level is 0.05. The test method is a one-sample t-test .

SE = s / sqrt(n) = 20 / sqrt(50) = 20/7.07 = 2.83

DF = n - 1 = 50 - 1 = 49

t = ( x - μ) / SE = (295 - 300)/2.83 = -1.77

where s is the standard deviation of the sample, x is the sample mean, μ is the hypothesized population mean, and n is the sample size.

Since we have a two-tailed test , the P-value is the probability that the t statistic having 49 degrees of freedom is less than -1.77 or greater than 1.77. We use the t Distribution Calculator to find P(t < -1.77) is about 0.04.

If you enter 1.77 as the sample mean in the t Distribution Calculator, you will find the that the P(t < 1.77) is about 0.04. Therefore, P(t > 1.77) is 1 minus 0.96 or 0.04. Thus, the P-value = 0.04 + 0.04 = 0.08.
Interpret results . Since the P-value (0.08) is greater than the significance level (0.05), we cannot reject the null hypothesis.

Note: If you use this approach on an exam, you may also want to mention why this approach is appropriate. Specifically, the approach is appropriate because the sampling method was simple random sampling, the population was normally distributed, and the sample size was small relative to the population size (less than 5%).

Problem 2: One-Tailed Test

Bon Air Elementary School has 1000 students. The principal of the school thinks that the average IQ of students at Bon Air is at least 110. To prove her point, she administers an IQ test to 20 randomly selected students. Among the sampled students, the average IQ is 108 with a standard deviation of 10. Based on these results, should the principal accept or reject her original hypothesis? Assume a significance level of 0.01. (Assume that test scores in the population of engines are normally distributed.)

Null hypothesis: μ >= 110

Alternative hypothesis: μ < 110

Formulate an analysis plan . For this analysis, the significance level is 0.01. The test method is a one-sample t-test .

SE = s / sqrt(n) = 10 / sqrt(20) = 10/4.472 = 2.236

DF = n - 1 = 20 - 1 = 19

t = ( x - μ) / SE = (108 - 110)/2.236 = -0.894

Here is the logic of the analysis: Given the alternative hypothesis (μ < 110), we want to know whether the observed sample mean is small enough to cause us to reject the null hypothesis.

The observed sample mean produced a t statistic test statistic of -0.894. We use the t Distribution Calculator to find P(t < -0.894) is about 0.19.

This means we would expect to find a sample mean of 108 or smaller in 19 percent of our samples, if the true population IQ were 110. Thus the P-value in this analysis is 0.19.
Interpret results . Since the P-value (0.19) is greater than the significance level (0.01), we cannot reject the null hypothesis.

13.1 Understanding Null Hypothesis Testing

Learning objectives.

Explain the purpose of null hypothesis testing, including the role of sampling error.
Describe the basic logic of null hypothesis testing.
Describe the role of relationship strength and sample size in determining statistical significance and make reasonable judgments about statistical significance based on these two factors.

The Purpose of Null Hypothesis Testing

As we have seen, psychological research typically involves measuring one or more variables in a sample and computing descriptive statistics for that sample. In general, however, the researcher’s goal is not to draw conclusions about that sample but to draw conclusions about the population that the sample was selected from. Thus researchers must use sample statistics to draw conclusions about the corresponding values in the population. These corresponding values in the population are called parameters . Imagine, for example, that a researcher measures the number of depressive symptoms exhibited by each of 50 adults with clinical depression and computes the mean number of symptoms. The researcher probably wants to use this sample statistic (the mean number of symptoms for the sample) to draw conclusions about the corresponding population parameter (the mean number of symptoms for adults with clinical depression).

Unfortunately, sample statistics are not perfect estimates of their corresponding population parameters. This is because there is a certain amount of random variability in any statistic from sample to sample. The mean number of depressive symptoms might be 8.73 in one sample of adults with clinical depression, 6.45 in a second sample, and 9.44 in a third—even though these samples are selected randomly from the same population. Similarly, the correlation (Pearson’s r ) between two variables might be +.24 in one sample, −.04 in a second sample, and +.15 in a third—again, even though these samples are selected randomly from the same population. This random variability in a statistic from sample to sample is called sampling error . (Note that the term error here refers to random variability and does not imply that anyone has made a mistake. No one “commits a sampling error.”)

One implication of this is that when there is a statistical relationship in a sample, it is not always clear that there is a statistical relationship in the population. A small difference between two group means in a sample might indicate that there is a small difference between the two group means in the population. But it could also be that there is no difference between the means in the population and that the difference in the sample is just a matter of sampling error. Similarly, a Pearson’s r value of −.29 in a sample might mean that there is a negative relationship in the population. But it could also be that there is no relationship in the population and that the relationship in the sample is just a matter of sampling error.

In fact, any statistical relationship in a sample can be interpreted in two ways:

There is a relationship in the population, and the relationship in the sample reflects this.
There is no relationship in the population, and the relationship in the sample reflects only sampling error.

The purpose of null hypothesis testing is simply to help researchers decide between these two interpretations.

The Logic of Null Hypothesis Testing

Null hypothesis testing is a formal approach to deciding between two interpretations of a statistical relationship in a sample. One interpretation is called the null hypothesis (often symbolized H 0 and read as “H-naught”). This is the idea that there is no relationship in the population and that the relationship in the sample reflects only sampling error. Informally, the null hypothesis is that the sample relationship “occurred by chance.” The other interpretation is called the alternative hypothesis (often symbolized as H 1 ). This is the idea that there is a relationship in the population and that the relationship in the sample reflects this relationship in the population.

Again, every statistical relationship in a sample can be interpreted in either of these two ways: It might have occurred by chance, or it might reflect a relationship in the population. So researchers need a way to decide between them. Although there are many specific null hypothesis testing techniques, they are all based on the same general logic. The steps are as follows:

Assume for the moment that the null hypothesis is true. There is no relationship between the variables in the population.
Determine how likely the sample relationship would be if the null hypothesis were true.
If the sample relationship would be extremely unlikely, then reject the null hypothesis in favor of the alternative hypothesis. If it would not be extremely unlikely, then retain the null hypothesis .

Following this logic, we can begin to understand why Mehl and his colleagues concluded that there is no difference in talkativeness between women and men in the population. In essence, they asked the following question: “If there were no difference in the population, how likely is it that we would find a small difference of d = 0.06 in our sample?” Their answer to this question was that this sample relationship would be fairly likely if the null hypothesis were true. Therefore, they retained the null hypothesis—concluding that there is no evidence of a sex difference in the population. We can also see why Kanner and his colleagues concluded that there is a correlation between hassles and symptoms in the population. They asked, “If the null hypothesis were true, how likely is it that we would find a strong correlation of +.60 in our sample?” Their answer to this question was that this sample relationship would be fairly unlikely if the null hypothesis were true. Therefore, they rejected the null hypothesis in favor of the alternative hypothesis—concluding that there is a positive correlation between these variables in the population.

A crucial step in null hypothesis testing is finding the likelihood of the sample result if the null hypothesis were true. This probability is called the p value . A low p value means that the sample result would be unlikely if the null hypothesis were true and leads to the rejection of the null hypothesis. A p value that is not low means that the sample result would be likely if the null hypothesis were true and leads to the retention of the null hypothesis. But how low must the p value be before the sample result is considered unlikely enough to reject the null hypothesis? In null hypothesis testing, this criterion is called α (alpha) and is almost always set to .05. If there is a 5% chance or less of a result as extreme as the sample result if the null hypothesis were true, then the null hypothesis is rejected. When this happens, the result is said to be statistically significant . If there is greater than a 5% chance of a result as extreme as the sample result when the null hypothesis is true, then the null hypothesis is retained. This does not necessarily mean that the researcher accepts the null hypothesis as true—only that there is not currently enough evidence to reject it. Researchers often use the expression “fail to reject the null hypothesis” rather than “retain the null hypothesis,” but they never use the expression “accept the null hypothesis.”

The Misunderstood p Value

The p value is one of the most misunderstood quantities in psychological research (Cohen, 1994) [1] . Even professional researchers misinterpret it, and it is not unusual for such misinterpretations to appear in statistics textbooks!

The most common misinterpretation is that the p value is the probability that the null hypothesis is true—that the sample result occurred by chance. For example, a misguided researcher might say that because the p value is .02, there is only a 2% chance that the result is due to chance and a 98% chance that it reflects a real relationship in the population. But this is incorrect . The p value is really the probability of a result at least as extreme as the sample result if the null hypothesis were true. So a p value of .02 means that if the null hypothesis were true, a sample result this extreme would occur only 2% of the time.

You can avoid this misunderstanding by remembering that the p value is not the probability that any particular hypothesis is true or false. Instead, it is the probability of obtaining the sample result if the null hypothesis were true.

“Null Hypothesis” retrieved from http://imgs.xkcd.com/comics/null_hypothesis.png (CC-BY-NC 2.5)

Role of Sample Size and Relationship Strength

Recall that null hypothesis testing involves answering the question, “If the null hypothesis were true, what is the probability of a sample result as extreme as this one?” In other words, “What is the p value?” It can be helpful to see that the answer to this question depends on just two considerations: the strength of the relationship and the size of the sample. Specifically, the stronger the sample relationship and the larger the sample, the less likely the result would be if the null hypothesis were true. That is, the lower the p value. This should make sense. Imagine a study in which a sample of 500 women is compared with a sample of 500 men in terms of some psychological characteristic, and Cohen’s d is a strong 0.50. If there were really no sex difference in the population, then a result this strong based on such a large sample should seem highly unlikely. Now imagine a similar study in which a sample of three women is compared with a sample of three men, and Cohen’s d is a weak 0.10. If there were no sex difference in the population, then a relationship this weak based on such a small sample should seem likely. And this is precisely why the null hypothesis would be rejected in the first example and retained in the second.

Of course, sometimes the result can be weak and the sample large, or the result can be strong and the sample small. In these cases, the two considerations trade off against each other so that a weak result can be statistically significant if the sample is large enough and a strong relationship can be statistically significant even if the sample is small. Table 13.1 shows roughly how relationship strength and sample size combine to determine whether a sample result is statistically significant. The columns of the table represent the three levels of relationship strength: weak, medium, and strong. The rows represent four sample sizes that can be considered small, medium, large, and extra large in the context of psychological research. Thus each cell in the table represents a combination of relationship strength and sample size. If a cell contains the word Yes , then this combination would be statistically significant for both Cohen’s d and Pearson’s r . If it contains the word No , then it would not be statistically significant for either. There is one cell where the decision for d and r would be different and another where it might be different depending on some additional considerations, which are discussed in Section 13.2 “Some Basic Null Hypothesis Tests”

Although Table 13.1 provides only a rough guideline, it shows very clearly that weak relationships based on medium or small samples are never statistically significant and that strong relationships based on medium or larger samples are always statistically significant. If you keep this lesson in mind, you will often know whether a result is statistically significant based on the descriptive statistics alone. It is extremely useful to be able to develop this kind of intuitive judgment. One reason is that it allows you to develop expectations about how your formal null hypothesis tests are going to come out, which in turn allows you to detect problems in your analyses. For example, if your sample relationship is strong and your sample is medium, then you would expect to reject the null hypothesis. If for some reason your formal null hypothesis test indicates otherwise, then you need to double-check your computations and interpretations. A second reason is that the ability to make this kind of intuitive judgment is an indication that you understand the basic logic of this approach in addition to being able to do the computations.

Statistical Significance Versus Practical Significance

Table 13.1 illustrates another extremely important point. A statistically significant result is not necessarily a strong one. Even a very weak result can be statistically significant if it is based on a large enough sample. This is closely related to Janet Shibley Hyde’s argument about sex differences (Hyde, 2007) [2] . The differences between women and men in mathematical problem solving and leadership ability are statistically significant. But the word significant can cause people to interpret these differences as strong and important—perhaps even important enough to influence the college courses they take or even who they vote for. As we have seen, however, these statistically significant differences are actually quite weak—perhaps even “trivial.”

This is why it is important to distinguish between the statistical significance of a result and the practical significance of that result. Practical significance refers to the importance or usefulness of the result in some real-world context. Many sex differences are statistically significant—and may even be interesting for purely scientific reasons—but they are not practically significant. In clinical practice, this same concept is often referred to as “clinical significance.” For example, a study on a new treatment for social phobia might show that it produces a statistically significant positive effect. Yet this effect still might not be strong enough to justify the time, effort, and other costs of putting it into practice—especially if easier and cheaper treatments that work almost as well already exist. Although statistically significant, this result would be said to lack practical or clinical significance.

“Conditional Risk” retrieved from http://imgs.xkcd.com/comics/conditional_risk.png (CC-BY-NC 2.5)

Key Takeaways

Null hypothesis testing is a formal approach to deciding whether a statistical relationship in a sample reflects a real relationship in the population or is just due to chance.
The logic of null hypothesis testing involves assuming that the null hypothesis is true, finding how likely the sample result would be if this assumption were correct, and then making a decision. If the sample result would be unlikely if the null hypothesis were true, then it is rejected in favor of the alternative hypothesis. If it would not be unlikely, then the null hypothesis is retained.
The probability of obtaining the sample result if the null hypothesis were true (the p value) is based on two considerations: relationship strength and sample size. Reasonable judgments about whether a sample relationship is statistically significant can often be made by quickly considering these two factors.
Statistical significance is not the same as relationship strength or importance. Even weak relationships can be statistically significant if the sample size is large enough. It is important to consider relationship strength and the practical significance of a result in addition to its statistical significance.
Discussion: Imagine a study showing that people who eat more broccoli tend to be happier. Explain for someone who knows nothing about statistics why the researchers would conduct a null hypothesis test.
The correlation between two variables is r = −.78 based on a sample size of 137.
The mean score on a psychological characteristic for women is 25 ( SD = 5) and the mean score for men is 24 ( SD = 5). There were 12 women and 10 men in this study.
In a memory experiment, the mean number of items recalled by the 40 participants in Condition A was 0.50 standard deviations greater than the mean number recalled by the 40 participants in Condition B.
In another memory experiment, the mean scores for participants in Condition A and Condition B came out exactly the same!
A student finds a correlation of r = .04 between the number of units the students in his research methods class are taking and the students’ level of stress.
Cohen, J. (1994). The world is round: p < .05. American Psychologist, 49 , 997–1003. ↵
Hyde, J. S. (2007). New directions in the study of gender similarities and differences. Current Directions in Psychological Science, 16 , 259–263. ↵

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Hypothesis Testing (cont...)

Hypothesis testing, the null and alternative hypothesis.

In order to undertake hypothesis testing you need to express your research hypothesis as a null and alternative hypothesis. The null hypothesis and alternative hypothesis are statements regarding the differences or effects that occur in the population. You will use your sample to test which statement (i.e., the null hypothesis or alternative hypothesis) is most likely (although technically, you test the evidence against the null hypothesis). So, with respect to our teaching example, the null and alternative hypothesis will reflect statements about all statistics students on graduate management courses.

The null hypothesis is essentially the "devil's advocate" position. That is, it assumes that whatever you are trying to prove did not happen ( hint: it usually states that something equals zero). For example, the two different teaching methods did not result in different exam performances (i.e., zero difference). Another example might be that there is no relationship between anxiety and athletic performance (i.e., the slope is zero). The alternative hypothesis states the opposite and is usually the hypothesis you are trying to prove (e.g., the two different teaching methods did result in different exam performances). Initially, you can state these hypotheses in more general terms (e.g., using terms like "effect", "relationship", etc.), as shown below for the teaching methods example:

Depending on how you want to "summarize" the exam performances will determine how you might want to write a more specific null and alternative hypothesis. For example, you could compare the mean exam performance of each group (i.e., the "seminar" group and the "lectures-only" group). This is what we will demonstrate here, but other options include comparing the distributions , medians , amongst other things. As such, we can state:

Now that you have identified the null and alternative hypotheses, you need to find evidence and develop a strategy for declaring your "support" for either the null or alternative hypothesis. We can do this using some statistical theory and some arbitrary cut-off points. Both these issues are dealt with next.

Significance levels

The level of statistical significance is often expressed as the so-called p -value . Depending on the statistical test you have chosen, you will calculate a probability (i.e., the p -value) of observing your sample results (or more extreme) given that the null hypothesis is true . Another way of phrasing this is to consider the probability that a difference in a mean score (or other statistic) could have arisen based on the assumption that there really is no difference. Let us consider this statement with respect to our example where we are interested in the difference in mean exam performance between two different teaching methods. If there really is no difference between the two teaching methods in the population (i.e., given that the null hypothesis is true), how likely would it be to see a difference in the mean exam performance between the two teaching methods as large as (or larger than) that which has been observed in your sample?

So, you might get a p -value such as 0.03 (i.e., p = .03). This means that there is a 3% chance of finding a difference as large as (or larger than) the one in your study given that the null hypothesis is true. However, you want to know whether this is "statistically significant". Typically, if there was a 5% or less chance (5 times in 100 or less) that the difference in the mean exam performance between the two teaching methods (or whatever statistic you are using) is as different as observed given the null hypothesis is true, you would reject the null hypothesis and accept the alternative hypothesis. Alternately, if the chance was greater than 5% (5 times in 100 or more), you would fail to reject the null hypothesis and would not accept the alternative hypothesis. As such, in this example where p = .03, we would reject the null hypothesis and accept the alternative hypothesis. We reject it because at a significance level of 0.03 (i.e., less than a 5% chance), the result we obtained could happen too frequently for us to be confident that it was the two teaching methods that had an effect on exam performance.

Whilst there is relatively little justification why a significance level of 0.05 is used rather than 0.01 or 0.10, for example, it is widely used in academic research. However, if you want to be particularly confident in your results, you can set a more stringent level of 0.01 (a 1% chance or less; 1 in 100 chance or less).

One- and two-tailed predictions

When considering whether we reject the null hypothesis and accept the alternative hypothesis, we need to consider the direction of the alternative hypothesis statement. For example, the alternative hypothesis that was stated earlier is:

The alternative hypothesis tells us two things. First, what predictions did we make about the effect of the independent variable(s) on the dependent variable(s)? Second, what was the predicted direction of this effect? Let's use our example to highlight these two points.

Sarah predicted that her teaching method (independent variable: teaching method), whereby she not only required her students to attend lectures, but also seminars, would have a positive effect (that is, increased) students' performance (dependent variable: exam marks). If an alternative hypothesis has a direction (and this is how you want to test it), the hypothesis is one-tailed. That is, it predicts direction of the effect. If the alternative hypothesis has stated that the effect was expected to be negative, this is also a one-tailed hypothesis.

Alternatively, a two-tailed prediction means that we do not make a choice over the direction that the effect of the experiment takes. Rather, it simply implies that the effect could be negative or positive. If Sarah had made a two-tailed prediction, the alternative hypothesis might have been:

In other words, we simply take out the word "positive", which implies the direction of our effect. In our example, making a two-tailed prediction may seem strange. After all, it would be logical to expect that "extra" tuition (going to seminar classes as well as lectures) would either have a positive effect on students' performance or no effect at all, but certainly not a negative effect. However, this is just our opinion (and hope) and certainly does not mean that we will get the effect we expect. Generally speaking, making a one-tail prediction (i.e., and testing for it this way) is frowned upon as it usually reflects the hope of a researcher rather than any certainty that it will happen. Notable exceptions to this rule are when there is only one possible way in which a change could occur. This can happen, for example, when biological activity/presence in measured. That is, a protein might be "dormant" and the stimulus you are using can only possibly "wake it up" (i.e., it cannot possibly reduce the activity of a "dormant" protein). In addition, for some statistical tests, one-tailed tests are not possible.

Rejecting or failing to reject the null hypothesis

Let's return finally to the question of whether we reject or fail to reject the null hypothesis.

If our statistical analysis shows that the significance level is below the cut-off value we have set (e.g., either 0.05 or 0.01), we reject the null hypothesis and accept the alternative hypothesis. Alternatively, if the significance level is above the cut-off value, we fail to reject the null hypothesis and cannot accept the alternative hypothesis. You should note that you cannot accept the null hypothesis, but only find evidence against it.

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AP®︎/College Statistics

Course: ap®︎/college statistics > unit 10.

Idea behind hypothesis testing

Examples of null and alternative hypotheses

Writing null and alternative hypotheses
P-values and significance tests
Comparing P-values to different significance levels
Estimating a P-value from a simulation
Estimating P-values from simulations
Using P-values to make conclusions

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Video transcript

Want to create or adapt books like this? Learn more about how Pressbooks supports open publishing practices.

Chapter 13: Inferential Statistics

Understanding Null Hypothesis Testing

Learning Objectives

Explain the purpose of null hypothesis testing, including the role of sampling error.
Describe the basic logic of null hypothesis testing.
Describe the role of relationship strength and sample size in determining statistical significance and make reasonable judgments about statistical significance based on these two factors.

The Purpose of Null Hypothesis Testing

As we have seen, psychological research typically involves measuring one or more variables for a sample and computing descriptive statistics for that sample. In general, however, the researcher’s goal is not to draw conclusions about that sample but to draw conclusions about the population that the sample was selected from. Thus researchers must use sample statistics to draw conclusions about the corresponding values in the population. These corresponding values in the population are called parameters . Imagine, for example, that a researcher measures the number of depressive symptoms exhibited by each of 50 clinically depressed adults and computes the mean number of symptoms. The researcher probably wants to use this sample statistic (the mean number of symptoms for the sample) to draw conclusions about the corresponding population parameter (the mean number of symptoms for clinically depressed adults).

Unfortunately, sample statistics are not perfect estimates of their corresponding population parameters. This is because there is a certain amount of random variability in any statistic from sample to sample. The mean number of depressive symptoms might be 8.73 in one sample of clinically depressed adults, 6.45 in a second sample, and 9.44 in a third—even though these samples are selected randomly from the same population. Similarly, the correlation (Pearson’s r ) between two variables might be +.24 in one sample, −.04 in a second sample, and +.15 in a third—again, even though these samples are selected randomly from the same population. This random variability in a statistic from sample to sample is called sampling error . (Note that the term error here refers to random variability and does not imply that anyone has made a mistake. No one “commits a sampling error.”)

In fact, any statistical relationship in a sample can be interpreted in two ways:

There is a relationship in the population, and the relationship in the sample reflects this.
There is no relationship in the population, and the relationship in the sample reflects only sampling error.

The purpose of null hypothesis testing is simply to help researchers decide between these two interpretations.

The Logic of Null Hypothesis Testing

Assume for the moment that the null hypothesis is true. There is no relationship between the variables in the population.
Determine how likely the sample relationship would be if the null hypothesis were true.
If the sample relationship would be extremely unlikely, then reject the null hypothesis in favour of the alternative hypothesis. If it would not be extremely unlikely, then retain the null hypothesis .

Following this logic, we can begin to understand why Mehl and his colleagues concluded that there is no difference in talkativeness between women and men in the population. In essence, they asked the following question: “If there were no difference in the population, how likely is it that we would find a small difference of d = 0.06 in our sample?” Their answer to this question was that this sample relationship would be fairly likely if the null hypothesis were true. Therefore, they retained the null hypothesis—concluding that there is no evidence of a sex difference in the population. We can also see why Kanner and his colleagues concluded that there is a correlation between hassles and symptoms in the population. They asked, “If the null hypothesis were true, how likely is it that we would find a strong correlation of +.60 in our sample?” Their answer to this question was that this sample relationship would be fairly unlikely if the null hypothesis were true. Therefore, they rejected the null hypothesis in favour of the alternative hypothesis—concluding that there is a positive correlation between these variables in the population.

A crucial step in null hypothesis testing is finding the likelihood of the sample result if the null hypothesis were true. This probability is called the p value . A low p value means that the sample result would be unlikely if the null hypothesis were true and leads to the rejection of the null hypothesis. A high p value means that the sample result would be likely if the null hypothesis were true and leads to the retention of the null hypothesis. But how low must the p value be before the sample result is considered unlikely enough to reject the null hypothesis? In null hypothesis testing, this criterion is called α (alpha) and is almost always set to .05. If there is less than a 5% chance of a result as extreme as the sample result if the null hypothesis were true, then the null hypothesis is rejected. When this happens, the result is said to be statistically significant . If there is greater than a 5% chance of a result as extreme as the sample result when the null hypothesis is true, then the null hypothesis is retained. This does not necessarily mean that the researcher accepts the null hypothesis as true—only that there is not currently enough evidence to conclude that it is true. Researchers often use the expression “fail to reject the null hypothesis” rather than “retain the null hypothesis,” but they never use the expression “accept the null hypothesis.”

The Misunderstood p Value

Role of Sample Size and Relationship Strength

Statistical Significance Versus Practical Significance

Key Takeaways

Null hypothesis testing is a formal approach to deciding whether a statistical relationship in a sample reflects a real relationship in the population or is just due to chance.
The logic of null hypothesis testing involves assuming that the null hypothesis is true, finding how likely the sample result would be if this assumption were correct, and then making a decision. If the sample result would be unlikely if the null hypothesis were true, then it is rejected in favour of the alternative hypothesis. If it would not be unlikely, then the null hypothesis is retained.
The probability of obtaining the sample result if the null hypothesis were true (the p value) is based on two considerations: relationship strength and sample size. Reasonable judgments about whether a sample relationship is statistically significant can often be made by quickly considering these two factors.
Statistical significance is not the same as relationship strength or importance. Even weak relationships can be statistically significant if the sample size is large enough. It is important to consider relationship strength and the practical significance of a result in addition to its statistical significance.
Discussion: Imagine a study showing that people who eat more broccoli tend to be happier. Explain for someone who knows nothing about statistics why the researchers would conduct a null hypothesis test.
The correlation between two variables is r = −.78 based on a sample size of 137.
The mean score on a psychological characteristic for women is 25 ( SD = 5) and the mean score for men is 24 ( SD = 5). There were 12 women and 10 men in this study.
In a memory experiment, the mean number of items recalled by the 40 participants in Condition A was 0.50 standard deviations greater than the mean number recalled by the 40 participants in Condition B.
In another memory experiment, the mean scores for participants in Condition A and Condition B came out exactly the same!
A student finds a correlation of r = .04 between the number of units the students in his research methods class are taking and the students’ level of stress.

Long Descriptions

“Null Hypothesis” long description: A comic depicting a man and a woman talking in the foreground. In the background is a child working at a desk. The man says to the woman, “I can’t believe schools are still teaching kids about the null hypothesis. I remember reading a big study that conclusively disproved it years ago.” [Return to “Null Hypothesis”]

“Conditional Risk” long description: A comic depicting two hikers beside a tree during a thunderstorm. A bolt of lightning goes “crack” in the dark sky as thunder booms. One of the hikers says, “Whoa! We should get inside!” The other hiker says, “It’s okay! Lightning only kills about 45 Americans a year, so the chances of dying are only one in 7,000,000. Let’s go on!” The comic’s caption says, “The annual death rate among people who know that statistic is one in six.” [Return to “Conditional Risk”]

Media Attributions

Null Hypothesis by XKCD CC BY-NC (Attribution NonCommercial)
Conditional Risk by XKCD CC BY-NC (Attribution NonCommercial)
Cohen, J. (1994). The world is round: p < .05. American Psychologist, 49 , 997–1003. ↵
Hyde, J. S. (2007). New directions in the study of gender similarities and differences. Current Directions in Psychological Science, 16 , 259–263. ↵

Values in a population that correspond to variables measured in a study.

The random variability in a statistic from sample to sample.

A formal approach to deciding between two interpretations of a statistical relationship in a sample.

The idea that there is no relationship in the population and that the relationship in the sample reflects only sampling error.

The idea that there is a relationship in the population and that the relationship in the sample reflects this relationship in the population.

When the relationship found in the sample would be extremely unlikely, the idea that the relationship occurred “by chance” is rejected.

When the relationship found in the sample is likely to have occurred by chance, the null hypothesis is not rejected.

The probability that, if the null hypothesis were true, the result found in the sample would occur.

How low the p value must be before the sample result is considered unlikely in null hypothesis testing.

When there is less than a 5% chance of a result as extreme as the sample result occurring and the null hypothesis is rejected.

Research Methods in Psychology - 2nd Canadian Edition Copyright © 2015 by Paul C. Price, Rajiv Jhangiani, & I-Chant A. Chiang is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License , except where otherwise noted.

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What is The Null Hypothesis & When Do You Reject The Null Hypothesis

Julia Simkus

Editor at Simply Psychology

BA (Hons) Psychology, Princeton University

Julia Simkus is a graduate of Princeton University with a Bachelor of Arts in Psychology. She is currently studying for a Master's Degree in Counseling for Mental Health and Wellness in September 2023. Julia's research has been published in peer reviewed journals.

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Editor-in-Chief for Simply Psychology

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Saul Mcleod, PhD., is a qualified psychology teacher with over 18 years of experience in further and higher education. He has been published in peer-reviewed journals, including the Journal of Clinical Psychology.

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On This Page:

A null hypothesis is a statistical concept suggesting no significant difference or relationship between measured variables. It’s the default assumption unless empirical evidence proves otherwise.

The null hypothesis states no relationship exists between the two variables being studied (i.e., one variable does not affect the other).

The null hypothesis is the statement that a researcher or an investigator wants to disprove.

Testing the null hypothesis can tell you whether your results are due to the effects of manipulating the dependent variable or due to random chance.

How to Write a Null Hypothesis

Null hypotheses (H0) start as research questions that the investigator rephrases as statements indicating no effect or relationship between the independent and dependent variables.

It is a default position that your research aims to challenge or confirm.

For example, if studying the impact of exercise on weight loss, your null hypothesis might be:

There is no significant difference in weight loss between individuals who exercise daily and those who do not.

Examples of Null Hypotheses

When do we reject the null hypothesis .

We reject the null hypothesis when the data provide strong enough evidence to conclude that it is likely incorrect. This often occurs when the p-value (probability of observing the data given the null hypothesis is true) is below a predetermined significance level.

If the collected data does not meet the expectation of the null hypothesis, a researcher can conclude that the data lacks sufficient evidence to back up the null hypothesis, and thus the null hypothesis is rejected.

Rejecting the null hypothesis means that a relationship does exist between a set of variables and the effect is statistically significant ( p > 0.05).

If the data collected from the random sample is not statistically significance , then the null hypothesis will be accepted, and the researchers can conclude that there is no relationship between the variables.

You need to perform a statistical test on your data in order to evaluate how consistent it is with the null hypothesis. A p-value is one statistical measurement used to validate a hypothesis against observed data.

Calculating the p-value is a critical part of null-hypothesis significance testing because it quantifies how strongly the sample data contradicts the null hypothesis.

The level of statistical significance is often expressed as a p -value between 0 and 1. The smaller the p-value, the stronger the evidence that you should reject the null hypothesis.

Probability and statistical significance in ab testing. Statistical significance in a b experiments

Usually, a researcher uses a confidence level of 95% or 99% (p-value of 0.05 or 0.01) as general guidelines to decide if you should reject or keep the null.

When your p-value is less than or equal to your significance level, you reject the null hypothesis.

In other words, smaller p-values are taken as stronger evidence against the null hypothesis. Conversely, when the p-value is greater than your significance level, you fail to reject the null hypothesis.

In this case, the sample data provides insufficient data to conclude that the effect exists in the population.

Because you can never know with complete certainty whether there is an effect in the population, your inferences about a population will sometimes be incorrect.

When you incorrectly reject the null hypothesis, it’s called a type I error. When you incorrectly fail to reject it, it’s called a type II error.

Why Do We Never Accept The Null Hypothesis?

The reason we do not say “accept the null” is because we are always assuming the null hypothesis is true and then conducting a study to see if there is evidence against it. And, even if we don’t find evidence against it, a null hypothesis is not accepted.

A lack of evidence only means that you haven’t proven that something exists. It does not prove that something doesn’t exist.

It is risky to conclude that the null hypothesis is true merely because we did not find evidence to reject it. It is always possible that researchers elsewhere have disproved the null hypothesis, so we cannot accept it as true, but instead, we state that we failed to reject the null.

One can either reject the null hypothesis, or fail to reject it, but can never accept it.

Why Do We Use The Null Hypothesis?

We can never prove with 100% certainty that a hypothesis is true; We can only collect evidence that supports a theory. However, testing a hypothesis can set the stage for rejecting or accepting this hypothesis within a certain confidence level.

The null hypothesis is useful because it can tell us whether the results of our study are due to random chance or the manipulation of a variable (with a certain level of confidence).

A null hypothesis is rejected if the measured data is significantly unlikely to have occurred and a null hypothesis is accepted if the observed outcome is consistent with the position held by the null hypothesis.

Rejecting the null hypothesis sets the stage for further experimentation to see if a relationship between two variables exists.

Hypothesis testing is a critical part of the scientific method as it helps decide whether the results of a research study support a particular theory about a given population. Hypothesis testing is a systematic way of backing up researchers’ predictions with statistical analysis.

It helps provide sufficient statistical evidence that either favors or rejects a certain hypothesis about the population parameter.

Purpose of a Null Hypothesis

The primary purpose of the null hypothesis is to disprove an assumption.
Whether rejected or accepted, the null hypothesis can help further progress a theory in many scientific cases.
A null hypothesis can be used to ascertain how consistent the outcomes of multiple studies are.

Do you always need both a Null Hypothesis and an Alternative Hypothesis?

The null (H0) and alternative (Ha or H1) hypotheses are two competing claims that describe the effect of the independent variable on the dependent variable. They are mutually exclusive, which means that only one of the two hypotheses can be true.

While the null hypothesis states that there is no effect in the population, an alternative hypothesis states that there is statistical significance between two variables.

The goal of hypothesis testing is to make inferences about a population based on a sample. In order to undertake hypothesis testing, you must express your research hypothesis as a null and alternative hypothesis. Both hypotheses are required to cover every possible outcome of the study.

What is the difference between a null hypothesis and an alternative hypothesis?

The alternative hypothesis is the complement to the null hypothesis. The null hypothesis states that there is no effect or no relationship between variables, while the alternative hypothesis claims that there is an effect or relationship in the population.

It is the claim that you expect or hope will be true. The null hypothesis and the alternative hypothesis are always mutually exclusive, meaning that only one can be true at a time.

What are some problems with the null hypothesis?

One major problem with the null hypothesis is that researchers typically will assume that accepting the null is a failure of the experiment. However, accepting or rejecting any hypothesis is a positive result. Even if the null is not refuted, the researchers will still learn something new.

Why can a null hypothesis not be accepted?

We can either reject or fail to reject a null hypothesis, but never accept it. If your test fails to detect an effect, this is not proof that the effect doesn’t exist. It just means that your sample did not have enough evidence to conclude that it exists.

We can’t accept a null hypothesis because a lack of evidence does not prove something that does not exist. Instead, we fail to reject it.

Failing to reject the null indicates that the sample did not provide sufficient enough evidence to conclude that an effect exists.

If the p-value is greater than the significance level, then you fail to reject the null hypothesis.

Is a null hypothesis directional or non-directional?

A hypothesis test can either contain an alternative directional hypothesis or a non-directional alternative hypothesis. A directional hypothesis is one that contains the less than (“<“) or greater than (“>”) sign.

A nondirectional hypothesis contains the not equal sign (“≠”). However, a null hypothesis is neither directional nor non-directional.

A null hypothesis is a prediction that there will be no change, relationship, or difference between two variables.

The directional hypothesis or nondirectional hypothesis would then be considered alternative hypotheses to the null hypothesis.

Gill, J. (1999). The insignificance of null hypothesis significance testing. Political research quarterly , 52 (3), 647-674.

Krueger, J. (2001). Null hypothesis significance testing: On the survival of a flawed method. American Psychologist , 56 (1), 16.

Masson, M. E. (2011). A tutorial on a practical Bayesian alternative to null-hypothesis significance testing. Behavior research methods , 43 , 679-690.

Nickerson, R. S. (2000). Null hypothesis significance testing: a review of an old and continuing controversy. Psychological methods , 5 (2), 241.

Rozeboom, W. W. (1960). The fallacy of the null-hypothesis significance test. Psychological bulletin , 57 (5), 416.

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10.29: Hypothesis Test for a Difference in Two Population Means (1 of 2)

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Learning Objectives

Under appropriate conditions, conduct a hypothesis test about a difference between two population means. State a conclusion in context.

Using the Hypothesis Test for a Difference in Two Population Means

The general steps of this hypothesis test are the same as always. As expected, the details of the conditions for use of the test and the test statistic are unique to this test (but similar in many ways to what we have seen before.)

Step 1: Determine the hypotheses.

The hypotheses for a difference in two population means are similar to those for a difference in two population proportions. The null hypothesis, H 0 , is again a statement of “no effect” or “no difference.”

H 0 : μ 1 – μ 2 = 0, which is the same as H 0 : μ 1 = μ 2

The alternative hypothesis, H a , can be any one of the following.

H a : μ 1 – μ 2 < 0, which is the same as H a : μ 1 < μ 2
H a : μ 1 – μ 2 > 0, which is the same as H a : μ 1 > μ 2
H a : μ 1 – μ 2 ≠ 0, which is the same as H a : μ 1 ≠ μ 2

Step 2: Collect the data.

As usual, how we collect the data determines whether we can use it in the inference procedure. We have our usual two requirements for data collection.

Samples must be random to remove or minimize bias.
Samples must be representative of the populations in question.

We use this hypothesis test when the data meets the following conditions.

The two random samples are independent .
The variable is normally distributed in both populations . If this variable is not known, samples of more than 30 will have a difference in sample means that can be modeled adequately by the t-distribution. As we discussed in “Hypothesis Test for a Population Mean,” t-procedures are robust even when the variable is not normally distributed in the population. If checking normality in the populations is impossible, then we look at the distribution in the samples. If a histogram or dotplot of the data does not show extreme skew or outliers, we take it as a sign that the variable is not heavily skewed in the populations, and we use the inference procedure. (Note: This is the same condition we used for the one-sample t-test in “Hypothesis Test for a Population Mean.”)

Step 3: Assess the evidence.

If the conditions are met, then we calculate the t-test statistic. The t-test statistic has a familiar form.

Since the null hypothesis assumes there is no difference in the population means, the expression (μ 1 – μ 2 ) is always zero.

As we learned in “Estimating a Population Mean,” the t-distribution depends on the degrees of freedom (df) . In the one-sample and matched-pair cases df = n – 1. For the two-sample t-test, determining the correct df is based on a complicated formula that we do not cover in this course. We will either give the df or use technology to find the df . With the t-test statistic and the degrees of freedom, we can use the appropriate t-model to find the P-value, just as we did in “Hypothesis Test for a Population Mean.” We can even use the same simulation.

Step 4: State a conclusion.

To state a conclusion, we follow what we have done with other hypothesis tests. We compare our P-value to a stated level of significance.

If the P-value ≤ α, we reject the null hypothesis in favor of the alternative hypothesis.
If the P-value > α, we fail to reject the null hypothesis. We do not have enough evidence to support the alternative hypothesis.

As always, we state our conclusion in context, usually by referring to the alternative hypothesis.

“Context and Calories”

Does the company you keep impact what you eat? This example comes from an article titled “Impact of Group Settings and Gender on Meals Purchased by College Students” (Allen-O’Donnell, M., T. C. Nowak, K. A. Snyder, and M. D. Cottingham, Journal of Applied Social Psychology 49(9), 2011, onlinelibrary.wiley.com/doi/10.1111/j.1559-1816.2011.00804.x/full) . In this study, researchers examined this issue in the context of gender-related theories in their field. For our purposes, we look at this research more narrowly.

Step 1: Stating the hypotheses.

In the article, the authors make the following hypothesis. “The attempt to appear feminine will be empirically demonstrated by the purchase of fewer calories by women in mixed-gender groups than by women in same-gender groups.” We translate this into a simpler and narrower research question: Do women purchase fewer calories when they eat with men compared to when they eat with women?

Here the two populations are “women eating with women” (population 1) and “women eating with men” (population 2). The variable is the calories in the meal. We test the following hypotheses at the 5% level of significance.

The null hypothesis is always H 0 : μ 1 – μ 2 = 0, which is the same as H 0 : μ 1 = μ 2 .

The alternative hypothesis H a : μ 1 – μ 2 > 0, which is the same as H a : μ 1 > μ 2 .

Here μ 1 represents the mean number of calories ordered by women when they were eating with other women, and μ 2 represents the mean number of calories ordered by women when they were eating with men.

Note: It does not matter which population we label as 1 or 2, but once we decide, we have to stay consistent throughout the hypothesis test. Since we expect the number of calories to be greater for the women eating with other women, the difference is positive if “women eating with women” is population 1. If you prefer to work with positive numbers, choose the group with the larger expected mean as population 1. This is a good general tip.

Step 2: Collect Data.

As usual, there are two major things to keep in mind when considering the collection of data.

Samples need to be representative of the population in question.
Samples need to be random in order to remove or minimize bias.

Representative Samples?

The researchers state their hypothesis in terms of “women.” We did the same. But the researchers gathered data by watching people eat at the HUB Rock Café II on the campus of Indiana University of Pennsylvania during the Spring semester of 2006. Almost all of the women in the data set were white undergraduates between the ages of 18 and 24, so there are some definite limitations on the scope of this study. These limitations will affect our conclusion (and the specific definition of the population means in our hypotheses.)

Random Samples?

The observations were collected on February 13, 2006, through February 22, 2006, between 11 a.m. and 7 p.m. We can see that the researchers included both lunch and dinner. They also made observations on all days of the week to ensure that weekly customer patterns did not confound their findings. The authors state that “since the time period for observations and the place where [they] observed students were limited, the sample was a convenience sample.” Despite these limitations, the researchers conducted inference procedures with the data, and the results were published in a reputable journal. We will also conduct inference with this data, but we also include a discussion of the limitations of the study with our conclusion. The authors did this, also.

Do the data met the conditions for use of a t-test?

The researchers reported the following sample statistics.

In a sample of 45 women dining with other women, the average number of calories ordered was 850, and the standard deviation was 252.
In a sample of 27 women dining with men, the average number of calories ordered was 719, and the standard deviation was 322.

One of the samples has fewer than 30 women. We need to make sure the distribution of calories in this sample is not heavily skewed and has no outliers, but we do not have access to a spreadsheet of the actual data. Since the researchers conducted a t-test with this data, we will assume that the conditions are met. This includes the assumption that the samples are independent.

As noted previously, the researchers reported the following sample statistics.

To compute the t-test statistic, make sure sample 1 corresponds to population 1. Here our population 1 is “women eating with other women.” So x 1 = 850, s 1 = 252, n 1 =45, and so on.

Using technology, we determined that the degrees of freedom are about 45 for this data. To find the P-value, we use our familiar simulation of the t-distribution. Since the alternative hypothesis is a “greater than” statement, we look for the area to the right of T = 1.81. The P-value is 0.0385.

The green area to the left of the t value = 0.9615. The blue area to the right of the T value = 0.0385.

Generic Conclusion

The hypotheses for this test are H 0 : μ 1 – μ 2 = 0 and H a : μ 1 – μ 2 > 0. Since the P-value is less than the significance level (0.0385 < 0.05), we reject H 0 and accept H a .

Conclusion in context

At Indiana University of Pennsylvania, the mean number of calories ordered by undergraduate women eating with other women is greater than the mean number of calories ordered by undergraduate women eating with men (P-value = 0.0385).

Comment about Conclusions

In the conclusion above, we did not generalize the findings to all women. Since the samples included only undergraduate women at one university, we included this information in our conclusion. But our conclusion is a cautious statement of the findings. The authors see the results more broadly in the context of theories in the field of social psychology. In the context of these theories, they write, “Our findings support the assertion that meal size is a tool for influencing the impressions of others. For traditional-age, predominantly White college women, diminished meal size appears to be an attempt to assert femininity in groups that include men.” This viewpoint is echoed in the following summary of the study for the general public on National Public Radio (npr.org).

Both men and women appear to choose larger portions when they eat with women, and both men and women choose smaller portions when they eat in the company of men, according to new research published in the Journal of Applied Social Psychology . The study, conducted among a sample of 127 college students, suggests that both men and women are influenced by unconscious scripts about how to behave in each other’s company. And these scripts change the way men and women eat when they eat together and when they eat apart.

Should we be concerned that the findings of this study are generalized in this way? Perhaps. But the authors of the article address this concern by including the following disclaimer with their findings: “While the results of our research are suggestive, they should be replicated with larger, representative samples. Studies should be done not only with primarily White, middle-class college students, but also with students who differ in terms of race/ethnicity, social class, age, sexual orientation, and so forth.” This is an example of good statistical practice. It is often very difficult to select truly random samples from the populations of interest. Researchers therefore discuss the limitations of their sampling design when they discuss their conclusions.

In the following activities, you will have the opportunity to practice parts of the hypothesis test for a difference in two population means. On the next page, the activities focus on the entire process and also incorporate technology.

National Health and Nutrition Survey

https://assessments.lumenlearning.co...sessments/3705

https://assessments.lumenlearning.co...sessments/3782

https://assessments.lumenlearning.co...sessments/3706

Contributors and Attributions

Concepts in Statistics. Provided by : Open Learning Initiative. Located at : http://oli.cmu.edu . License : CC BY: Attribution

Hypothesis Testing for Means & Proportions

Lisa Sullivan, PhD

Professor of Biostatistics

Boston University School of Public Health

Introduction

This is the first of three modules that will addresses the second area of statistical inference, which is hypothesis testing, in which a specific statement or hypothesis is generated about a population parameter, and sample statistics are used to assess the likelihood that the hypothesis is true. The hypothesis is based on available information and the investigator's belief about the population parameters. The process of hypothesis testing involves setting up two competing hypotheses, the null hypothesis and the alternate hypothesis. One selects a random sample (or multiple samples when there are more comparison groups), computes summary statistics and then assesses the likelihood that the sample data support the research or alternative hypothesis. Similar to estimation, the process of hypothesis testing is based on probability theory and the Central Limit Theorem.

This module will focus on hypothesis testing for means and proportions. The next two modules in this series will address analysis of variance and chi-squared tests.

Learning Objectives

After completing this module, the student will be able to:

Define null and research hypothesis, test statistic, level of significance and decision rule
Distinguish between Type I and Type II errors and discuss the implications of each
Explain the difference between one and two sided tests of hypothesis
Estimate and interpret p-values
Explain the relationship between confidence interval estimates and p-values in drawing inferences
Differentiate hypothesis testing procedures based on type of outcome variable and number of sample

Introduction to Hypothesis Testing

Techniques for hypothesis testing .

The techniques for hypothesis testing depend on

the type of outcome variable being analyzed (continuous, dichotomous, discrete)
the number of comparison groups in the investigation
whether the comparison groups are independent (i.e., physically separate such as men versus women) or dependent (i.e., matched or paired such as pre- and post-assessments on the same participants).

In estimation we focused explicitly on techniques for one and two samples and discussed estimation for a specific parameter (e.g., the mean or proportion of a population), for differences (e.g., difference in means, the risk difference) and ratios (e.g., the relative risk and odds ratio). Here we will focus on procedures for one and two samples when the outcome is either continuous (and we focus on means) or dichotomous (and we focus on proportions).

General Approach: A Simple Example

The Centers for Disease Control (CDC) reported on trends in weight, height and body mass index from the 1960's through 2002. 1 The general trend was that Americans were much heavier and slightly taller in 2002 as compared to 1960; both men and women gained approximately 24 pounds, on average, between 1960 and 2002. In 2002, the mean weight for men was reported at 191 pounds. Suppose that an investigator hypothesizes that weights are even higher in 2006 (i.e., that the trend continued over the subsequent 4 years). The research hypothesis is that the mean weight in men in 2006 is more than 191 pounds. The null hypothesis is that there is no change in weight, and therefore the mean weight is still 191 pounds in 2006.

In order to test the hypotheses, we select a random sample of American males in 2006 and measure their weights. Suppose we have resources available to recruit n=100 men into our sample. We weigh each participant and compute summary statistics on the sample data. Suppose in the sample we determine the following:

Do the sample data support the null or research hypothesis? The sample mean of 197.1 is numerically higher than 191. However, is this difference more than would be expected by chance? In hypothesis testing, we assume that the null hypothesis holds until proven otherwise. We therefore need to determine the likelihood of observing a sample mean of 197.1 or higher when the true population mean is 191 (i.e., if the null hypothesis is true or under the null hypothesis). We can compute this probability using the Central Limit Theorem. Specifically,

(Notice that we use the sample standard deviation in computing the Z score. This is generally an appropriate substitution as long as the sample size is large, n > 30. Thus, there is less than a 1% probability of observing a sample mean as large as 197.1 when the true population mean is 191. Do you think that the null hypothesis is likely true? Based on how unlikely it is to observe a sample mean of 197.1 under the null hypothesis (i.e., <1% probability), we might infer, from our data, that the null hypothesis is probably not true.

Suppose that the sample data had turned out differently. Suppose that we instead observed the following in 2006:

How likely it is to observe a sample mean of 192.1 or higher when the true population mean is 191 (i.e., if the null hypothesis is true)? We can again compute this probability using the Central Limit Theorem. Specifically,

There is a 33.4% probability of observing a sample mean as large as 192.1 when the true population mean is 191. Do you think that the null hypothesis is likely true?

Neither of the sample means that we obtained allows us to know with certainty whether the null hypothesis is true or not. However, our computations suggest that, if the null hypothesis were true, the probability of observing a sample mean >197.1 is less than 1%. In contrast, if the null hypothesis were true, the probability of observing a sample mean >192.1 is about 33%. We can't know whether the null hypothesis is true, but the sample that provided a mean value of 197.1 provides much stronger evidence in favor of rejecting the null hypothesis, than the sample that provided a mean value of 192.1. Note that this does not mean that a sample mean of 192.1 indicates that the null hypothesis is true; it just doesn't provide compelling evidence to reject it.

In essence, hypothesis testing is a procedure to compute a probability that reflects the strength of the evidence (based on a given sample) for rejecting the null hypothesis. In hypothesis testing, we determine a threshold or cut-off point (called the critical value) to decide when to believe the null hypothesis and when to believe the research hypothesis. It is important to note that it is possible to observe any sample mean when the true population mean is true (in this example equal to 191), but some sample means are very unlikely. Based on the two samples above it would seem reasonable to believe the research hypothesis when x̄ = 197.1, but to believe the null hypothesis when x̄ =192.1. What we need is a threshold value such that if x̄ is above that threshold then we believe that H 1 is true and if x̄ is below that threshold then we believe that H 0 is true. The difficulty in determining a threshold for x̄ is that it depends on the scale of measurement. In this example, the threshold, sometimes called the critical value, might be 195 (i.e., if the sample mean is 195 or more then we believe that H 1 is true and if the sample mean is less than 195 then we believe that H 0 is true). Suppose we are interested in assessing an increase in blood pressure over time, the critical value will be different because blood pressures are measured in millimeters of mercury (mmHg) as opposed to in pounds. In the following we will explain how the critical value is determined and how we handle the issue of scale.

First, to address the issue of scale in determining the critical value, we convert our sample data (in particular the sample mean) into a Z score. We know from the module on probability that the center of the Z distribution is zero and extreme values are those that exceed 2 or fall below -2. Z scores above 2 and below -2 represent approximately 5% of all Z values. If the observed sample mean is close to the mean specified in H 0 (here m =191), then Z will be close to zero. If the observed sample mean is much larger than the mean specified in H 0 , then Z will be large.

In hypothesis testing, we select a critical value from the Z distribution. This is done by first determining what is called the level of significance, denoted α ("alpha"). What we are doing here is drawing a line at extreme values. The level of significance is the probability that we reject the null hypothesis (in favor of the alternative) when it is actually true and is also called the Type I error rate.

α = Level of significance = P(Type I error) = P(Reject H 0 | H 0 is true).

Because α is a probability, it ranges between 0 and 1. The most commonly used value in the medical literature for α is 0.05, or 5%. Thus, if an investigator selects α=0.05, then they are allowing a 5% probability of incorrectly rejecting the null hypothesis in favor of the alternative when the null is in fact true. Depending on the circumstances, one might choose to use a level of significance of 1% or 10%. For example, if an investigator wanted to reject the null only if there were even stronger evidence than that ensured with α=0.05, they could choose a =0.01as their level of significance. The typical values for α are 0.01, 0.05 and 0.10, with α=0.05 the most commonly used value.

Suppose in our weight study we select α=0.05. We need to determine the value of Z that holds 5% of the values above it (see below).

Standard normal distribution curve showing an upper tail at z=1.645 where alpha=0.05

The critical value of Z for α =0.05 is Z = 1.645 (i.e., 5% of the distribution is above Z=1.645). With this value we can set up what is called our decision rule for the test. The rule is to reject H 0 if the Z score is 1.645 or more.

With the first sample we have

Because 2.38 > 1.645, we reject the null hypothesis. (The same conclusion can be drawn by comparing the 0.0087 probability of observing a sample mean as extreme as 197.1 to the level of significance of 0.05. If the observed probability is smaller than the level of significance we reject H 0 ). Because the Z score exceeds the critical value, we conclude that the mean weight for men in 2006 is more than 191 pounds, the value reported in 2002. If we observed the second sample (i.e., sample mean =192.1), we would not be able to reject the null hypothesis because the Z score is 0.43 which is not in the rejection region (i.e., the region in the tail end of the curve above 1.645). With the second sample we do not have sufficient evidence (because we set our level of significance at 5%) to conclude that weights have increased. Again, the same conclusion can be reached by comparing probabilities. The probability of observing a sample mean as extreme as 192.1 is 33.4% which is not below our 5% level of significance.

Hypothesis Testing: Upper-, Lower, and Two Tailed Tests

The procedure for hypothesis testing is based on the ideas described above. Specifically, we set up competing hypotheses, select a random sample from the population of interest and compute summary statistics. We then determine whether the sample data supports the null or alternative hypotheses. The procedure can be broken down into the following five steps.

Step 1. Set up hypotheses and select the level of significance α.

H 0 : Null hypothesis (no change, no difference);

H 1 : Research hypothesis (investigator's belief); α =0.05

Step 2. Select the appropriate test statistic.

The test statistic is a single number that summarizes the sample information. An example of a test statistic is the Z statistic computed as follows:

When the sample size is small, we will use t statistics (just as we did when constructing confidence intervals for small samples). As we present each scenario, alternative test statistics are provided along with conditions for their appropriate use.

Step 3. Set up decision rule.

The decision rule is a statement that tells under what circumstances to reject the null hypothesis. The decision rule is based on specific values of the test statistic (e.g., reject H 0 if Z > 1.645). The decision rule for a specific test depends on 3 factors: the research or alternative hypothesis, the test statistic and the level of significance. Each is discussed below.

The decision rule depends on whether an upper-tailed, lower-tailed, or two-tailed test is proposed. In an upper-tailed test the decision rule has investigators reject H 0 if the test statistic is larger than the critical value. In a lower-tailed test the decision rule has investigators reject H 0 if the test statistic is smaller than the critical value. In a two-tailed test the decision rule has investigators reject H 0 if the test statistic is extreme, either larger than an upper critical value or smaller than a lower critical value.
The exact form of the test statistic is also important in determining the decision rule. If the test statistic follows the standard normal distribution (Z), then the decision rule will be based on the standard normal distribution. If the test statistic follows the t distribution, then the decision rule will be based on the t distribution. The appropriate critical value will be selected from the t distribution again depending on the specific alternative hypothesis and the level of significance.
The third factor is the level of significance. The level of significance which is selected in Step 1 (e.g., α =0.05) dictates the critical value. For example, in an upper tailed Z test, if α =0.05 then the critical value is Z=1.645.

The following figures illustrate the rejection regions defined by the decision rule for upper-, lower- and two-tailed Z tests with α=0.05. Notice that the rejection regions are in the upper, lower and both tails of the curves, respectively. The decision rules are written below each figure.

Standard normal distribution with lower tail at -1.645 and alpha=0.05

Rejection Region for Lower-Tailed Z Test (H 1 : μ < μ 0 ) with α =0.05

The decision rule is: Reject H 0 if Z < 1.645.

Standard normal distribution with two tails

Rejection Region for Two-Tailed Z Test (H 1 : μ ≠ μ 0 ) with α =0.05

The decision rule is: Reject H 0 if Z < -1.960 or if Z > 1.960.

The complete table of critical values of Z for upper, lower and two-tailed tests can be found in the table of Z values to the right in "Other Resources."

Critical values of t for upper, lower and two-tailed tests can be found in the table of t values in "Other Resources."

Step 4. Compute the test statistic.

Here we compute the test statistic by substituting the observed sample data into the test statistic identified in Step 2.

Step 5. Conclusion.

The final conclusion is made by comparing the test statistic (which is a summary of the information observed in the sample) to the decision rule. The final conclusion will be either to reject the null hypothesis (because the sample data are very unlikely if the null hypothesis is true) or not to reject the null hypothesis (because the sample data are not very unlikely).

If the null hypothesis is rejected, then an exact significance level is computed to describe the likelihood of observing the sample data assuming that the null hypothesis is true. The exact level of significance is called the p-value and it will be less than the chosen level of significance if we reject H 0 .

Statistical computing packages provide exact p-values as part of their standard output for hypothesis tests. In fact, when using a statistical computing package, the steps outlined about can be abbreviated. The hypotheses (step 1) should always be set up in advance of any analysis and the significance criterion should also be determined (e.g., α =0.05). Statistical computing packages will produce the test statistic (usually reporting the test statistic as t) and a p-value. The investigator can then determine statistical significance using the following: If p < α then reject H 0 .

Step 1. Set up hypotheses and determine level of significance

H 0 : μ = 191 H 1 : μ > 191 α =0.05

The research hypothesis is that weights have increased, and therefore an upper tailed test is used.

Step 2. Select the appropriate test statistic.

Because the sample size is large (n > 30) the appropriate test statistic is

Step 3. Set up decision rule.

In this example, we are performing an upper tailed test (H 1 : μ> 191), with a Z test statistic and selected α =0.05. Reject H 0 if Z > 1.645.

We now substitute the sample data into the formula for the test statistic identified in Step 2.

We reject H 0 because 2.38 > 1.645. We have statistically significant evidence at a =0.05, to show that the mean weight in men in 2006 is more than 191 pounds. Because we rejected the null hypothesis, we now approximate the p-value which is the likelihood of observing the sample data if the null hypothesis is true. An alternative definition of the p-value is the smallest level of significance where we can still reject H 0 . In this example, we observed Z=2.38 and for α=0.05, the critical value was 1.645. Because 2.38 exceeded 1.645 we rejected H 0 . In our conclusion we reported a statistically significant increase in mean weight at a 5% level of significance. Using the table of critical values for upper tailed tests, we can approximate the p-value. If we select α=0.025, the critical value is 1.96, and we still reject H 0 because 2.38 > 1.960. If we select α=0.010 the critical value is 2.326, and we still reject H 0 because 2.38 > 2.326. However, if we select α=0.005, the critical value is 2.576, and we cannot reject H 0 because 2.38 < 2.576. Therefore, the smallest α where we still reject H 0 is 0.010. This is the p-value. A statistical computing package would produce a more precise p-value which would be in between 0.005 and 0.010. Here we are approximating the p-value and would report p < 0.010.

Type I and Type II Errors

In all tests of hypothesis, there are two types of errors that can be committed. The first is called a Type I error and refers to the situation where we incorrectly reject H 0 when in fact it is true. This is also called a false positive result (as we incorrectly conclude that the research hypothesis is true when in fact it is not). When we run a test of hypothesis and decide to reject H 0 (e.g., because the test statistic exceeds the critical value in an upper tailed test) then either we make a correct decision because the research hypothesis is true or we commit a Type I error. The different conclusions are summarized in the table below. Note that we will never know whether the null hypothesis is really true or false (i.e., we will never know which row of the following table reflects reality).

Table - Conclusions in Test of Hypothesis

In the first step of the hypothesis test, we select a level of significance, α, and α= P(Type I error). Because we purposely select a small value for α, we control the probability of committing a Type I error. For example, if we select α=0.05, and our test tells us to reject H 0 , then there is a 5% probability that we commit a Type I error. Most investigators are very comfortable with this and are confident when rejecting H 0 that the research hypothesis is true (as it is the more likely scenario when we reject H 0 ).

When we run a test of hypothesis and decide not to reject H 0 (e.g., because the test statistic is below the critical value in an upper tailed test) then either we make a correct decision because the null hypothesis is true or we commit a Type II error. Beta (β) represents the probability of a Type II error and is defined as follows: β=P(Type II error) = P(Do not Reject H 0 | H 0 is false). Unfortunately, we cannot choose β to be small (e.g., 0.05) to control the probability of committing a Type II error because β depends on several factors including the sample size, α, and the research hypothesis. When we do not reject H 0 , it may be very likely that we are committing a Type II error (i.e., failing to reject H 0 when in fact it is false). Therefore, when tests are run and the null hypothesis is not rejected we often make a weak concluding statement allowing for the possibility that we might be committing a Type II error. If we do not reject H 0 , we conclude that we do not have significant evidence to show that H 1 is true. We do not conclude that H 0 is true.

The most common reason for a Type II error is a small sample size.

Tests with One Sample, Continuous Outcome

Hypothesis testing applications with a continuous outcome variable in a single population are performed according to the five-step procedure outlined above. A key component is setting up the null and research hypotheses. The objective is to compare the mean in a single population to known mean (μ 0 ). The known value is generally derived from another study or report, for example a study in a similar, but not identical, population or a study performed some years ago. The latter is called a historical control. It is important in setting up the hypotheses in a one sample test that the mean specified in the null hypothesis is a fair and reasonable comparator. This will be discussed in the examples that follow.

Test Statistics for Testing H 0 : μ= μ 0

if n > 30
if n < 30

Note that statistical computing packages will use the t statistic exclusively and make the necessary adjustments for comparing the test statistic to appropriate values from probability tables to produce a p-value.

The National Center for Health Statistics (NCHS) published a report in 2005 entitled Health, United States, containing extensive information on major trends in the health of Americans. Data are provided for the US population as a whole and for specific ages, sexes and races. The NCHS report indicated that in 2002 Americans paid an average of $3,302 per year on health care and prescription drugs. An investigator hypothesizes that in 2005 expenditures have decreased primarily due to the availability of generic drugs. To test the hypothesis, a sample of 100 Americans are selected and their expenditures on health care and prescription drugs in 2005 are measured. The sample data are summarized as follows: n=100, x̄

=$3,190 and s=$890. Is there statistical evidence of a reduction in expenditures on health care and prescription drugs in 2005? Is the sample mean of $3,190 evidence of a true reduction in the mean or is it within chance fluctuation? We will run the test using the five-step approach.

Step 1. Set up hypotheses and determine level of significance

H 0 : μ = 3,302 H 1 : μ < 3,302 α =0.05

The research hypothesis is that expenditures have decreased, and therefore a lower-tailed test is used.

This is a lower tailed test, using a Z statistic and a 5% level of significance. Reject H 0 if Z < -1.645.

Step 4. Compute the test statistic.

We do not reject H 0 because -1.26 > -1.645. We do not have statistically significant evidence at α=0.05 to show that the mean expenditures on health care and prescription drugs are lower in 2005 than the mean of $3,302 reported in 2002.

Recall that when we fail to reject H 0 in a test of hypothesis that either the null hypothesis is true (here the mean expenditures in 2005 are the same as those in 2002 and equal to $3,302) or we committed a Type II error (i.e., we failed to reject H 0 when in fact it is false). In summarizing this test, we conclude that we do not have sufficient evidence to reject H 0 . We do not conclude that H 0 is true, because there may be a moderate to high probability that we committed a Type II error. It is possible that the sample size is not large enough to detect a difference in mean expenditures.

The NCHS reported that the mean total cholesterol level in 2002 for all adults was 203. Total cholesterol levels in participants who attended the seventh examination of the Offspring in the Framingham Heart Study are summarized as follows: n=3,310, x̄ =200.3, and s=36.8. Is there statistical evidence of a difference in mean cholesterol levels in the Framingham Offspring?

Here we want to assess whether the sample mean of 200.3 in the Framingham sample is statistically significantly different from 203 (i.e., beyond what we would expect by chance). We will run the test using the five-step approach.

H 0 : μ= 203 H 1 : μ≠ 203 α=0.05

The research hypothesis is that cholesterol levels are different in the Framingham Offspring, and therefore a two-tailed test is used.

Step 3. Set up decision rule.

This is a two-tailed test, using a Z statistic and a 5% level of significance. Reject H 0 if Z < -1.960 or is Z > 1.960.

We reject H 0 because -4.22 ≤ -1. .960. We have statistically significant evidence at α=0.05 to show that the mean total cholesterol level in the Framingham Offspring is different from the national average of 203 reported in 2002. Because we reject H 0 , we also approximate a p-value. Using the two-sided significance levels, p < 0.0001.

Statistical Significance versus Clinical (Practical) Significance

This example raises an important concept of statistical versus clinical or practical significance. From a statistical standpoint, the total cholesterol levels in the Framingham sample are highly statistically significantly different from the national average with p < 0.0001 (i.e., there is less than a 0.01% chance that we are incorrectly rejecting the null hypothesis). However, the sample mean in the Framingham Offspring study is 200.3, less than 3 units different from the national mean of 203. The reason that the data are so highly statistically significant is due to the very large sample size. It is always important to assess both statistical and clinical significance of data. This is particularly relevant when the sample size is large. Is a 3 unit difference in total cholesterol a meaningful difference?

Consider again the NCHS-reported mean total cholesterol level in 2002 for all adults of 203. Suppose a new drug is proposed to lower total cholesterol. A study is designed to evaluate the efficacy of the drug in lowering cholesterol. Fifteen patients are enrolled in the study and asked to take the new drug for 6 weeks. At the end of 6 weeks, each patient's total cholesterol level is measured and the sample statistics are as follows: n=15, x̄ =195.9 and s=28.7. Is there statistical evidence of a reduction in mean total cholesterol in patients after using the new drug for 6 weeks? We will run the test using the five-step approach.

H 0 : μ= 203 H 1 : μ< 203 α=0.05

Step 2. Select the appropriate test statistic.

Because the sample size is small (n<30) the appropriate test statistic is

This is a lower tailed test, using a t statistic and a 5% level of significance. In order to determine the critical value of t, we need degrees of freedom, df, defined as df=n-1. In this example df=15-1=14. The critical value for a lower tailed test with df=14 and a =0.05 is -2.145 and the decision rule is as follows: Reject H 0 if t < -2.145.

We do not reject H 0 because -0.96 > -2.145. We do not have statistically significant evidence at α=0.05 to show that the mean total cholesterol level is lower than the national mean in patients taking the new drug for 6 weeks. Again, because we failed to reject the null hypothesis we make a weaker concluding statement allowing for the possibility that we may have committed a Type II error (i.e., failed to reject H 0 when in fact the drug is efficacious).

This example raises an important issue in terms of study design. In this example we assume in the null hypothesis that the mean cholesterol level is 203. This is taken to be the mean cholesterol level in patients without treatment. Is this an appropriate comparator? Alternative and potentially more efficient study designs to evaluate the effect of the new drug could involve two treatment groups, where one group receives the new drug and the other does not, or we could measure each patient's baseline or pre-treatment cholesterol level and then assess changes from baseline to 6 weeks post-treatment. These designs are also discussed here.

Video - Comparing a Sample Mean to Known Population Mean (8:20)

Link to transcript of the video

Tests with One Sample, Dichotomous Outcome

Hypothesis testing applications with a dichotomous outcome variable in a single population are also performed according to the five-step procedure. Similar to tests for means, a key component is setting up the null and research hypotheses. The objective is to compare the proportion of successes in a single population to a known proportion (p 0 ). That known proportion is generally derived from another study or report and is sometimes called a historical control. It is important in setting up the hypotheses in a one sample test that the proportion specified in the null hypothesis is a fair and reasonable comparator.

In one sample tests for a dichotomous outcome, we set up our hypotheses against an appropriate comparator. We select a sample and compute descriptive statistics on the sample data. Specifically, we compute the sample size (n) and the sample proportion which is computed by taking the ratio of the number of successes to the sample size,

We then determine the appropriate test statistic (Step 2) for the hypothesis test. The formula for the test statistic is given below.

Test Statistic for Testing H 0 : p = p 0

if min(np 0 , n(1-p 0 )) > 5

The formula above is appropriate for large samples, defined when the smaller of np 0 and n(1-p 0 ) is at least 5. This is similar, but not identical, to the condition required for appropriate use of the confidence interval formula for a population proportion, i.e.,

Here we use the proportion specified in the null hypothesis as the true proportion of successes rather than the sample proportion. If we fail to satisfy the condition, then alternative procedures, called exact methods must be used to test the hypothesis about the population proportion.

Example:

The NCHS report indicated that in 2002 the prevalence of cigarette smoking among American adults was 21.1%. Data on prevalent smoking in n=3,536 participants who attended the seventh examination of the Offspring in the Framingham Heart Study indicated that 482/3,536 = 13.6% of the respondents were currently smoking at the time of the exam. Suppose we want to assess whether the prevalence of smoking is lower in the Framingham Offspring sample given the focus on cardiovascular health in that community. Is there evidence of a statistically lower prevalence of smoking in the Framingham Offspring study as compared to the prevalence among all Americans?

H 0 : p = 0.211 H 1 : p < 0.211 α=0.05

We must first check that the sample size is adequate. Specifically, we need to check min(np 0 , n(1-p 0 )) = min( 3,536(0.211), 3,536(1-0.211))=min(746, 2790)=746. The sample size is more than adequate so the following formula can be used:

This is a lower tailed test, using a Z statistic and a 5% level of significance. Reject H 0 if Z < -1.645.

We reject H 0 because -10.93 < -1.645. We have statistically significant evidence at α=0.05 to show that the prevalence of smoking in the Framingham Offspring is lower than the prevalence nationally (21.1%). Here, p < 0.0001.

The NCHS report indicated that in 2002, 75% of children aged 2 to 17 saw a dentist in the past year. An investigator wants to assess whether use of dental services is similar in children living in the city of Boston. A sample of 125 children aged 2 to 17 living in Boston are surveyed and 64 reported seeing a dentist over the past 12 months. Is there a significant difference in use of dental services between children living in Boston and the national data?

Calculate this on your own before checking the answer.

Video - Hypothesis Test for One Sample and a Dichotomous Outcome (3:55)

Tests with Two Independent Samples, Continuous Outcome

There are many applications where it is of interest to compare two independent groups with respect to their mean scores on a continuous outcome. Here we compare means between groups, but rather than generating an estimate of the difference, we will test whether the observed difference (increase, decrease or difference) is statistically significant or not. Remember, that hypothesis testing gives an assessment of statistical significance, whereas estimation gives an estimate of effect and both are important.

Here we discuss the comparison of means when the two comparison groups are independent or physically separate. The two groups might be determined by a particular attribute (e.g., sex, diagnosis of cardiovascular disease) or might be set up by the investigator (e.g., participants assigned to receive an experimental treatment or placebo). The first step in the analysis involves computing descriptive statistics on each of the two samples. Specifically, we compute the sample size, mean and standard deviation in each sample and we denote these summary statistics as follows:

for sample 1:

for sample 2:

The designation of sample 1 and sample 2 is arbitrary. In a clinical trial setting the convention is to call the treatment group 1 and the control group 2. However, when comparing men and women, for example, either group can be 1 or 2.

In the two independent samples application with a continuous outcome, the parameter of interest in the test of hypothesis is the difference in population means, μ 1 -μ 2 . The null hypothesis is always that there is no difference between groups with respect to means, i.e.,

The null hypothesis can also be written as follows: H 0 : μ 1 = μ 2 . In the research hypothesis, an investigator can hypothesize that the first mean is larger than the second (H 1 : μ 1 > μ 2 ), that the first mean is smaller than the second (H 1 : μ 1 < μ 2 ), or that the means are different (H 1 : μ 1 ≠ μ 2 ). The three different alternatives represent upper-, lower-, and two-tailed tests, respectively. The following test statistics are used to test these hypotheses.

Test Statistics for Testing H 0 : μ 1 = μ 2

if n 1 > 30 and n 2 > 30
if n 1 < 30 or n 2 < 30

NOTE: The formulas above assume equal variability in the two populations (i.e., the population variances are equal, or s 1 2 = s 2 2 ). This means that the outcome is equally variable in each of the comparison populations. For analysis, we have samples from each of the comparison populations. If the sample variances are similar, then the assumption about variability in the populations is probably reasonable. As a guideline, if the ratio of the sample variances, s 1 2 /s 2 2 is between 0.5 and 2 (i.e., if one variance is no more than double the other), then the formulas above are appropriate. If the ratio of the sample variances is greater than 2 or less than 0.5 then alternative formulas must be used to account for the heterogeneity in variances.

The test statistics include Sp, which is the pooled estimate of the common standard deviation (again assuming that the variances in the populations are similar) computed as the weighted average of the standard deviations in the samples as follows:

Because we are assuming equal variances between groups, we pool the information on variability (sample variances) to generate an estimate of the variability in the population. Note: Because Sp is a weighted average of the standard deviations in the sample, Sp will always be in between s 1 and s 2 .)

Data measured on n=3,539 participants who attended the seventh examination of the Offspring in the Framingham Heart Study are shown below.

Suppose we now wish to assess whether there is a statistically significant difference in mean systolic blood pressures between men and women using a 5% level of significance.

H 0 : μ 1 = μ 2

H 1 : μ 1 ≠ μ 2 α=0.05

Because both samples are large ( > 30), we can use the Z test statistic as opposed to t. Note that statistical computing packages use t throughout. Before implementing the formula, we first check whether the assumption of equality of population variances is reasonable. The guideline suggests investigating the ratio of the sample variances, s 1 2 /s 2 2 . Suppose we call the men group 1 and the women group 2. Again, this is arbitrary; it only needs to be noted when interpreting the results. The ratio of the sample variances is 17.5 2 /20.1 2 = 0.76, which falls between 0.5 and 2 suggesting that the assumption of equality of population variances is reasonable. The appropriate test statistic is

We now substitute the sample data into the formula for the test statistic identified in Step 2. Before substituting, we will first compute Sp, the pooled estimate of the common standard deviation.

Notice that the pooled estimate of the common standard deviation, Sp, falls in between the standard deviations in the comparison groups (i.e., 17.5 and 20.1). Sp is slightly closer in value to the standard deviation in the women (20.1) as there were slightly more women in the sample. Recall, Sp is a weight average of the standard deviations in the comparison groups, weighted by the respective sample sizes.

Now the test statistic:

We reject H 0 because 2.66 > 1.960. We have statistically significant evidence at α=0.05 to show that there is a difference in mean systolic blood pressures between men and women. The p-value is p < 0.010.

Here again we find that there is a statistically significant difference in mean systolic blood pressures between men and women at p < 0.010. Notice that there is a very small difference in the sample means (128.2-126.5 = 1.7 units), but this difference is beyond what would be expected by chance. Is this a clinically meaningful difference? The large sample size in this example is driving the statistical significance. A 95% confidence interval for the difference in mean systolic blood pressures is: 1.7 + 1.26 or (0.44, 2.96). The confidence interval provides an assessment of the magnitude of the difference between means whereas the test of hypothesis and p-value provide an assessment of the statistical significance of the difference.

Above we performed a study to evaluate a new drug designed to lower total cholesterol. The study involved one sample of patients, each patient took the new drug for 6 weeks and had their cholesterol measured. As a means of evaluating the efficacy of the new drug, the mean total cholesterol following 6 weeks of treatment was compared to the NCHS-reported mean total cholesterol level in 2002 for all adults of 203. At the end of the example, we discussed the appropriateness of the fixed comparator as well as an alternative study design to evaluate the effect of the new drug involving two treatment groups, where one group receives the new drug and the other does not. Here, we revisit the example with a concurrent or parallel control group, which is very typical in randomized controlled trials or clinical trials (refer to the EP713 module on Clinical Trials).

A new drug is proposed to lower total cholesterol. A randomized controlled trial is designed to evaluate the efficacy of the medication in lowering cholesterol. Thirty participants are enrolled in the trial and are randomly assigned to receive either the new drug or a placebo. The participants do not know which treatment they are assigned. Each participant is asked to take the assigned treatment for 6 weeks. At the end of 6 weeks, each patient's total cholesterol level is measured and the sample statistics are as follows.

Is there statistical evidence of a reduction in mean total cholesterol in patients taking the new drug for 6 weeks as compared to participants taking placebo? We will run the test using the five-step approach.

H 0 : μ 1 = μ 2 H 1 : μ 1 < μ 2 α=0.05

Because both samples are small (< 30), we use the t test statistic. Before implementing the formula, we first check whether the assumption of equality of population variances is reasonable. The ratio of the sample variances, s 1 2 /s 2 2 =28.7 2 /30.3 2 = 0.90, which falls between 0.5 and 2, suggesting that the assumption of equality of population variances is reasonable. The appropriate test statistic is:

This is a lower-tailed test, using a t statistic and a 5% level of significance. The appropriate critical value can be found in the t Table (in More Resources to the right). In order to determine the critical value of t we need degrees of freedom, df, defined as df=n 1 +n 2 -2 = 15+15-2=28. The critical value for a lower tailed test with df=28 and α=0.05 is -1.701 and the decision rule is: Reject H 0 if t < -1.701.

Now the test statistic,

We reject H 0 because -2.92 < -1.701. We have statistically significant evidence at α=0.05 to show that the mean total cholesterol level is lower in patients taking the new drug for 6 weeks as compared to patients taking placebo, p < 0.005.

The clinical trial in this example finds a statistically significant reduction in total cholesterol, whereas in the previous example where we had a historical control (as opposed to a parallel control group) we did not demonstrate efficacy of the new drug. Notice that the mean total cholesterol level in patients taking placebo is 217.4 which is very different from the mean cholesterol reported among all Americans in 2002 of 203 and used as the comparator in the prior example. The historical control value may not have been the most appropriate comparator as cholesterol levels have been increasing over time. In the next section, we present another design that can be used to assess the efficacy of the new drug.

Video - Comparison of Two Independent Samples With a Continuous Outcome (8:02)

Tests with Matched Samples, Continuous Outcome

In the previous section we compared two groups with respect to their mean scores on a continuous outcome. An alternative study design is to compare matched or paired samples. The two comparison groups are said to be dependent, and the data can arise from a single sample of participants where each participant is measured twice (possibly before and after an intervention) or from two samples that are matched on specific characteristics (e.g., siblings). When the samples are dependent, we focus on difference scores in each participant or between members of a pair and the test of hypothesis is based on the mean difference, μ d . The null hypothesis again reflects "no difference" and is stated as H 0 : μ d =0 . Note that there are some instances where it is of interest to test whether there is a difference of a particular magnitude (e.g., μ d =5) but in most instances the null hypothesis reflects no difference (i.e., μ d =0).

The appropriate formula for the test of hypothesis depends on the sample size. The formulas are shown below and are identical to those we presented for estimating the mean of a single sample presented (e.g., when comparing against an external or historical control), except here we focus on difference scores.

Test Statistics for Testing H 0 : μ d =0

A new drug is proposed to lower total cholesterol and a study is designed to evaluate the efficacy of the drug in lowering cholesterol. Fifteen patients agree to participate in the study and each is asked to take the new drug for 6 weeks. However, before starting the treatment, each patient's total cholesterol level is measured. The initial measurement is a pre-treatment or baseline value. After taking the drug for 6 weeks, each patient's total cholesterol level is measured again and the data are shown below. The rightmost column contains difference scores for each patient, computed by subtracting the 6 week cholesterol level from the baseline level. The differences represent the reduction in total cholesterol over 4 weeks. (The differences could have been computed by subtracting the baseline total cholesterol level from the level measured at 6 weeks. The way in which the differences are computed does not affect the outcome of the analysis only the interpretation.)

Because the differences are computed by subtracting the cholesterols measured at 6 weeks from the baseline values, positive differences indicate reductions and negative differences indicate increases (e.g., participant 12 increases by 2 units over 6 weeks). The goal here is to test whether there is a statistically significant reduction in cholesterol. Because of the way in which we computed the differences, we want to look for an increase in the mean difference (i.e., a positive reduction). In order to conduct the test, we need to summarize the differences. In this sample, we have

The calculations are shown below.

Is there statistical evidence of a reduction in mean total cholesterol in patients after using the new medication for 6 weeks? We will run the test using the five-step approach.

H 0 : μ d = 0 H 1 : μ d > 0 α=0.05

NOTE: If we had computed differences by subtracting the baseline level from the level measured at 6 weeks then negative differences would have reflected reductions and the research hypothesis would have been H 1 : μ d < 0.

Step 2 . Select the appropriate test statistic.

This is an upper-tailed test, using a t statistic and a 5% level of significance. The appropriate critical value can be found in the t Table at the right, with df=15-1=14. The critical value for an upper-tailed test with df=14 and α=0.05 is 2.145 and the decision rule is Reject H 0 if t > 2.145.

We now substitute the sample data into the formula for the test statistic identified in Step 2.

We reject H 0 because 4.61 > 2.145. We have statistically significant evidence at α=0.05 to show that there is a reduction in cholesterol levels over 6 weeks.

Here we illustrate the use of a matched design to test the efficacy of a new drug to lower total cholesterol. We also considered a parallel design (randomized clinical trial) and a study using a historical comparator. It is extremely important to design studies that are best suited to detect a meaningful difference when one exists. There are often several alternatives and investigators work with biostatisticians to determine the best design for each application. It is worth noting that the matched design used here can be problematic in that observed differences may only reflect a "placebo" effect. All participants took the assigned medication, but is the observed reduction attributable to the medication or a result of these participation in a study.

Video - Hypothesis Testing With a Matched Sample and a Continuous Outcome (3:11)

Tests with Two Independent Samples, Dichotomous Outcome

There are several approaches that can be used to test hypotheses concerning two independent proportions. Here we present one approach - the chi-square test of independence is an alternative, equivalent, and perhaps more popular approach to the same analysis. Hypothesis testing with the chi-square test is addressed in the third module in this series: BS704_HypothesisTesting-ChiSquare.

In tests of hypothesis comparing proportions between two independent groups, one test is performed and results can be interpreted to apply to a risk difference, relative risk or odds ratio. As a reminder, the risk difference is computed by taking the difference in proportions between comparison groups, the risk ratio is computed by taking the ratio of proportions, and the odds ratio is computed by taking the ratio of the odds of success in the comparison groups. Because the null values for the risk difference, the risk ratio and the odds ratio are different, the hypotheses in tests of hypothesis look slightly different depending on which measure is used. When performing tests of hypothesis for the risk difference, relative risk or odds ratio, the convention is to label the exposed or treated group 1 and the unexposed or control group 2.

For example, suppose a study is designed to assess whether there is a significant difference in proportions in two independent comparison groups. The test of interest is as follows:

H 0 : p 1 = p 2 versus H 1 : p 1 ≠ p 2 .

The following are the hypothesis for testing for a difference in proportions using the risk difference, the risk ratio and the odds ratio. First, the hypotheses above are equivalent to the following:

For the risk difference, H 0 : p 1 - p 2 = 0 versus H 1 : p 1 - p 2 ≠ 0 which are, by definition, equal to H 0 : RD = 0 versus H 1 : RD ≠ 0.
If an investigator wants to focus on the risk ratio, the equivalent hypotheses are H 0 : RR = 1 versus H 1 : RR ≠ 1.
If the investigator wants to focus on the odds ratio, the equivalent hypotheses are H 0 : OR = 1 versus H 1 : OR ≠ 1.

Suppose a test is performed to test H 0 : RD = 0 versus H 1 : RD ≠ 0 and the test rejects H 0 at α=0.05. Based on this test we can conclude that there is significant evidence, α=0.05, of a difference in proportions, significant evidence that the risk difference is not zero, significant evidence that the risk ratio and odds ratio are not one. The risk difference is analogous to the difference in means when the outcome is continuous. Here the parameter of interest is the difference in proportions in the population, RD = p 1 -p 2 and the null value for the risk difference is zero. In a test of hypothesis for the risk difference, the null hypothesis is always H 0 : RD = 0. This is equivalent to H 0 : RR = 1 and H 0 : OR = 1. In the research hypothesis, an investigator can hypothesize that the first proportion is larger than the second (H 1 : p 1 > p 2 , which is equivalent to H 1 : RD > 0, H 1 : RR > 1 and H 1 : OR > 1), that the first proportion is smaller than the second (H 1 : p 1 < p 2 , which is equivalent to H 1 : RD < 0, H 1 : RR < 1 and H 1 : OR < 1), or that the proportions are different (H 1 : p 1 ≠ p 2 , which is equivalent to H 1 : RD ≠ 0, H 1 : RR ≠ 1 and H 1 : OR ≠

1). The three different alternatives represent upper-, lower- and two-tailed tests, respectively.

The formula for the test of hypothesis for the difference in proportions is given below.

Test Statistics for Testing H 0 : p 1 = p

The formula above is appropriate for large samples, defined as at least 5 successes (np > 5) and at least 5 failures (n(1-p > 5)) in each of the two samples. If there are fewer than 5 successes or failures in either comparison group, then alternative procedures, called exact methods must be used to estimate the difference in population proportions.

The following table summarizes data from n=3,799 participants who attended the fifth examination of the Offspring in the Framingham Heart Study. The outcome of interest is prevalent CVD and we want to test whether the prevalence of CVD is significantly higher in smokers as compared to non-smokers.

The prevalence of CVD (or proportion of participants with prevalent CVD) among non-smokers is 298/3,055 = 0.0975 and the prevalence of CVD among current smokers is 81/744 = 0.1089. Here smoking status defines the comparison groups and we will call the current smokers group 1 (exposed) and the non-smokers (unexposed) group 2. The test of hypothesis is conducted below using the five step approach.

H 0 : p 1 = p 2 H 1 : p 1 ≠ p 2 α=0.05

Step 2. Select the appropriate test statistic.

We must first check that the sample size is adequate. Specifically, we need to ensure that we have at least 5 successes and 5 failures in each comparison group. In this example, we have more than enough successes (cases of prevalent CVD) and failures (persons free of CVD) in each comparison group. The sample size is more than adequate so the following formula can be used:

Reject H 0 if Z < -1.960 or if Z > 1.960.

We now substitute the sample data into the formula for the test statistic identified in Step 2. We first compute the overall proportion of successes:

We now substitute to compute the test statistic.

Step 5. Conclusion.

We do not reject H 0 because -1.960 < 0.927 < 1.960. We do not have statistically significant evidence at α=0.05 to show that there is a difference in prevalent CVD between smokers and non-smokers.

A 95% confidence interval for the difference in prevalent CVD (or risk difference) between smokers and non-smokers as 0.0114 + 0.0247, or between -0.0133 and 0.0361. Because the 95% confidence interval for the risk difference includes zero we again conclude that there is no statistically significant difference in prevalent CVD between smokers and non-smokers.

Smoking has been shown over and over to be a risk factor for cardiovascular disease. What might explain the fact that we did not observe a statistically significant difference using data from the Framingham Heart Study? HINT: Here we consider prevalent CVD, would the results have been different if we considered incident CVD?

A randomized trial is designed to evaluate the effectiveness of a newly developed pain reliever designed to reduce pain in patients following joint replacement surgery. The trial compares the new pain reliever to the pain reliever currently in use (called the standard of care). A total of 100 patients undergoing joint replacement surgery agreed to participate in the trial. Patients were randomly assigned to receive either the new pain reliever or the standard pain reliever following surgery and were blind to the treatment assignment. Before receiving the assigned treatment, patients were asked to rate their pain on a scale of 0-10 with higher scores indicative of more pain. Each patient was then given the assigned treatment and after 30 minutes was again asked to rate their pain on the same scale. The primary outcome was a reduction in pain of 3 or more scale points (defined by clinicians as a clinically meaningful reduction). The following data were observed in the trial.

We now test whether there is a statistically significant difference in the proportions of patients reporting a meaningful reduction (i.e., a reduction of 3 or more scale points) using the five step approach.

H 0 : p 1 = p 2 H 1 : p 1 ≠ p 2 α=0.05

Here the new or experimental pain reliever is group 1 and the standard pain reliever is group 2.

We must first check that the sample size is adequate. Specifically, we need to ensure that we have at least 5 successes and 5 failures in each comparison group, i.e.,

In this example, we have min(50(0.46), 50(1-0.46), 50(0.22), 50(1-0.22)) = min(23, 27, 11, 39) = 11. The sample size is adequate so the following formula can be used

We reject H 0 because 2.526 > 1960. We have statistically significant evidence at a =0.05 to show that there is a difference in the proportions of patients on the new pain reliever reporting a meaningful reduction (i.e., a reduction of 3 or more scale points) as compared to patients on the standard pain reliever.

A 95% confidence interval for the difference in proportions of patients on the new pain reliever reporting a meaningful reduction (i.e., a reduction of 3 or more scale points) as compared to patients on the standard pain reliever is 0.24 + 0.18 or between 0.06 and 0.42. Because the 95% confidence interval does not include zero we concluded that there was a statistically significant difference in proportions which is consistent with the test of hypothesis result.

Again, the procedures discussed here apply to applications where there are two independent comparison groups and a dichotomous outcome. There are other applications in which it is of interest to compare a dichotomous outcome in matched or paired samples. For example, in a clinical trial we might wish to test the effectiveness of a new antibiotic eye drop for the treatment of bacterial conjunctivitis. Participants use the new antibiotic eye drop in one eye and a comparator (placebo or active control treatment) in the other. The success of the treatment (yes/no) is recorded for each participant for each eye. Because the two assessments (success or failure) are paired, we cannot use the procedures discussed here. The appropriate test is called McNemar's test (sometimes called McNemar's test for dependent proportions).

Vide0 - Hypothesis Testing With Two Independent Samples and a Dichotomous Outcome (2:55)

Here we presented hypothesis testing techniques for means and proportions in one and two sample situations. Tests of hypothesis involve several steps, including specifying the null and alternative or research hypothesis, selecting and computing an appropriate test statistic, setting up a decision rule and drawing a conclusion. There are many details to consider in hypothesis testing. The first is to determine the appropriate test. We discussed Z and t tests here for different applications. The appropriate test depends on the distribution of the outcome variable (continuous or dichotomous), the number of comparison groups (one, two) and whether the comparison groups are independent or dependent. The following table summarizes the different tests of hypothesis discussed here.

Continuous Outcome, One Sample: H0: μ = μ0
Continuous Outcome, Two Independent Samples: H0: μ1 = μ2
Continuous Outcome, Two Matched Samples: H0: μd = 0
Dichotomous Outcome, One Sample: H0: p = p 0
Dichotomous Outcome, Two Independent Samples: H0: p1 = p2, RD=0, RR=1, OR=1

Once the type of test is determined, the details of the test must be specified. Specifically, the null and alternative hypotheses must be clearly stated. The null hypothesis always reflects the "no change" or "no difference" situation. The alternative or research hypothesis reflects the investigator's belief. The investigator might hypothesize that a parameter (e.g., a mean, proportion, difference in means or proportions) will increase, will decrease or will be different under specific conditions (sometimes the conditions are different experimental conditions and other times the conditions are simply different groups of participants). Once the hypotheses are specified, data are collected and summarized. The appropriate test is then conducted according to the five step approach. If the test leads to rejection of the null hypothesis, an approximate p-value is computed to summarize the significance of the findings. When tests of hypothesis are conducted using statistical computing packages, exact p-values are computed. Because the statistical tables in this textbook are limited, we can only approximate p-values. If the test fails to reject the null hypothesis, then a weaker concluding statement is made for the following reason.

In hypothesis testing, there are two types of errors that can be committed. A Type I error occurs when a test incorrectly rejects the null hypothesis. This is referred to as a false positive result, and the probability that this occurs is equal to the level of significance, α. The investigator chooses the level of significance in Step 1, and purposely chooses a small value such as α=0.05 to control the probability of committing a Type I error. A Type II error occurs when a test fails to reject the null hypothesis when in fact it is false. The probability that this occurs is equal to β. Unfortunately, the investigator cannot specify β at the outset because it depends on several factors including the sample size (smaller samples have higher b), the level of significance (β decreases as a increases), and the difference in the parameter under the null and alternative hypothesis.

We noted in several examples in this chapter, the relationship between confidence intervals and tests of hypothesis. The approaches are different, yet related. It is possible to draw a conclusion about statistical significance by examining a confidence interval. For example, if a 95% confidence interval does not contain the null value (e.g., zero when analyzing a mean difference or risk difference, one when analyzing relative risks or odds ratios), then one can conclude that a two-sided test of hypothesis would reject the null at α=0.05. It is important to note that the correspondence between a confidence interval and test of hypothesis relates to a two-sided test and that the confidence level corresponds to a specific level of significance (e.g., 95% to α=0.05, 90% to α=0.10 and so on). The exact significance of the test, the p-value, can only be determined using the hypothesis testing approach and the p-value provides an assessment of the strength of the evidence and not an estimate of the effect.

Answers to Selected Problems

Dental services problem - bottom of page 5.

Step 1: Set up hypotheses and determine the level of significance.

α=0.05

Step 2: Select the appropriate test statistic.

First, determine whether the sample size is adequate.

Therefore the sample size is adequate, and we can use the following formula:

Step 3: Set up the decision rule.

Reject H0 if Z is less than or equal to -1.96 or if Z is greater than or equal to 1.96.

Step 4: Compute the test statistic
Step 5: Conclusion.

We reject the null hypothesis because -6.15<-1.96. Therefore there is a statistically significant difference in the proportion of children in Boston using dental services compated to the national proportion.

Statistics Made Easy

How to Write a Null Hypothesis (5 Examples)

A hypothesis test uses sample data to determine whether or not some claim about a population parameter is true.

Whenever we perform a hypothesis test, we always write a null hypothesis and an alternative hypothesis, which take the following forms:

H 0 (Null Hypothesis): Population parameter =, ≤, ≥ some value

H A (Alternative Hypothesis): Population parameter <, >, ≠ some value

Note that the null hypothesis always contains the equal sign .

We interpret the hypotheses as follows:

Null hypothesis: The sample data provides no evidence to support some claim being made by an individual.

Alternative hypothesis: The sample data does provide sufficient evidence to support the claim being made by an individual.

For example, suppose it’s assumed that the average height of a certain species of plant is 20 inches tall. However, one botanist claims the true average height is greater than 20 inches.

To test this claim, she may go out and collect a random sample of plants. She can then use this sample data to perform a hypothesis test using the following two hypotheses:

H 0 : μ ≤ 20 (the true mean height of plants is equal to or even less than 20 inches)

H A : μ > 20 (the true mean height of plants is greater than 20 inches)

If the sample data gathered by the botanist shows that the mean height of this species of plants is significantly greater than 20 inches, she can reject the null hypothesis and conclude that the mean height is greater than 20 inches.

Read through the following examples to gain a better understanding of how to write a null hypothesis in different situations.

Example 1: Weight of Turtles

A biologist wants to test whether or not the true mean weight of a certain species of turtles is 300 pounds. To test this, he goes out and measures the weight of a random sample of 40 turtles.

Here is how to write the null and alternative hypotheses for this scenario:

H 0 : μ = 300 (the true mean weight is equal to 300 pounds)

H A : μ ≠ 300 (the true mean weight is not equal to 300 pounds)

Example 2: Height of Males

It’s assumed that the mean height of males in a certain city is 68 inches. However, an independent researcher believes the true mean height is greater than 68 inches. To test this, he goes out and collects the height of 50 males in the city.

H 0 : μ ≤ 68 (the true mean height is equal to or even less than 68 inches)

H A : μ > 68 (the true mean height is greater than 68 inches)

Example 3: Graduation Rates

A university states that 80% of all students graduate on time. However, an independent researcher believes that less than 80% of all students graduate on time. To test this, she collects data on the proportion of students who graduated on time last year at the university.

H 0 : p ≥ 0.80 (the true proportion of students who graduate on time is 80% or higher)

H A : μ < 0.80 (the true proportion of students who graduate on time is less than 80%)

Example 4: Burger Weights

A food researcher wants to test whether or not the true mean weight of a burger at a certain restaurant is 7 ounces. To test this, he goes out and measures the weight of a random sample of 20 burgers from this restaurant.

H 0 : μ = 7 (the true mean weight is equal to 7 ounces)

H A : μ ≠ 7 (the true mean weight is not equal to 7 ounces)

Example 5: Citizen Support

A politician claims that less than 30% of citizens in a certain town support a certain law. To test this, he goes out and surveys 200 citizens on whether or not they support the law.

H 0 : p ≥ .30 (the true proportion of citizens who support the law is greater than or equal to 30%)

H A : μ < 0.30 (the true proportion of citizens who support the law is less than 30%)

Additional Resources

Introduction to Hypothesis Testing Introduction to Confidence Intervals An Explanation of P-Values and Statistical Significance

Published by Zach

8.4.3 Hypothesis Testing for the Mean

$\quad$ $H_0$: $\mu=\mu_0$, $\quad$ $H_1$: $\mu \neq \mu_0$.

$\quad$ $H_0$: $\mu \leq \mu_0$, $\quad$ $H_1$: $\mu > \mu_0$.

$\quad$ $H_0$: $\mu \geq \mu_0$, $\quad$ $H_1$: $\mu \lt \mu_0$.

Two-sided Tests for the Mean:

Therefore, we can suggest the following test. Choose a threshold, and call it $c$. If $|W| \leq c$, accept $H_0$, and if $|W|>c$, accept $H_1$. How do we choose $c$? If $\alpha$ is the required significance level, we must have

As discussed above, we let \begin{align}%\label{} W(X_1,X_2, \cdots,X_n)=\frac{\overline{X}-\mu_0}{\sigma / \sqrt{n}}. \end{align} Note that, assuming $H_0$, $W \sim N(0,1)$. We will choose a threshold, $c$. If $|W| \leq c$, we accept $H_0$, and if $|W|>c$, accept $H_1$. To choose $c$, we let \begin{align} P(|W| > c \; | \; H_0) =\alpha. \end{align} Since the standard normal PDF is symmetric around $0$, we have \begin{align} P(|W| > c \; | \; H_0) = 2 P(W>c | \; H_0). \end{align} Thus, we conclude $P(W>c | \; H_0)=\frac{\alpha}{2}$. Therefore, \begin{align} c=z_{\frac{\alpha}{2}}. \end{align} Therefore, we accept $H_0$ if \begin{align} \left|\frac{\overline{X}-\mu_0}{\sigma / \sqrt{n}} \right| \leq z_{\frac{\alpha}{2}}, \end{align} and reject it otherwise.
We have \begin{align} \beta (\mu) &=P(\textrm{type II error}) = P(\textrm{accept }H_0 \; | \; \mu) \\ &= P\left(\left|\frac{\overline{X}-\mu_0}{\sigma / \sqrt{n}} \right| \lt z_{\frac{\alpha}{2}}\; | \; \mu \right). \end{align} If $X_i \sim N(\mu,\sigma^2)$, then $\overline{X} \sim N(\mu, \frac{\sigma^2}{n})$. Thus, \begin{align} \beta (\mu)&=P\left(\left|\frac{\overline{X}-\mu_0}{\sigma / \sqrt{n}} \right| \lt z_{\frac{\alpha}{2}}\; | \; \mu \right)\\ &=P\left(\mu_0- z_{\frac{\alpha}{2}} \frac{\sigma}{\sqrt{n}} \leq \overline{X} \leq \mu_0+ z_{\frac{\alpha}{2}} \frac{\sigma}{\sqrt{n}}\right)\\ &=\Phi\left(z_{\frac{\alpha}{2}}+\frac{\mu_0-\mu}{\sigma / \sqrt{n}}\right)-\Phi\left(-z_{\frac{\alpha}{2}}+\frac{\mu_0-\mu}{\sigma / \sqrt{n}}\right). \end{align}
Let $S^2$ be the sample variance for this random sample. Then, the random variable $W$ defined as \begin{equation} W(X_1,X_2, \cdots, X_n)=\frac{\overline{X}-\mu_0}{S / \sqrt{n}} \end{equation} has a $t$-distribution with $n-1$ degrees of freedom, i.e., $W \sim T(n-1)$. Thus, we can repeat the analysis of Example 8.24 here. The only difference is that we need to replace $\sigma$ by $S$ and $z_{\frac{\alpha}{2}}$ by $t_{\frac{\alpha}{2},n-1}$. Therefore, we accept $H_0$ if \begin{align} |W| \leq t_{\frac{\alpha}{2},n-1}, \end{align} and reject it otherwise. Let us look at a numerical example of this case.

$\quad$ $H_0$: $\mu=170$, $\quad$ $H_1$: $\mu \neq 170$.

Let's first compute the sample mean and the sample standard deviation. The sample mean is \begin{align}%\label{} \overline{X}&=\frac{X_1+X_2+X_3+X_4+X_5+X_6+X_7+X_8+X_9}{9}\\ &=165.8 \end{align} The sample variance is given by \begin{align}%\label{} {S}^2=\frac{1}{9-1} \sum_{k=1}^9 (X_k-\overline{X})^2&=68.01 \end{align} The sample standard deviation is given by \begin{align}%\label{} S&= \sqrt{S^2}=8.25 \end{align} The following MATLAB code can be used to obtain these values: x=[176.2,157.9,160.1,180.9,165.1,167.2,162.9,155.7,166.2]; m=mean(x); v=var(x); s=std(x); Now, our test statistic is \begin{align} W(X_1,X_2, \cdots, X_9)&=\frac{\overline{X}-\mu_0}{S / \sqrt{n}}\\ &=\frac{165.8-170}{8.25 / 3}=-1.52 \end{align} Thus, $|W|=1.52$. Also, we have \begin{align} t_{\frac{\alpha}{2},n-1} = t_{0.025,8} \approx 2.31 \end{align} The above value can be obtained in MATLAB using the command $\mathtt{tinv(0.975,8)}$. Thus, we conclude \begin{align} |W| \leq t_{\frac{\alpha}{2},n-1}. \end{align} Therefore, we accept $H_0$. In other words, we do not have enough evidence to conclude that the average height in the city is different from the average height in the country.

Let us summarize what we have obtained for the two-sided test for the mean.

One-sided Tests for the Mean:

As before, we define the test statistic as \begin{align}%\label{} W(X_1,X_2, \cdots,X_n)=\frac{\overline{X}-\mu_0}{\sigma / \sqrt{n}}. \end{align} If $H_0$ is true (i.e., $\mu \leq \mu_0$), we expect $\overline{X}$ (and thus $W$) to be relatively small, while if $H_1$ is true, we expect $\overline{X}$ (and thus $W$) to be larger. This suggests the following test: Choose a threshold, and call it $c$. If $W \leq c$, accept $H_0$, and if $W>c$, accept $H_1$. How do we choose $c$? If $\alpha$ is the required significance level, we must have \begin{align} P(\textrm{type I error}) &= P(\textrm{Reject }H_0 \; | \; H_0) \\ &= P(W > c \; | \; \mu \leq \mu_0) \leq \alpha. \end{align} Here, the probability of type I error depends on $\mu$. More specifically, for any $\mu \leq \mu_0$, we can write \begin{align} P(\textrm{type I error} \; | \; \mu) &= P(\textrm{Reject }H_0 \; | \; \mu) \\ &= P(W > c \; | \; \mu)\\ &=P \left(\frac{\overline{X}-\mu_0}{\sigma / \sqrt{n}}> c \; | \; \mu\right)\\ &=P \left(\frac{\overline{X}-\mu}{\sigma / \sqrt{n}}+\frac{\mu-\mu_0}{\sigma / \sqrt{n}}> c \; | \; \mu\right)\\ &=P \left(\frac{\overline{X}-\mu}{\sigma / \sqrt{n}}> c+\frac{\mu_0-\mu}{\sigma / \sqrt{n}} \; | \; \mu\right)\\ &\leq P \left(\frac{\overline{X}-\mu}{\sigma / \sqrt{n}}> c \; | \; \mu\right) \quad (\textrm{ since }\mu \leq \mu_0)\\ &=1-\Phi(c) \quad \big(\textrm{ since given }\mu, \frac{\overline{X}-\mu}{\sigma / \sqrt{n}} \sim N(0,1) \big). \end{align} Thus, we can choose $\alpha=1-\Phi(c)$, which results in \begin{align} c=z_{\alpha}. \end{align} Therefore, we accept $H_0$ if \begin{align} \frac{\overline{X}-\mu_0}{\sigma / \sqrt{n}} \leq z_{\alpha}, \end{align} and reject it otherwise.

$\quad$ $H_0$: $\mu \geq \mu_0$, $\quad$ $H_1$: $\mu \lt \mu_0$,

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Null Hypothesis

Null Hypothesis , often denoted as H 0, is a foundational concept in statistical hypothesis testing. It represents an assumption that no significant difference, effect, or relationship exists between variables within a population. It serves as a baseline assumption, positing no observed change or effect occurring. The null is t he truth or falsity of an idea in analysis.

In this article, we will discuss the null hypothesis in detail, along with some solved examples and questions on the null hypothesis.

Table of Content

What Is a Null Hypothesis?

Symbol of Null Hypothesis

Formula of null hypothesis, types of null hypothesis, principle of null hypothesis, how do you find null hypothesis, what is a null hypothesis.

Null Hypothesis in statistical analysis suggests the absence of statistical significance within a specific set of observed data. Hypothesis testing, using sample data, evaluates the validity of this hypothesis. Commonly denoted as H 0 or simply “null,” it plays an important role in quantitative analysis, examining theories related to markets, investment strategies, or economies to determine their validity.

Definition of Null Hypothesis

Null Hypothesis represent a default position, often suggesting no effect or difference, against which researchers compare their experimental results. The Null Hypothesis, often denoted as H 0 , asserts a default assumption in statistical analysis. It posits no significant difference or effect, serving as a baseline for comparison in hypothesis testing.

Null Hypothesis is represented as H 0 , the Null Hypothesis symbolizes the absence of a measurable effect or difference in the variables under examination.

Certainly, a simple example would be asserting that the mean score of a group is equal to a specified value like stating that the average IQ of a population is 100.

The Null Hypothesis is typically formulated as a statement of equality or absence of a specific parameter in the population being studied. It provides a clear and testable prediction for comparison with the alternative hypothesis. The formulation of the Null Hypothesis typically follows a concise structure, stating the equality or absence of a specific parameter in the population.

Mean Comparison (Two-sample t-test)

H 0 : μ 1 = μ 2

This asserts that there is no significant difference between the means of two populations or groups.

Proportion Comparison

H 0 : p 1 − p 2 = 0

This suggests no significant difference in proportions between two populations or conditions.

Equality in Variance (F-test in ANOVA)

H 0 : σ 1 = σ 2

This states that there’s no significant difference in variances between groups or populations.

Independence (Chi-square Test of Independence):

H 0 : Variables are independent

This asserts that there’s no association or relationship between categorical variables.

Null Hypotheses vary including simple and composite forms, each tailored to the complexity of the research question. Understanding these types is pivotal for effective hypothesis testing.

Equality Null Hypothesis (Simple Null Hypothesis)

The Equality Null Hypothesis, also known as the Simple Null Hypothesis, is a fundamental concept in statistical hypothesis testing that assumes no difference, effect or relationship between groups, conditions or populations being compared.

Non-Inferiority Null Hypothesis

In some studies, the focus might be on demonstrating that a new treatment or method is not significantly worse than the standard or existing one.

Superiority Null Hypothesis

The concept of a superiority null hypothesis comes into play when a study aims to demonstrate that a new treatment, method, or intervention is significantly better than an existing or standard one.

Independence Null Hypothesis

In certain statistical tests, such as chi-square tests for independence, the null hypothesis assumes no association or independence between categorical variables.

Homogeneity Null Hypothesis

In tests like ANOVA (Analysis of Variance), the null hypothesis suggests that there’s no difference in population means across different groups.

Examples of Null Hypothesis

Medicine: Null Hypothesis: “No significant difference exists in blood pressure levels between patients given the experimental drug versus those given a placebo.”
Education: Null Hypothesis: “There’s no significant variation in test scores between students using a new teaching method and those using traditional teaching.”
Economics: Null Hypothesis: “There’s no significant change in consumer spending pre- and post-implementation of a new taxation policy.”
Environmental Science: Null Hypothesis: “There’s no substantial difference in pollution levels before and after a water treatment plant’s establishment.”

The principle of the null hypothesis is a fundamental concept in statistical hypothesis testing. It involves making an assumption about the population parameter or the absence of an effect or relationship between variables.

In essence, the null hypothesis (H 0 ) proposes that there is no significant difference, effect, or relationship between variables. It serves as a starting point or a default assumption that there is no real change, no effect or no difference between groups or conditions.

$\alpha$

Null Hypothesis Rejection

Rejecting the Null Hypothesis occurs when statistical evidence suggests a significant departure from the assumed baseline. It implies that there is enough evidence to support the alternative hypothesis, indicating a meaningful effect or difference. Null Hypothesis rejection occurs when statistical evidence suggests a deviation from the assumed baseline, prompting a reconsideration of the initial hypothesis.

Identifying the Null Hypothesis involves defining the status quotient, asserting no effect and formulating a statement suitable for statistical analysis.

When is Null Hypothesis Rejected?

The Null Hypothesis is rejected when statistical tests indicate a significant departure from the expected outcome, leading to the consideration of alternative hypotheses. It occurs when statistical evidence suggests a deviation from the assumed baseline, prompting a reconsideration of the initial hypothesis.

Null Hypothesis and Alternative Hypothesis

In the realm of hypothesis testing, the null hypothesis (H 0 ) and alternative hypothesis (H₁ or Ha) play critical roles. The null hypothesis generally assumes no difference, effect, or relationship between variables, suggesting that any observed change or effect is due to random chance. Its counterpart, the alternative hypothesis, asserts the presence of a significant difference, effect, or relationship between variables, challenging the null hypothesis. These hypotheses are formulated based on the research question and guide statistical analyses.

Null Hypothesis vs Alternative Hypothesis

The null hypothesis (H 0 ) serves as the baseline assumption in statistical testing, suggesting no significant effect, relationship, or difference within the data. It often proposes that any observed change or correlation is merely due to chance or random variation. Conversely, the alternative hypothesis (H 1 or Ha) contradicts the null hypothesis, positing the existence of a genuine effect, relationship or difference in the data. It represents the researcher’s intended focus, seeking to provide evidence against the null hypothesis and support for a specific outcome or theory. These hypotheses form the crux of hypothesis testing, guiding the assessment of data to draw conclusions about the population being studied.

Example of Alternative and Null Hypothesis

Let’s envision a scenario where a researcher aims to examine the impact of a new medication on reducing blood pressure among patients. In this context:

Null Hypothesis (H 0 ): “The new medication does not produce a significant effect in reducing blood pressure levels among patients.”

Alternative Hypothesis (H 1 or Ha): “The new medication yields a significant effect in reducing blood pressure levels among patients.”

The null hypothesis implies that any observed alterations in blood pressure subsequent to the medication’s administration are a result of random fluctuations rather than a consequence of the medication itself. Conversely, the alternative hypothesis contends that the medication does indeed generate a meaningful alteration in blood pressure levels, distinct from what might naturally occur or by random chance.

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Solved Examples on Null Hypothesis

Example 1: A researcher claims that the average time students spend on homework is 2 hours per night.

Null Hypothesis (H 0 ): The average time students spend on homework is equal to 2 hours per night. Data: A random sample of 30 students has an average homework time of 1.8 hours with a standard deviation of 0.5 hours. Test Statistic and Decision: Using a t-test, if the calculated t-statistic falls within the acceptance region, we fail to reject the null hypothesis. If it falls in the rejection region, we reject the null hypothesis. Conclusion: Based on the statistical analysis, we fail to reject the null hypothesis, suggesting that there is not enough evidence to dispute the claim of the average homework time being 2 hours per night.

Example 2: A company asserts that the error rate in its production process is less than 1%.

Null Hypothesis (H 0 ): The error rate in the production process is 1% or higher. Data: A sample of 500 products shows an error rate of 0.8%. Test Statistic and Decision: Using a z-test, if the calculated z-statistic falls within the acceptance region, we fail to reject the null hypothesis. If it falls in the rejection region, we reject the null hypothesis. Conclusion: The statistical analysis supports rejecting the null hypothesis, indicating that there is enough evidence to dispute the company’s claim of an error rate of 1% or higher.

Null Hypothesis – Practice Problems

Q1. A researcher claims that the average time spent by students on homework is less than 2 hours per day. Formulate the null hypothesis for this claim?

Q2. A manufacturing company states that their new machine produces widgets with a defect rate of less than 5%. Write the null hypothesis to test this claim?

Q3. An educational institute believes that their online course completion rate is at least 60%. Develop the null hypothesis to validate this assertion?

Q4. A restaurant claims that the waiting time for customers during peak hours is not more than 15 minutes. Formulate the null hypothesis for this claim?

Q5. A study suggests that the mean weight loss after following a specific diet plan for a month is more than 8 pounds. Construct the null hypothesis to evaluate this statement?

Null Hypothesis – Frequently Asked Questions

How to form a null hypothesis.

A null hypothesis is formed based on the assumption that there is no significant difference or effect between the groups being compared or no association between variables being tested. It often involves stating that there is no relationship, no change, or no effect in the population being studied.

When Do we reject the Null Hypothesis?

In statistical hypothesis testing, if the p-value (the probability of obtaining the observed results) is lower than the chosen significance level (commonly 0.05), we reject the null hypothesis. This suggests that the data provides enough evidence to refute the assumption made in the null hypothesis.

What is a Null Hypothesis in Research?

In research, the null hypothesis represents the default assumption or position that there is no significant difference or effect. Researchers often try to test this hypothesis by collecting data and performing statistical analyses to see if the observed results contradict the assumption.

What Are Alternative and Null Hypotheses?

The null hypothesis (H0) is the default assumption that there is no significant difference or effect. The alternative hypothesis (H1 or Ha) is the opposite, suggesting there is a significant difference, effect or relationship.

What Does it Mean to Reject the Null Hypothesis?

Rejecting the null hypothesis implies that there is enough evidence in the data to support the alternative hypothesis. In simpler terms, it suggests that there might be a significant difference, effect or relationship between the groups or variables being studied.

How to Find Null Hypothesis?

Formulating a null hypothesis often involves considering the research question and assuming that no difference or effect exists. It should be a statement that can be tested through data collection and statistical analysis, typically stating no relationship or no change between variables or groups.

How is Null Hypothesis denoted?

The null hypothesis is commonly symbolized as H 0 in statistical notation.

What is the Purpose of the Null hypothesis in Statistical Analysis?

The null hypothesis serves as a starting point for hypothesis testing, enabling researchers to assess if there’s enough evidence to reject it in favor of an alternative hypothesis.

What happens if we Reject the Null hypothesis?

Rejecting the null hypothesis implies that there is sufficient evidence to support an alternative hypothesis, suggesting a significant effect or relationship between variables.

Is it Possible to Prove the Null Hypothesis?

No, statistical testing aims to either reject or fail to reject the null hypothesis based on evidence from sample data. It does not prove the null hypothesis to be true.

What are Test for Null Hypothesis?

Various statistical tests, such as t-tests or chi-square tests, are employed to evaluate the validity of the Null Hypothesis in different scenarios.

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Null Hypothesis: Definition, Rejecting & Examples
Null Hypothesis H 0: The correlation in the population is zero: ρ = 0. Alternative Hypothesis H A: The correlation in the population is not zero: ρ ≠ 0. For all these cases, the analysts define the hypotheses before the study. After collecting the data, they perform a hypothesis test to determine whether they can reject the null hypothesis.
Hypothesis Test: Difference in Means
The first step is to state the null hypothesis and an alternative hypothesis. Null hypothesis: μ 1 - μ 2 = 0. Alternative hypothesis: μ 1 - μ 2 ≠ 0. Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the difference between sample means is too big or if it is too small.
Null & Alternative Hypotheses
Research question: Null hypothesis (H 0): General: Test-specific: Does tooth flossing affect the number of cavities? Tooth flossing has no effect on the number of cavities.: t test:. The mean number of cavities per person does not differ between the flossing group (µ 1) and the non-flossing group (µ 2) in the population; µ 1 = µ 2.: Does the amount of text highlighted in the textbook ...
9.1 Null and Alternative Hypotheses
The actual test begins by considering two hypotheses.They are called the null hypothesis and the alternative hypothesis.These hypotheses contain opposing viewpoints. H 0, the —null hypothesis: a statement of no difference between sample means or proportions or no difference between a sample mean or proportion and a population mean or proportion. In other words, the difference equals 0.
Hypothesis Test for a Mean
The first step is to state the null hypothesis and an alternative hypothesis. Null hypothesis: μ >= 110. Alternative hypothesis: μ < 110. Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the sample mean is too small. Formulate an analysis plan. For this analysis, the significance level is 0.01.
9.1: Null and Alternative Hypotheses
The actual test begins by considering two hypotheses.They are called the null hypothesis and the alternative hypothesis.These hypotheses contain opposing viewpoints. $H_0$: The null hypothesis: It is a statement of no difference between the variables—they are not related. This can often be considered the status quo and as a result if you cannot accept the null it requires some action.
An Introduction to t Tests
Revised on June 22, 2023. A t test is a statistical test that is used to compare the means of two groups. It is often used in hypothesis testing to determine whether a process or treatment actually has an effect on the population of interest, or whether two groups are different from one another. t test example.
Null hypothesis
A possible null hypothesis is that the mean male score is the same as the mean female score: H 0: ... Testing the null hypothesis is a central task in statistical hypothesis testing in the modern practice of science. There are precise criteria for excluding or not excluding a null hypothesis at a certain confidence level. The confidence level ...
Hypothesis Testing
Table of contents. Step 1: State your null and alternate hypothesis. Step 2: Collect data. Step 3: Perform a statistical test. Step 4: Decide whether to reject or fail to reject your null hypothesis. Step 5: Present your findings. Other interesting articles. Frequently asked questions about hypothesis testing.
16.3: The Process of Null Hypothesis Testing
As expected under the null hypothesis, this distribution is centered at zero (the mean of the distribution is -0.016. From the figure we can also see that the distribution of t values after shuffling roughly follows the theoretical t distribution under the null hypothesis (with mean=0), showing that randomization worked to generate null data.
13.1 Understanding Null Hypothesis Testing
A crucial step in null hypothesis testing is finding the likelihood of the sample result if the null hypothesis were true. This probability is called the p value. A low p value means that the sample result would be unlikely if the null hypothesis were true and leads to the rejection of the null hypothesis. A p value that is not low means that ...
Hypothesis Testing
Let's return finally to the question of whether we reject or fail to reject the null hypothesis. If our statistical analysis shows that the significance level is below the cut-off value we have set (e.g., either 0.05 or 0.01), we reject the null hypothesis and accept the alternative hypothesis. Alternatively, if the significance level is above ...
Examples of null and alternative hypotheses
It is the opposite of your research hypothesis. The alternative hypothesis--that is, the research hypothesis--is the idea, phenomenon, observation that you want to prove. If you suspect that girls take longer to get ready for school than boys, then: Alternative: girls time > boys time. Null: girls time <= boys time.
Understanding Null Hypothesis Testing
A crucial step in null hypothesis testing is finding the likelihood of the sample result if the null hypothesis were true. This probability is called the p value. A low p value means that the sample result would be unlikely if the null hypothesis were true and leads to the rejection of the null hypothesis. A high p value means that the sample ...
What Is The Null Hypothesis & When To Reject It
Rejecting the null hypothesis means that a relationship does exist between a set of variables and the effect is statistically significant ... R. S. (2000). Null hypothesis significance testing: a review of an old and continuing controversy. Psychological methods, 5(2), 241. Rozeboom, W. W. (1960). The fallacy of the null-hypothesis significance ...
10.29: Hypothesis Test for a Difference in Two Population Means (1 of 2
Step 1: Determine the hypotheses. The hypotheses for a difference in two population means are similar to those for a difference in two population proportions. The null hypothesis, H 0, is again a statement of "no effect" or "no difference.". H 0: μ 1 - μ 2 = 0, which is the same as H 0: μ 1 = μ 2. The alternative hypothesis, H a ...
Hypothesis Testing for Means & Proportions
In the two independent samples application with a continuous outcome, the parameter of interest in the test of hypothesis is the difference in population means, μ 1-μ 2. The null hypothesis is always that there is no difference between groups with respect to means, i.e., The null hypothesis can also be written as follows: H 0: μ 1 = μ 2.
How to Write a Null Hypothesis (5 Examples)
Whenever we perform a hypothesis test, we always write a null hypothesis and an alternative hypothesis, which take the following forms: H0 (Null Hypothesis): Population parameter =, ≤, ≥ some value. HA (Alternative Hypothesis): Population parameter <, >, ≠ some value. Note that the null hypothesis always contains the equal sign.
T-test and Hypothesis Testing (Explained Simply)
Aug 5, 2022. 6. Photo by Andrew George on Unsplash. Student's t-tests are commonly used in inferential statistics for testing a hypothesis on the basis of a difference between sample means. However, people often misinterpret the results of t-tests, which leads to false research findings and a lack of reproducibility of studies.
Hypothesis Testing for the Mean
Table 8.3: One-sided hypothesis testing for the mean: H0: μ ≤ μ0, H1: μ > μ0. Note that the tests mentioned in Table 8.3 remain valid if we replace the null hypothesis by μ = μ0. The reason for this is that in choosing the threshold c, we assumed the worst case scenario, i.e, μ = μ0 .
Null Hypothesis
Mean Comparison (Two-sample t-test) H 0: ... In the realm of hypothesis testing, the null hypothesis (H 0) and alternative hypothesis (H₁ or Ha) play critical roles. The null hypothesis generally assumes no difference, effect, or relationship between variables, suggesting that any observed change or effect is due to random chance. ...
Choosing the Right Statistical Test
The p-value estimates how likely it is that you would see the difference described by the test statistic if the null hypothesis of no relationship were true. ... The most common threshold is p < 0.05, which means that the data is likely to occur less than 5% of the time under the null hypothesis. When the p-value falls below the chosen alpha ...
How to Find P Value from a Test Statistic
Hypothesis tests are used to test the validity of a claim that is made about a population. This claim that's on trial, in essence, is called the null hypothesis (H 0).The alternative hypothesis (H a) is the one you would believe if the null hypothesis is concluded to be untrue.Learning how to find the p-value in statistics is a fundamental skill in testing, helping you weigh the evidence ...