## Have a language expert improve your writing

Run a free plagiarism check in 10 minutes, generate accurate citations for free.

- Knowledge Base

## Hypothesis Testing | A Step-by-Step Guide with Easy Examples

Published on November 8, 2019 by Rebecca Bevans . Revised on June 22, 2023.

Hypothesis testing is a formal procedure for investigating our ideas about the world using statistics . It is most often used by scientists to test specific predictions, called hypotheses, that arise from theories.

There are 5 main steps in hypothesis testing:

- State your research hypothesis as a null hypothesis and alternate hypothesis (H o ) and (H a or H 1 ).
- Collect data in a way designed to test the hypothesis.
- Perform an appropriate statistical test .
- Decide whether to reject or fail to reject your null hypothesis.
- Present the findings in your results and discussion section.

Though the specific details might vary, the procedure you will use when testing a hypothesis will always follow some version of these steps.

## 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.

After developing your initial research hypothesis (the prediction that you want to investigate), it is important to restate it as a null (H o ) and alternate (H a ) hypothesis so that you can test it mathematically.

The alternate hypothesis is usually your initial hypothesis that predicts a relationship between variables. The null hypothesis is a prediction of no relationship between the variables you are interested in.

- H 0 : Men are, on average, not taller than women. H a : Men are, on average, taller than women.

## Prevent plagiarism. Run a free check.

For a statistical test to be valid , it is important to perform sampling and collect data in a way that is designed to test your hypothesis. If your data are not representative, then you cannot make statistical inferences about the population you are interested in.

There are a variety of statistical tests available, but they are all based on the comparison of within-group variance (how spread out the data is within a category) versus between-group variance (how different the categories are from one another).

If the between-group variance is large enough that there is little or no overlap between groups, then your statistical test will reflect that by showing a low p -value . This means it is unlikely that the differences between these groups came about by chance.

Alternatively, if there is high within-group variance and low between-group variance, then your statistical test will reflect that with a high p -value. This means it is likely that any difference you measure between groups is due to chance.

Your choice of statistical test will be based on the type of variables and the level of measurement of your collected data .

- an estimate of the difference in average height between the two groups.
- a p -value showing how likely you are to see this difference if the null hypothesis of no difference is true.

Based on the outcome of your statistical test, you will have to decide whether to reject or fail to reject your null hypothesis.

In most cases you will use the p -value generated by your statistical test to guide your decision. And in most cases, your predetermined level of significance for rejecting the null hypothesis will be 0.05 – that is, when there is a less than 5% chance that you would see these results if the null hypothesis were true.

In some cases, researchers choose a more conservative level of significance, such as 0.01 (1%). This minimizes the risk of incorrectly rejecting the null hypothesis ( Type I error ).

## Receive feedback on language, structure, and formatting

Professional editors proofread and edit your paper by focusing on:

- Academic style
- Vague sentences
- Style consistency

See an example

The results of hypothesis testing will be presented in the results and discussion sections of your research paper , dissertation or thesis .

In the results section you should give a brief summary of the data and a summary of the results of your statistical test (for example, the estimated difference between group means and associated p -value). In the discussion , you can discuss whether your initial hypothesis was supported by your results or not.

In the formal language of hypothesis testing, we talk about rejecting or failing to reject the null hypothesis. You will probably be asked to do this in your statistics assignments.

However, when presenting research results in academic papers we rarely talk this way. Instead, we go back to our alternate hypothesis (in this case, the hypothesis that men are on average taller than women) and state whether the result of our test did or did not support the alternate hypothesis.

If your null hypothesis was rejected, this result is interpreted as “supported the alternate hypothesis.”

These are superficial differences; you can see that they mean the same thing.

You might notice that we don’t say that we reject or fail to reject the alternate hypothesis . This is because hypothesis testing is not designed to prove or disprove anything. It is only designed to test whether a pattern we measure could have arisen spuriously, or by chance.

If we reject the null hypothesis based on our research (i.e., we find that it is unlikely that the pattern arose by chance), then we can say our test lends support to our hypothesis . But if the pattern does not pass our decision rule, meaning that it could have arisen by chance, then we say the test is inconsistent with our hypothesis .

If you want to know more about statistics , methodology , or research bias , make sure to check out some of our other articles with explanations and examples.

- Normal distribution
- Descriptive statistics
- Measures of central tendency
- Correlation coefficient

Methodology

- Cluster sampling
- Stratified sampling
- Types of interviews
- Cohort study
- Thematic analysis

Research bias

- Implicit bias
- Cognitive bias
- Survivorship bias
- Availability heuristic
- Nonresponse bias
- Regression to the mean

Hypothesis testing is a formal procedure for investigating our ideas about the world using statistics. It is used by scientists to test specific predictions, called hypotheses , by calculating how likely it is that a pattern or relationship between variables could have arisen by chance.

A hypothesis states your predictions about what your research will find. It is a tentative answer to your research question that has not yet been tested. For some research projects, you might have to write several hypotheses that address different aspects of your research question.

A hypothesis is not just a guess — it should be based on existing theories and knowledge. It also has to be testable, which means you can support or refute it through scientific research methods (such as experiments, observations and statistical analysis of data).

Null and alternative hypotheses are used in statistical hypothesis testing . The null hypothesis of a test always predicts no effect or no relationship between variables, while the alternative hypothesis states your research prediction of an effect or relationship.

## Cite this Scribbr article

If you want to cite this source, you can copy and paste the citation or click the “Cite this Scribbr article” button to automatically add the citation to our free Citation Generator.

Bevans, R. (2023, June 22). Hypothesis Testing | A Step-by-Step Guide with Easy Examples. Scribbr. Retrieved March 14, 2024, from https://www.scribbr.com/statistics/hypothesis-testing/

## Is this article helpful?

## Rebecca Bevans

Other students also liked, choosing the right statistical test | types & examples, understanding p values | definition and examples, what is your plagiarism score.

## Hypothesis Testing

This page looks at Hypothesis testing. Topics include null hypothesis, alternative hypothesis, testing and critical regions.

The parameters of a distribution are those quantities that you need to specify when describing the distribution. For example, a normal distribution has parameters μ and σ 2 and a Poisson distribution has parameter λ.

If we know that some data comes from a certain distribution, but the parameter is unknown, we might try to predict what the parameter is. Hypothesis testing is about working out how likely our predictions are.

The null hypothesis , denoted by H 0 , is a prediction about a parameter (so if we are dealing with a normal distribution, we might predict the mean or the variance of the distribution).

We also have an alternative hypothesis , denoted by H 1 . We then perform a test to decide whether or not we should reject the null hypothesis in favour of the alternative.

Suppose we are given a value and told that it comes from a certain distribution, but we don"t know what the parameter of that distribution is.

Suppose we make a null hypothesis about the parameter. We test how likely it is that the value we were given could have come from the distribution with this predicted parameter.

For example, suppose we are told that the value of 3 has come from a Poisson distribution. We might want to test the null hypothesis that the parameter (which is the mean) of the Poisson distribution is 9. So we work out how likely it is that the value of 3 could have come from a Poisson distribution with parameter 9. If it"s not very likely, we reject the null hypothesis in favour of the alternative.

Critical Region

But what exactly is "not very likely"?

We choose a region known as the critical region . If the result of our test lies in this region, then we reject the null hypothesis in favour of the alternative.

## AS Level Mechanics and Statistics - Hypothesis Testing

A Level Maths

Maths A-Level Resources for AQA, OCR and Edexcel

## A-Level Maths Statistical Hypothesis Testing

Everything you need to know about statistical hypothesis testing for A-Level Maths. Save countless hours of time!

## Statistical Hypothesis Testing Topics for A-Level Maths

This module will teach you the following:

## Hypothesis testing in a binomial distribution

Hypothesis testing in a normal distribution, hypothesis test using pearson’s correlation coefficient, what’s included.

We’ve created 52 modules covering every Maths topic needed for A level, and each module contains:

- An editable PowerPoint lesson presentation
- Editable revision handouts
- A glossary which covers the key terminologies of the module
- Topical mind maps for visualising the key concepts
- Printable flashcards to help students engage active recall and confidence-based repetition
- A quiz with accompanying answer key to test knowledge and understanding of the module

As a premium member, once rolled out you get access to the entire library of A-Level Maths resources. For now, we have made the first five topics completely free of charge for you to get a taste of what’s to come.

## A Level Maths Resources Mapped by Exam Board

Once completed our modules can be used with both UK and international A Level examination board specifications.

We will put together comprehensive mapping documents which will show you exactly which modules align to the specification you are teaching or learning.

## Statistical Hypothesis Testing Topics:

Want to pass your maths a-levels.

Join the waiting list now, get the 5 free chapters of our COMPREHENSIVE A-LEVEL MATHS COURSE and we’ll help you to pass your exams!

## 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.

Statistics Made Easy

## Introduction to Hypothesis Testing

A statistical hypothesis is an assumption about a population parameter .

For example, we may assume that the mean height of a male in the U.S. is 70 inches.

The assumption about the height is the statistical hypothesis and the true mean height of a male in the U.S. is the population parameter .

A hypothesis test is a formal statistical test we use to reject or fail to reject a statistical hypothesis.

## The Two Types of Statistical Hypotheses

To test whether a statistical hypothesis about a population parameter is true, we obtain a random sample from the population and perform a hypothesis test on the sample data.

There are two types of statistical hypotheses:

The null hypothesis , denoted as H 0 , is the hypothesis that the sample data occurs purely from chance.

The alternative hypothesis , denoted as H 1 or H a , is the hypothesis that the sample data is influenced by some non-random cause.

## Hypothesis Tests

A hypothesis test consists of five steps:

1. State the hypotheses.

State the null and alternative hypotheses. These two hypotheses need to be mutually exclusive, so if one is true then the other must be false.

2. Determine a significance level to use for the hypothesis.

Decide on a significance level. Common choices are .01, .05, and .1.

3. Find the test statistic.

Find the test statistic and the corresponding p-value. Often we are analyzing a population mean or proportion and the general formula to find the test statistic is: (sample statistic – population parameter) / (standard deviation of statistic)

4. Reject or fail to reject the null hypothesis.

Using the test statistic or the p-value, determine if you can reject or fail to reject the null hypothesis based on the significance level.

The p-value tells us the strength of evidence in support of a null hypothesis. If the p-value is less than the significance level, we reject the null hypothesis.

5. Interpret the results.

Interpret the results of the hypothesis test in the context of the question being asked.

## The Two Types of Decision Errors

There are two types of decision errors that one can make when doing a hypothesis test:

Type I error: You reject the null hypothesis when it is actually true. The probability of committing a Type I error is equal to the significance level, often called alpha , and denoted as α.

Type II error: You fail to reject the null hypothesis when it is actually false. The probability of committing a Type II error is called the Power of the test or Beta , denoted as β.

## One-Tailed and Two-Tailed Tests

A statistical hypothesis can be one-tailed or two-tailed.

A one-tailed hypothesis involves making a “greater than” or “less than ” statement.

For example, suppose we assume the mean height of a male in the U.S. is greater than or equal to 70 inches. The null hypothesis would be H0: µ ≥ 70 inches and the alternative hypothesis would be Ha: µ < 70 inches.

A two-tailed hypothesis involves making an “equal to” or “not equal to” statement.

For example, suppose we assume the mean height of a male in the U.S. is equal to 70 inches. The null hypothesis would be H0: µ = 70 inches and the alternative hypothesis would be Ha: µ ≠ 70 inches.

Note: The “equal” sign is always included in the null hypothesis, whether it is =, ≥, or ≤.

Related: What is a Directional Hypothesis?

## Types of Hypothesis Tests

There are many different types of hypothesis tests you can perform depending on the type of data you’re working with and the goal of your analysis.

The following tutorials provide an explanation of the most common types of hypothesis tests:

Introduction to the One Sample t-test Introduction to the Two Sample t-test Introduction to the Paired Samples t-test Introduction to the One Proportion Z-Test Introduction to the Two Proportion Z-Test

## Published by Zach

Leave a reply cancel reply.

Your email address will not be published. Required fields are marked *

- school Campus Bookshelves
- menu_book Bookshelves
- perm_media Learning Objects
- login Login
- how_to_reg Request Instructor Account
- hub Instructor Commons
- Download Page (PDF)
- Download Full Book (PDF)
- Periodic Table
- Physics Constants
- Scientific Calculator
- Reference & Cite
- Tools expand_more
- Readability

selected template will load here

This action is not available.

## 9.E: Hypothesis Testing with One Sample (Exercises)

- Last updated
- Save as PDF
- Page ID 1146

These are homework exercises to accompany the Textmap created for "Introductory Statistics" by OpenStax.

## 9.1: Introduction

9.2: null and alternative hypotheses.

Some of the following statements refer to the null hypothesis, some to the alternate hypothesis.

State the null hypothesis, \(H_{0}\), and the alternative hypothesis. \(H_{a}\), in terms of the appropriate parameter \((\mu \text{or} p)\).

- The mean number of years Americans work before retiring is 34.
- At most 60% of Americans vote in presidential elections.
- The mean starting salary for San Jose State University graduates is at least $100,000 per year.
- Twenty-nine percent of high school seniors get drunk each month.
- Fewer than 5% of adults ride the bus to work in Los Angeles.
- The mean number of cars a person owns in her lifetime is not more than ten.
- About half of Americans prefer to live away from cities, given the choice.
- Europeans have a mean paid vacation each year of six weeks.
- The chance of developing breast cancer is under 11% for women.
- Private universities' mean tuition cost is more than $20,000 per year.
- \(H_{0}: \mu = 34; H_{a}: \mu \neq 34\)
- \(H_{0}: p \leq 0.60; H_{a}: p > 0.60\)
- \(H_{0}: \mu \geq 100,000; H_{a}: \mu < 100,000\)
- \(H_{0}: p = 0.29; H_{a}: p \neq 0.29\)
- \(H_{0}: p = 0.05; H_{a}: p < 0.05\)
- \(H_{0}: \mu \leq 10; H_{a}: \mu > 10\)
- \(H_{0}: p = 0.50; H_{a}: p \neq 0.50\)
- \(H_{0}: \mu = 6; H_{a}: \mu \neq 6\)
- \(H_{0}: p ≥ 0.11; H_{a}: p < 0.11\)
- \(H_{0}: \mu \leq 20,000; H_{a}: \mu > 20,000\)

Over the past few decades, public health officials have examined the link between weight concerns and teen girls' smoking. Researchers surveyed a group of 273 randomly selected teen girls living in Massachusetts (between 12 and 15 years old). After four years the girls were surveyed again. Sixty-three said they smoked to stay thin. Is there good evidence that more than thirty percent of the teen girls smoke to stay thin? The alternative hypothesis is:

- \(p < 0.30\)
- \(p \leq 0.30\)
- \(p \geq 0.30\)
- \(p > 0.30\)

A statistics instructor believes that fewer than 20% of Evergreen Valley College (EVC) students attended the opening night midnight showing of the latest Harry Potter movie. She surveys 84 of her students and finds that 11 attended the midnight showing. An appropriate alternative hypothesis is:

- \(p = 0.20\)
- \(p > 0.20\)
- \(p < 0.20\)
- \(p \leq 0.20\)

Previously, an organization reported that teenagers spent 4.5 hours per week, on average, on the phone. The organization thinks that, currently, the mean is higher. Fifteen randomly chosen teenagers were asked how many hours per week they spend on the phone. The sample mean was 4.75 hours with a sample standard deviation of 2.0. Conduct a hypothesis test. The null and alternative hypotheses are:

- \(H_{0}: \bar{x} = 4.5, H_{a}: \bar{x} > 4.5\)
- \(H_{0}: \mu \geq 4.5, H_{a}: \mu < 4.5\)
- \(H_{0}: \mu = 4.75, H_{a}: \mu > 4.75\)
- \(H_{0}: \mu = 4.5, H_{a}: \mu > 4.5\)

## 9.3: Outcomes and the Type I and Type II Errors

State the Type I and Type II errors in complete sentences given the following statements.

- The mean number of cars a person owns in his or her lifetime is not more than ten.
- Private universities mean tuition cost is more than $20,000 per year.
- Type I error: We conclude that the mean is not 34 years, when it really is 34 years. Type II error: We conclude that the mean is 34 years, when in fact it really is not 34 years.
- Type I error: We conclude that more than 60% of Americans vote in presidential elections, when the actual percentage is at most 60%.Type II error: We conclude that at most 60% of Americans vote in presidential elections when, in fact, more than 60% do.
- Type I error: We conclude that the mean starting salary is less than $100,000, when it really is at least $100,000. Type II error: We conclude that the mean starting salary is at least $100,000 when, in fact, it is less than $100,000.
- Type I error: We conclude that the proportion of high school seniors who get drunk each month is not 29%, when it really is 29%. Type II error: We conclude that the proportion of high school seniors who get drunk each month is 29% when, in fact, it is not 29%.
- Type I error: We conclude that fewer than 5% of adults ride the bus to work in Los Angeles, when the percentage that do is really 5% or more. Type II error: We conclude that 5% or more adults ride the bus to work in Los Angeles when, in fact, fewer that 5% do.
- Type I error: We conclude that the mean number of cars a person owns in his or her lifetime is more than 10, when in reality it is not more than 10. Type II error: We conclude that the mean number of cars a person owns in his or her lifetime is not more than 10 when, in fact, it is more than 10.
- Type I error: We conclude that the proportion of Americans who prefer to live away from cities is not about half, though the actual proportion is about half. Type II error: We conclude that the proportion of Americans who prefer to live away from cities is half when, in fact, it is not half.
- Type I error: We conclude that the duration of paid vacations each year for Europeans is not six weeks, when in fact it is six weeks. Type II error: We conclude that the duration of paid vacations each year for Europeans is six weeks when, in fact, it is not.
- Type I error: We conclude that the proportion is less than 11%, when it is really at least 11%. Type II error: We conclude that the proportion of women who develop breast cancer is at least 11%, when in fact it is less than 11%.
- Type I error: We conclude that the average tuition cost at private universities is more than $20,000, though in reality it is at most $20,000. Type II error: We conclude that the average tuition cost at private universities is at most $20,000 when, in fact, it is more than $20,000.

For statements a-j in Exercise 9.109 , answer the following in complete sentences.

- State a consequence of committing a Type I error.
- State a consequence of committing a Type II error.

When a new drug is created, the pharmaceutical company must subject it to testing before receiving the necessary permission from the Food and Drug Administration (FDA) to market the drug. Suppose the null hypothesis is “the drug is unsafe.” What is the Type II Error?

- To conclude the drug is safe when in, fact, it is unsafe.
- Not to conclude the drug is safe when, in fact, it is safe.
- To conclude the drug is safe when, in fact, it is safe.
- Not to conclude the drug is unsafe when, in fact, it is unsafe.

A statistics instructor believes that fewer than 20% of Evergreen Valley College (EVC) students attended the opening midnight showing of the latest Harry Potter movie. She surveys 84 of her students and finds that 11 of them attended the midnight showing. The Type I error is to conclude that the percent of EVC students who attended is ________.

- at least 20%, when in fact, it is less than 20%.
- 20%, when in fact, it is 20%.
- less than 20%, when in fact, it is at least 20%.
- less than 20%, when in fact, it is less than 20%.

It is believed that Lake Tahoe Community College (LTCC) Intermediate Algebra students get less than seven hours of sleep per night, on average. A survey of 22 LTCC Intermediate Algebra students generated a mean of 7.24 hours with a standard deviation of 1.93 hours. At a level of significance of 5%, do LTCC Intermediate Algebra students get less than seven hours of sleep per night, on average?

The Type II error is not to reject that the mean number of hours of sleep LTCC students get per night is at least seven when, in fact, the mean number of hours

- is more than seven hours.
- is at most seven hours.
- is at least seven hours.
- is less than seven hours.

Previously, an organization reported that teenagers spent 4.5 hours per week, on average, on the phone. The organization thinks that, currently, the mean is higher. Fifteen randomly chosen teenagers were asked how many hours per week they spend on the phone. The sample mean was 4.75 hours with a sample standard deviation of 2.0. Conduct a hypothesis test, the Type I error is:

- to conclude that the current mean hours per week is higher than 4.5, when in fact, it is higher
- to conclude that the current mean hours per week is higher than 4.5, when in fact, it is the same
- to conclude that the mean hours per week currently is 4.5, when in fact, it is higher
- to conclude that the mean hours per week currently is no higher than 4.5, when in fact, it is not higher

## 9.4: Distribution Needed for Hypothesis Testing

It is believed that Lake Tahoe Community College (LTCC) Intermediate Algebra students get less than seven hours of sleep per night, on average. A survey of 22 LTCC Intermediate Algebra students generated a mean of 7.24 hours with a standard deviation of 1.93 hours. At a level of significance of 5%, do LTCC Intermediate Algebra students get less than seven hours of sleep per night, on average? The distribution to be used for this test is \(\bar{X} \sim\) ________________

- \(N\left(7.24, \frac{1.93}{\sqrt{22}}\right)\)
- \(N\left(7.24, 1.93\right)\)

## 9.5: Rare Events, the Sample, Decision and Conclusion

The National Institute of Mental Health published an article stating that in any one-year period, approximately 9.5 percent of American adults suffer from depression or a depressive illness. Suppose that in a survey of 100 people in a certain town, seven of them suffered from depression or a depressive illness. Conduct a hypothesis test to determine if the true proportion of people in that town suffering from depression or a depressive illness is lower than the percent in the general adult American population.

- Is this a test of one mean or proportion?
- State the null and alternative hypotheses. \(H_{0}\) : ____________________ \(H_{a}\) : ____________________
- Is this a right-tailed, left-tailed, or two-tailed test?
- What symbol represents the random variable for this test?
- In words, define the random variable for this test.
- \(x =\) ________________
- \(n =\) ________________
- \(p′ =\) _____________
- Calculate \(\sigma_{x} =\) __________. Show the formula set-up.
- State the distribution to use for the hypothesis test.
- Find the \(p\text{-value}\).
- Reason for the decision:
- Conclusion (write out in a complete sentence):

## 9.6: Additional Information and Full Hypothesis Test Examples

For each of the word problems, use a solution sheet to do the hypothesis test. The solution sheet is found in [link] . Please feel free to make copies of the solution sheets. For the online version of the book, it is suggested that you copy the .doc or the .pdf files.

If you are using a Student's \(t\) - distribution for one of the following homework problems, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, however.)

A particular brand of tires claims that its deluxe tire averages at least 50,000 miles before it needs to be replaced. From past studies of this tire, the standard deviation is known to be 8,000. A survey of owners of that tire design is conducted. From the 28 tires surveyed, the mean lifespan was 46,500 miles with a standard deviation of 9,800 miles. Using \(\alpha = 0.05\), is the data highly inconsistent with the claim?

- \(H_{0}: \mu \geq 50,000\)
- \(H_{a}: \mu < 50,000\)
- Let \(\bar{X} =\) the average lifespan of a brand of tires.
- normal distribution
- \(z = -2.315\)
- \(p\text{-value} = 0.0103\)
- Check student’s solution.
- alpha: 0.05
- Decision: Reject the null hypothesis.
- Reason for decision: The \(p\text{-value}\) is less than 0.05.
- Conclusion: There is sufficient evidence to conclude that the mean lifespan of the tires is less than 50,000 miles.
- \((43,537, 49,463)\)

From generation to generation, the mean age when smokers first start to smoke varies. However, the standard deviation of that age remains constant of around 2.1 years. A survey of 40 smokers of this generation was done to see if the mean starting age is at least 19. The sample mean was 18.1 with a sample standard deviation of 1.3. Do the data support the claim at the 5% level?

The cost of a daily newspaper varies from city to city. However, the variation among prices remains steady with a standard deviation of 20¢. A study was done to test the claim that the mean cost of a daily newspaper is $1.00. Twelve costs yield a mean cost of 95¢ with a standard deviation of 18¢. Do the data support the claim at the 1% level?

- \(H_{0}: \mu = $1.00\)
- \(H_{a}: \mu \neq $1.00\)
- Let \(\bar{X} =\) the average cost of a daily newspaper.
- \(z = –0.866\)
- \(p\text{-value} = 0.3865\)
- \(\alpha: 0.01\)
- Decision: Do not reject the null hypothesis.
- Reason for decision: The \(p\text{-value}\) is greater than 0.01.
- Conclusion: There is sufficient evidence to support the claim that the mean cost of daily papers is $1. The mean cost could be $1.
- \(($0.84, $1.06)\)

An article in the San Jose Mercury News stated that students in the California state university system take 4.5 years, on average, to finish their undergraduate degrees. Suppose you believe that the mean time is longer. You conduct a survey of 49 students and obtain a sample mean of 5.1 with a sample standard deviation of 1.2. Do the data support your claim at the 1% level?

The mean number of sick days an employee takes per year is believed to be about ten. Members of a personnel department do not believe this figure. They randomly survey eight employees. The number of sick days they took for the past year are as follows: 12; 4; 15; 3; 11; 8; 6; 8. Let \(x =\) the number of sick days they took for the past year. Should the personnel team believe that the mean number is ten?

- \(H_{0}: \mu = 10\)
- \(H_{a}: \mu \neq 10\)
- Let \(\bar{X}\) the mean number of sick days an employee takes per year.
- Student’s t -distribution
- \(t = –1.12\)
- \(p\text{-value} = 0.300\)
- \(\alpha: 0.05\)
- Reason for decision: The \(p\text{-value}\) is greater than 0.05.
- Conclusion: At the 5% significance level, there is insufficient evidence to conclude that the mean number of sick days is not ten.
- \((4.9443, 11.806)\)

In 1955, Life Magazine reported that the 25 year-old mother of three worked, on average, an 80 hour week. Recently, many groups have been studying whether or not the women's movement has, in fact, resulted in an increase in the average work week for women (combining employment and at-home work). Suppose a study was done to determine if the mean work week has increased. 81 women were surveyed with the following results. The sample mean was 83; the sample standard deviation was ten. Does it appear that the mean work week has increased for women at the 5% level?

Your statistics instructor claims that 60 percent of the students who take her Elementary Statistics class go through life feeling more enriched. For some reason that she can't quite figure out, most people don't believe her. You decide to check this out on your own. You randomly survey 64 of her past Elementary Statistics students and find that 34 feel more enriched as a result of her class. Now, what do you think?

- \(H_{0}: p \geq 0.6\)
- \(H_{a}: p < 0.6\)
- Let \(P′ =\) the proportion of students who feel more enriched as a result of taking Elementary Statistics.
- normal for a single proportion
- \(p\text{-value} = 0.1308\)
- Conclusion: There is insufficient evidence to conclude that less than 60 percent of her students feel more enriched.

The “plus-4s” confidence interval is \((0.411, 0.648)\)

A Nissan Motor Corporation advertisement read, “The average man’s I.Q. is 107. The average brown trout’s I.Q. is 4. So why can’t man catch brown trout?” Suppose you believe that the brown trout’s mean I.Q. is greater than four. You catch 12 brown trout. A fish psychologist determines the I.Q.s as follows: 5; 4; 7; 3; 6; 4; 5; 3; 6; 3; 8; 5. Conduct a hypothesis test of your belief.

Refer to Exercise 9.119 . Conduct a hypothesis test to see if your decision and conclusion would change if your belief were that the brown trout’s mean I.Q. is not four.

- \(H_{0}: \mu = 4\)
- \(H_{a}: \mu \neq 4\)
- Let \(\bar{X}\) the average I.Q. of a set of brown trout.
- two-tailed Student's t-test
- \(t = 1.95\)
- \(p\text{-value} = 0.076\)
- Reason for decision: The \(p\text{-value}\) is greater than 0.05
- Conclusion: There is insufficient evidence to conclude that the average IQ of brown trout is not four.
- \((3.8865,5.9468)\)

According to an article in Newsweek , the natural ratio of girls to boys is 100:105. In China, the birth ratio is 100: 114 (46.7% girls). Suppose you don’t believe the reported figures of the percent of girls born in China. You conduct a study. In this study, you count the number of girls and boys born in 150 randomly chosen recent births. There are 60 girls and 90 boys born of the 150. Based on your study, do you believe that the percent of girls born in China is 46.7?

A poll done for Newsweek found that 13% of Americans have seen or sensed the presence of an angel. A contingent doubts that the percent is really that high. It conducts its own survey. Out of 76 Americans surveyed, only two had seen or sensed the presence of an angel. As a result of the contingent’s survey, would you agree with the Newsweek poll? In complete sentences, also give three reasons why the two polls might give different results.

- \(H_{a}: p < 0.13\)
- Let \(P′ =\) the proportion of Americans who have seen or sensed angels
- –2.688
- \(p\text{-value} = 0.0036\)
- Reason for decision: The \(p\text{-value}\)e is less than 0.05.
- Conclusion: There is sufficient evidence to conclude that the percentage of Americans who have seen or sensed an angel is less than 13%.

The“plus-4s” confidence interval is (0.0022, 0.0978)

The mean work week for engineers in a start-up company is believed to be about 60 hours. A newly hired engineer hopes that it’s shorter. She asks ten engineering friends in start-ups for the lengths of their mean work weeks. Based on the results that follow, should she count on the mean work week to be shorter than 60 hours?

Data (length of mean work week): 70; 45; 55; 60; 65; 55; 55; 60; 50; 55.

Use the “Lap time” data for Lap 4 (see [link] ) to test the claim that Terri finishes Lap 4, on average, in less than 129 seconds. Use all twenty races given.

- \(H_{0}: \mu \geq 129\)
- \(H_{a}: \mu < 129\)
- Let \(\bar{X} =\) the average time in seconds that Terri finishes Lap 4.
- Student's t -distribution
- \(t = 1.209\)
- Conclusion: There is insufficient evidence to conclude that Terri’s mean lap time is less than 129 seconds.
- \((128.63, 130.37)\)

Use the “Initial Public Offering” data (see [link] ) to test the claim that the mean offer price was $18 per share. Do not use all the data. Use your random number generator to randomly survey 15 prices.

The following questions were written by past students. They are excellent problems!

"Asian Family Reunion," by Chau Nguyen

Every two years it comes around.

We all get together from different towns.

In my honest opinion,

It's not a typical family reunion.

Not forty, or fifty, or sixty,

But how about seventy companions!

The kids would play, scream, and shout

One minute they're happy, another they'll pout.

The teenagers would look, stare, and compare

From how they look to what they wear.

The men would chat about their business

That they make more, but never less.

Money is always their subject

And there's always talk of more new projects.

The women get tired from all of the chats

They head to the kitchen to set out the mats.

Some would sit and some would stand

Eating and talking with plates in their hands.

Then come the games and the songs

And suddenly, everyone gets along!

With all that laughter, it's sad to say

That it always ends in the same old way.

They hug and kiss and say "good-bye"

And then they all begin to cry!

I say that 60 percent shed their tears

But my mom counted 35 people this year.

She said that boys and men will always have their pride,

So we won't ever see them cry.

I myself don't think she's correct,

So could you please try this problem to see if you object?

- \(H_{0}: p = 0.60\)
- \(H_{a}: p < 0.60\)
- Let \(P′ =\) the proportion of family members who shed tears at a reunion.
- –1.71
- Reason for decision: \(p\text{-value} < \alpha\)
- Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the proportion of family members who shed tears at a reunion is less than 0.60. However, the test is weak because the \(p\text{-value}\) and alpha are quite close, so other tests should be done.
- We are 95% confident that between 38.29% and 61.71% of family members will shed tears at a family reunion. \((0.3829, 0.6171)\). The“plus-4s” confidence interval (see chapter 8) is \((0.3861, 0.6139)\)

Note that here the “large-sample” \(1 - \text{PropZTest}\) provides the approximate \(p\text{-value}\) of 0.0438. Whenever a \(p\text{-value}\) based on a normal approximation is close to the level of significance, the exact \(p\text{-value}\) based on binomial probabilities should be calculated whenever possible. This is beyond the scope of this course.

"The Problem with Angels," by Cyndy Dowling

Although this problem is wholly mine,

The catalyst came from the magazine, Time.

On the magazine cover I did find

The realm of angels tickling my mind.

Inside, 69% I found to be

In angels, Americans do believe.

Then, it was time to rise to the task,

Ninety-five high school and college students I did ask.

Viewing all as one group,

Random sampling to get the scoop.

So, I asked each to be true,

"Do you believe in angels?" Tell me, do!

Hypothesizing at the start,

Totally believing in my heart

That the proportion who said yes

Would be equal on this test.

Lo and behold, seventy-three did arrive,

Out of the sample of ninety-five.

Now your job has just begun,

Solve this problem and have some fun.

"Blowing Bubbles," by Sondra Prull

Studying stats just made me tense,

I had to find some sane defense.

Some light and lifting simple play

To float my math anxiety away.

Blowing bubbles lifts me high

Takes my troubles to the sky.

POIK! They're gone, with all my stress

Bubble therapy is the best.

The label said each time I blew

The average number of bubbles would be at least 22.

I blew and blew and this I found

From 64 blows, they all are round!

But the number of bubbles in 64 blows

Varied widely, this I know.

20 per blow became the mean

They deviated by 6, and not 16.

From counting bubbles, I sure did relax

But now I give to you your task.

Was 22 a reasonable guess?

Find the answer and pass this test!

- \(H_{0}: \mu \geq 22\)
- \(H_{a}: \mu < 22\)
- Let \(\bar{X} =\) the mean number of bubbles per blow.
- –2.667
- \(p\text{-value} = 0.00486\)
- Conclusion: There is sufficient evidence to conclude that the mean number of bubbles per blow is less than 22.
- \((18.501, 21.499)\)

"Dalmatian Darnation," by Kathy Sparling

A greedy dog breeder named Spreckles

Bred puppies with numerous freckles

The Dalmatians he sought

Possessed spot upon spot

The more spots, he thought, the more shekels.

His competitors did not agree

That freckles would increase the fee.

They said, “Spots are quite nice

But they don't affect price;

One should breed for improved pedigree.”

The breeders decided to prove

This strategy was a wrong move.

Breeding only for spots

Would wreak havoc, they thought.

His theory they want to disprove.

They proposed a contest to Spreckles

Comparing dog prices to freckles.

In records they looked up

One hundred one pups:

Dalmatians that fetched the most shekels.

They asked Mr. Spreckles to name

An average spot count he'd claim

To bring in big bucks.

Said Spreckles, “Well, shucks,

It's for one hundred one that I aim.”

Said an amateur statistician

Who wanted to help with this mission.

“Twenty-one for the sample

Standard deviation's ample:

They examined one hundred and one

Dalmatians that fetched a good sum.

They counted each spot,

Mark, freckle and dot

And tallied up every one.

Instead of one hundred one spots

They averaged ninety six dots

Can they muzzle Spreckles’

Obsession with freckles

Based on all the dog data they've got?

"Macaroni and Cheese, please!!" by Nedda Misherghi and Rachelle Hall

As a poor starving student I don't have much money to spend for even the bare necessities. So my favorite and main staple food is macaroni and cheese. It's high in taste and low in cost and nutritional value.

One day, as I sat down to determine the meaning of life, I got a serious craving for this, oh, so important, food of my life. So I went down the street to Greatway to get a box of macaroni and cheese, but it was SO expensive! $2.02 !!! Can you believe it? It made me stop and think. The world is changing fast. I had thought that the mean cost of a box (the normal size, not some super-gigantic-family-value-pack) was at most $1, but now I wasn't so sure. However, I was determined to find out. I went to 53 of the closest grocery stores and surveyed the prices of macaroni and cheese. Here are the data I wrote in my notebook:

Price per box of Mac and Cheese:

- 5 stores @ $2.02
- 15 stores @ $0.25
- 3 stores @ $1.29
- 6 stores @ $0.35
- 4 stores @ $2.27
- 7 stores @ $1.50
- 5 stores @ $1.89
- 8 stores @ 0.75.

I could see that the cost varied but I had to sit down to figure out whether or not I was right. If it does turn out that this mouth-watering dish is at most $1, then I'll throw a big cheesy party in our next statistics lab, with enough macaroni and cheese for just me. (After all, as a poor starving student I can't be expected to feed our class of animals!)

- \(H_{0}: \mu \leq 1\)
- \(H_{a}: \mu > 1\)
- Let \(\bar{X} =\) the mean cost in dollars of macaroni and cheese in a certain town.
- Student's \(t\)-distribution
- \(t = 0.340\)
- \(p\text{-value} = 0.36756\)
- Conclusion: The mean cost could be $1, or less. At the 5% significance level, there is insufficient evidence to conclude that the mean price of a box of macaroni and cheese is more than $1.
- \((0.8291, 1.241)\)

"William Shakespeare: The Tragedy of Hamlet, Prince of Denmark," by Jacqueline Ghodsi

THE CHARACTERS (in order of appearance):

- HAMLET, Prince of Denmark and student of Statistics
- POLONIUS, Hamlet’s tutor
- HOROTIO, friend to Hamlet and fellow student

Scene: The great library of the castle, in which Hamlet does his lessons

(The day is fair, but the face of Hamlet is clouded. He paces the large room. His tutor, Polonius, is reprimanding Hamlet regarding the latter’s recent experience. Horatio is seated at the large table at right stage.)

POLONIUS: My Lord, how cans’t thou admit that thou hast seen a ghost! It is but a figment of your imagination!

HAMLET: I beg to differ; I know of a certainty that five-and-seventy in one hundred of us, condemned to the whips and scorns of time as we are, have gazed upon a spirit of health, or goblin damn’d, be their intents wicked or charitable.

POLONIUS If thou doest insist upon thy wretched vision then let me invest your time; be true to thy work and speak to me through the reason of the null and alternate hypotheses. (He turns to Horatio.) Did not Hamlet himself say, “What piece of work is man, how noble in reason, how infinite in faculties? Then let not this foolishness persist. Go, Horatio, make a survey of three-and-sixty and discover what the true proportion be. For my part, I will never succumb to this fantasy, but deem man to be devoid of all reason should thy proposal of at least five-and-seventy in one hundred hold true.

HORATIO (to Hamlet): What should we do, my Lord?

HAMLET: Go to thy purpose, Horatio.

HORATIO: To what end, my Lord?

HAMLET: That you must teach me. But let me conjure you by the rights of our fellowship, by the consonance of our youth, but the obligation of our ever-preserved love, be even and direct with me, whether I am right or no.

(Horatio exits, followed by Polonius, leaving Hamlet to ponder alone.)

(The next day, Hamlet awaits anxiously the presence of his friend, Horatio. Polonius enters and places some books upon the table just a moment before Horatio enters.)

POLONIUS: So, Horatio, what is it thou didst reveal through thy deliberations?

HORATIO: In a random survey, for which purpose thou thyself sent me forth, I did discover that one-and-forty believe fervently that the spirits of the dead walk with us. Before my God, I might not this believe, without the sensible and true avouch of mine own eyes.

POLONIUS: Give thine own thoughts no tongue, Horatio. (Polonius turns to Hamlet.) But look to’t I charge you, my Lord. Come Horatio, let us go together, for this is not our test. (Horatio and Polonius leave together.)

HAMLET: To reject, or not reject, that is the question: whether ‘tis nobler in the mind to suffer the slings and arrows of outrageous statistics, or to take arms against a sea of data, and, by opposing, end them. (Hamlet resignedly attends to his task.)

(Curtain falls)

"Untitled," by Stephen Chen

I've often wondered how software is released and sold to the public. Ironically, I work for a company that sells products with known problems. Unfortunately, most of the problems are difficult to create, which makes them difficult to fix. I usually use the test program X, which tests the product, to try to create a specific problem. When the test program is run to make an error occur, the likelihood of generating an error is 1%.

So, armed with this knowledge, I wrote a new test program Y that will generate the same error that test program X creates, but more often. To find out if my test program is better than the original, so that I can convince the management that I'm right, I ran my test program to find out how often I can generate the same error. When I ran my test program 50 times, I generated the error twice. While this may not seem much better, I think that I can convince the management to use my test program instead of the original test program. Am I right?

- \(H_{0}: p = 0.01\)
- \(H_{a}: p > 0.01\)
- Let \(P′ =\) the proportion of errors generated
- Normal for a single proportion
- Decision: Reject the null hypothesis
- Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the proportion of errors generated is more than 0.01.

The“plus-4s” confidence interval is \((0.004, 0.144)\).

"Japanese Girls’ Names"

by Kumi Furuichi

It used to be very typical for Japanese girls’ names to end with “ko.” (The trend might have started around my grandmothers’ generation and its peak might have been around my mother’s generation.) “Ko” means “child” in Chinese characters. Parents would name their daughters with “ko” attaching to other Chinese characters which have meanings that they want their daughters to become, such as Sachiko—happy child, Yoshiko—a good child, Yasuko—a healthy child, and so on.

However, I noticed recently that only two out of nine of my Japanese girlfriends at this school have names which end with “ko.” More and more, parents seem to have become creative, modernized, and, sometimes, westernized in naming their children.

I have a feeling that, while 70 percent or more of my mother’s generation would have names with “ko” at the end, the proportion has dropped among my peers. I wrote down all my Japanese friends’, ex-classmates’, co-workers, and acquaintances’ names that I could remember. Following are the names. (Some are repeats.) Test to see if the proportion has dropped for this generation.

Ai, Akemi, Akiko, Ayumi, Chiaki, Chie, Eiko, Eri, Eriko, Fumiko, Harumi, Hitomi, Hiroko, Hiroko, Hidemi, Hisako, Hinako, Izumi, Izumi, Junko, Junko, Kana, Kanako, Kanayo, Kayo, Kayoko, Kazumi, Keiko, Keiko, Kei, Kumi, Kumiko, Kyoko, Kyoko, Madoka, Maho, Mai, Maiko, Maki, Miki, Miki, Mikiko, Mina, Minako, Miyako, Momoko, Nana, Naoko, Naoko, Naoko, Noriko, Rieko, Rika, Rika, Rumiko, Rei, Reiko, Reiko, Sachiko, Sachiko, Sachiyo, Saki, Sayaka, Sayoko, Sayuri, Seiko, Shiho, Shizuka, Sumiko, Takako, Takako, Tomoe, Tomoe, Tomoko, Touko, Yasuko, Yasuko, Yasuyo, Yoko, Yoko, Yoko, Yoshiko, Yoshiko, Yoshiko, Yuka, Yuki, Yuki, Yukiko, Yuko, Yuko.

"Phillip’s Wish," by Suzanne Osorio

My nephew likes to play

Chasing the girls makes his day.

He asked his mother

If it is okay

To get his ear pierced.

She said, “No way!”

To poke a hole through your ear,

Is not what I want for you, dear.

He argued his point quite well,

Says even my macho pal, Mel,

Has gotten this done.

It’s all just for fun.

C’mon please, mom, please, what the hell.

Again Phillip complained to his mother,

Saying half his friends (including their brothers)

Are piercing their ears

And they have no fears

He wants to be like the others.

She said, “I think it’s much less.

We must do a hypothesis test.

And if you are right,

I won’t put up a fight.

But, if not, then my case will rest.”

We proceeded to call fifty guys

To see whose prediction would fly.

Nineteen of the fifty

Said piercing was nifty

And earrings they’d occasionally buy.

Then there’s the other thirty-one,

Who said they’d never have this done.

So now this poem’s finished.

Will his hopes be diminished,

Or will my nephew have his fun?

- \(H_{0}: p = 0.50\)
- \(H_{a}: p < 0.50\)
- Let \(P′ =\) the proportion of friends that has a pierced ear.
- –1.70
- \(p\text{-value} = 0.0448\)
- Reason for decision: The \(p\text{-value}\) is less than 0.05. (However, they are very close.)
- Conclusion: There is sufficient evidence to support the claim that less than 50% of his friends have pierced ears.
- Confidence Interval: \((0.245, 0.515)\): The “plus-4s” confidence interval is \((0.259, 0.519)\).

"The Craven," by Mark Salangsang

Once upon a morning dreary

In stats class I was weak and weary.

Pondering over last night’s homework

Whose answers were now on the board

This I did and nothing more.

While I nodded nearly napping

Suddenly, there came a tapping.

As someone gently rapping,

Rapping my head as I snore.

Quoth the teacher, “Sleep no more.”

“In every class you fall asleep,”

The teacher said, his voice was deep.

“So a tally I’ve begun to keep

Of every class you nap and snore.

The percentage being forty-four.”

“My dear teacher I must confess,

While sleeping is what I do best.

The percentage, I think, must be less,

A percentage less than forty-four.”

This I said and nothing more.

“We’ll see,” he said and walked away,

And fifty classes from that day

He counted till the month of May

The classes in which I napped and snored.

The number he found was twenty-four.

At a significance level of 0.05,

Please tell me am I still alive?

Or did my grade just take a dive

Plunging down beneath the floor?

Upon thee I hereby implore.

Toastmasters International cites a report by Gallop Poll that 40% of Americans fear public speaking. A student believes that less than 40% of students at her school fear public speaking. She randomly surveys 361 schoolmates and finds that 135 report they fear public speaking. Conduct a hypothesis test to determine if the percent at her school is less than 40%.

- \(H_{0}: p = 0.40\)
- \(H_{a}: p < 0.40\)
- Let \(P′ =\) the proportion of schoolmates who fear public speaking.
- –1.01
- \(p\text{-value} = 0.1563\)
- Conclusion: There is insufficient evidence to support the claim that less than 40% of students at the school fear public speaking.
- Confidence Interval: \((0.3241, 0.4240)\): The “plus-4s” confidence interval is \((0.3257, 0.4250)\).

Sixty-eight percent of online courses taught at community colleges nationwide were taught by full-time faculty. To test if 68% also represents California’s percent for full-time faculty teaching the online classes, Long Beach City College (LBCC) in California, was randomly selected for comparison. In the same year, 34 of the 44 online courses LBCC offered were taught by full-time faculty. Conduct a hypothesis test to determine if 68% represents California. NOTE: For more accurate results, use more California community colleges and this past year's data.

According to an article in Bloomberg Businessweek , New York City's most recent adult smoking rate is 14%. Suppose that a survey is conducted to determine this year’s rate. Nine out of 70 randomly chosen N.Y. City residents reply that they smoke. Conduct a hypothesis test to determine if the rate is still 14% or if it has decreased.

- \(H_{0}: p = 0.14\)
- \(H_{a}: p < 0.14\)
- Let \(P′ =\) the proportion of NYC residents that smoke.
- –0.2756
- \(p\text{-value} = 0.3914\)
- At the 5% significance level, there is insufficient evidence to conclude that the proportion of NYC residents who smoke is less than 0.14.
- Confidence Interval: \((0.0502, 0.2070)\): The “plus-4s” confidence interval (see chapter 8) is \((0.0676, 0.2297)\).

The mean age of De Anza College students in a previous term was 26.6 years old. An instructor thinks the mean age for online students is older than 26.6. She randomly surveys 56 online students and finds that the sample mean is 29.4 with a standard deviation of 2.1. Conduct a hypothesis test.

Registered nurses earned an average annual salary of $69,110. For that same year, a survey was conducted of 41 California registered nurses to determine if the annual salary is higher than $69,110 for California nurses. The sample average was $71,121 with a sample standard deviation of $7,489. Conduct a hypothesis test.

- \(H_{0}: \mu = 69,110\)
- \(H_{0}: \mu > 69,110\)
- Let \(\bar{X} =\) the mean salary in dollars for California registered nurses.
- \(t = 1.719\)
- \(p\text{-value}: 0.0466\)
- Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the mean salary of California registered nurses exceeds $69,110.
- \(($68,757, $73,485)\)

La Leche League International reports that the mean age of weaning a child from breastfeeding is age four to five worldwide. In America, most nursing mothers wean their children much earlier. Suppose a random survey is conducted of 21 U.S. mothers who recently weaned their children. The mean weaning age was nine months (3/4 year) with a standard deviation of 4 months. Conduct a hypothesis test to determine if the mean weaning age in the U.S. is less than four years old.

Over the past few decades, public health officials have examined the link between weight concerns and teen girls' smoking. Researchers surveyed a group of 273 randomly selected teen girls living in Massachusetts (between 12 and 15 years old). After four years the girls were surveyed again. Sixty-three said they smoked to stay thin. Is there good evidence that more than thirty percent of the teen girls smoke to stay thin?

After conducting the test, your decision and conclusion are

- Reject \(H_{0}\): There is sufficient evidence to conclude that more than 30% of teen girls smoke to stay thin.
- Do not reject \(H_{0}\): There is not sufficient evidence to conclude that less than 30% of teen girls smoke to stay thin.
- Do not reject \(H_{0}\): There is not sufficient evidence to conclude that more than 30% of teen girls smoke to stay thin.
- Reject \(H_{0}\): There is sufficient evidence to conclude that less than 30% of teen girls smoke to stay thin.

A statistics instructor believes that fewer than 20% of Evergreen Valley College (EVC) students attended the opening night midnight showing of the latest Harry Potter movie. She surveys 84 of her students and finds that 11 of them attended the midnight showing.

At a 1% level of significance, an appropriate conclusion is:

- There is insufficient evidence to conclude that the percent of EVC students who attended the midnight showing of Harry Potter is less than 20%.
- There is sufficient evidence to conclude that the percent of EVC students who attended the midnight showing of Harry Potter is more than 20%.
- There is sufficient evidence to conclude that the percent of EVC students who attended the midnight showing of Harry Potter is less than 20%.
- There is insufficient evidence to conclude that the percent of EVC students who attended the midnight showing of Harry Potter is at least 20%.

Previously, an organization reported that teenagers spent 4.5 hours per week, on average, on the phone. The organization thinks that, currently, the mean is higher. Fifteen randomly chosen teenagers were asked how many hours per week they spend on the phone. The sample mean was 4.75 hours with a sample standard deviation of 2.0. Conduct a hypothesis test.

At a significance level of \(a = 0.05\), what is the correct conclusion?

- There is enough evidence to conclude that the mean number of hours is more than 4.75
- There is enough evidence to conclude that the mean number of hours is more than 4.5
- There is not enough evidence to conclude that the mean number of hours is more than 4.5
- There is not enough evidence to conclude that the mean number of hours is more than 4.75

Instructions: For the following ten exercises,

Hypothesis testing: For the following ten exercises, answer each question.

State the null and alternate hypothesis.

State the \(p\text{-value}\).

State \(\alpha\).

What is your decision?

Write a conclusion.

Answer any other questions asked in the problem.

According to the Center for Disease Control website, in 2011 at least 18% of high school students have smoked a cigarette. An Introduction to Statistics class in Davies County, KY conducted a hypothesis test at the local high school (a medium sized–approximately 1,200 students–small city demographic) to determine if the local high school’s percentage was lower. One hundred fifty students were chosen at random and surveyed. Of the 150 students surveyed, 82 have smoked. Use a significance level of 0.05 and using appropriate statistical evidence, conduct a hypothesis test and state the conclusions.

A recent survey in the N.Y. Times Almanac indicated that 48.8% of families own stock. A broker wanted to determine if this survey could be valid. He surveyed a random sample of 250 families and found that 142 owned some type of stock. At the 0.05 significance level, can the survey be considered to be accurate?

- \(H_{0}: p = 0.488\) \(H_{a}: p \neq 0.488\)
- \(p\text{-value} = 0.0114\)
- \(\alpha = 0.05\)
- Reject the null hypothesis.
- At the 5% level of significance, there is enough evidence to conclude that 48.8% of families own stocks.
- The survey does not appear to be accurate.

Driver error can be listed as the cause of approximately 54% of all fatal auto accidents, according to the American Automobile Association. Thirty randomly selected fatal accidents are examined, and it is determined that 14 were caused by driver error. Using \(\alpha = 0.05\), is the AAA proportion accurate?

The US Department of Energy reported that 51.7% of homes were heated by natural gas. A random sample of 221 homes in Kentucky found that 115 were heated by natural gas. Does the evidence support the claim for Kentucky at the \(\alpha = 0.05\) level in Kentucky? Are the results applicable across the country? Why?

- \(H_{0}: p = 0.517\) \(H_{0}: p \neq 0.517\)
- \(p\text{-value} = 0.9203\).
- \(\alpha = 0.05\).
- Do not reject the null hypothesis.
- At the 5% significance level, there is not enough evidence to conclude that the proportion of homes in Kentucky that are heated by natural gas is 0.517.
- However, we cannot generalize this result to the entire nation. First, the sample’s population is only the state of Kentucky. Second, it is reasonable to assume that homes in the extreme north and south will have extreme high usage and low usage, respectively. We would need to expand our sample base to include these possibilities if we wanted to generalize this claim to the entire nation.

For Americans using library services, the American Library Association claims that at most 67% of patrons borrow books. The library director in Owensboro, Kentucky feels this is not true, so she asked a local college statistic class to conduct a survey. The class randomly selected 100 patrons and found that 82 borrowed books. Did the class demonstrate that the percentage was higher in Owensboro, KY? Use \(\alpha = 0.01\) level of significance. What is the possible proportion of patrons that do borrow books from the Owensboro Library?

The Weather Underground reported that the mean amount of summer rainfall for the northeastern US is at least 11.52 inches. Ten cities in the northeast are randomly selected and the mean rainfall amount is calculated to be 7.42 inches with a standard deviation of 1.3 inches. At the \(\alpha = 0.05 level\), can it be concluded that the mean rainfall was below the reported average? What if \(\alpha = 0.01\)? Assume the amount of summer rainfall follows a normal distribution.

- \(H_{0}: \mu \geq 11.52\) \(H_{a}: \mu < 11.52\)
- \(p\text{-value} = 0.000002\) which is almost 0.
- At the 5% significance level, there is enough evidence to conclude that the mean amount of summer rain in the northeaster US is less than 11.52 inches, on average.
- We would make the same conclusion if alpha was 1% because the \(p\text{-value}\) is almost 0.

A survey in the N.Y. Times Almanac finds the mean commute time (one way) is 25.4 minutes for the 15 largest US cities. The Austin, TX chamber of commerce feels that Austin’s commute time is less and wants to publicize this fact. The mean for 25 randomly selected commuters is 22.1 minutes with a standard deviation of 5.3 minutes. At the \(\alpha = 0.10\) level, is the Austin, TX commute significantly less than the mean commute time for the 15 largest US cities?

A report by the Gallup Poll found that a woman visits her doctor, on average, at most 5.8 times each year. A random sample of 20 women results in these yearly visit totals

3; 2; 1; 3; 7; 2; 9; 4; 6; 6; 8; 0; 5; 6; 4; 2; 1; 3; 4; 1

At the \(\alpha = 0.05\) level can it be concluded that the sample mean is higher than 5.8 visits per year?

- \(H_{0}: \mu \leq 5.8\) \(H_{a}: \mu > 5.8\)
- \(p\text{-value} = 0.9987\)
- At the 5% level of significance, there is not enough evidence to conclude that a woman visits her doctor, on average, more than 5.8 times a year.

According to the N.Y. Times Almanac the mean family size in the U.S. is 3.18. A sample of a college math class resulted in the following family sizes:

5; 4; 5; 4; 4; 3; 6; 4; 3; 3; 5; 5; 6; 3; 3; 2; 7; 4; 5; 2; 2; 2; 3; 2

At \(\alpha = 0.05\) level, is the class’ mean family size greater than the national average? Does the Almanac result remain valid? Why?

The student academic group on a college campus claims that freshman students study at least 2.5 hours per day, on average. One Introduction to Statistics class was skeptical. The class took a random sample of 30 freshman students and found a mean study time of 137 minutes with a standard deviation of 45 minutes. At α = 0.01 level, is the student academic group’s claim correct?

- \(H_{0}: \mu \geq 150\) \(H_{0}: \mu < 150\)
- \(p\text{-value} = 0.0622\)
- \(\alpha = 0.01\)
- At the 1% significance level, there is not enough evidence to conclude that freshmen students study less than 2.5 hours per day, on average.
- The student academic group’s claim appears to be correct.

## 9.7: Hypothesis Testing of a Single Mean and Single Proportion

## Tutorial Playlist

Statistics tutorial, everything you need to know about the probability density function in statistics, the best guide to understand central limit theorem, an in-depth guide to measures of central tendency : mean, median and mode, the ultimate guide to understand conditional probability.

A Comprehensive Look at Percentile in Statistics

## The Best Guide to Understand Bayes Theorem

Everything you need to know about the normal distribution, an in-depth explanation of cumulative distribution function, a complete guide to chi-square test, a complete guide on hypothesis testing in statistics, understanding the fundamentals of arithmetic and geometric progression, the definitive guide to understand spearman’s rank correlation, a comprehensive guide to understand mean squared error, all you need to know about the empirical rule in statistics, the complete guide to skewness and kurtosis, a holistic look at bernoulli distribution.

All You Need to Know About Bias in Statistics

## A Complete Guide to Get a Grasp of Time Series Analysis

The Key Differences Between Z-Test Vs. T-Test

## The Complete Guide to Understand Pearson's Correlation

A complete guide on the types of statistical studies, everything you need to know about poisson distribution, your best guide to understand correlation vs. regression, the most comprehensive guide for beginners on what is correlation, what is hypothesis testing in statistics types and examples.

Lesson 10 of 24 By Avijeet Biswal

## Table of Contents

In today’s data-driven world , decisions are based on data all the time. Hypothesis plays a crucial role in that process, whether it may be making business decisions, in the health sector, academia, or in quality improvement. Without hypothesis & hypothesis tests, you risk drawing the wrong conclusions and making bad decisions. In this tutorial, you will look at Hypothesis Testing in Statistics.

## What Is Hypothesis Testing in Statistics?

Hypothesis Testing is a type of statistical analysis in which you put your assumptions about a population parameter to the test. It is used to estimate the relationship between 2 statistical variables.

Let's discuss few examples of statistical hypothesis from real-life -

- A teacher assumes that 60% of his college's students come from lower-middle-class families.
- A doctor believes that 3D (Diet, Dose, and Discipline) is 90% effective for diabetic patients.

Now that you know about hypothesis testing, look at the two types of hypothesis testing in statistics.

## Hypothesis Testing Formula

Z = ( x̅ – μ0 ) / (σ /√n)

- Here, x̅ is the sample mean,
- μ0 is the population mean,
- σ is the standard deviation,
- n is the sample size.

## How Hypothesis Testing Works?

An analyst performs hypothesis testing on a statistical sample to present evidence of the plausibility of the null hypothesis. Measurements and analyses are conducted on a random sample of the population to test a theory. Analysts use a random population sample to test two hypotheses: the null and alternative hypotheses.

The null hypothesis is typically an equality hypothesis between population parameters; for example, a null hypothesis may claim that the population means return equals zero. The alternate hypothesis is essentially the inverse of the null hypothesis (e.g., the population means the return is not equal to zero). As a result, they are mutually exclusive, and only one can be correct. One of the two possibilities, however, will always be correct.

## Your Dream Career is Just Around The Corner!

## Null Hypothesis and Alternate Hypothesis

The Null Hypothesis is the assumption that the event will not occur. A null hypothesis has no bearing on the study's outcome unless it is rejected.

H0 is the symbol for it, and it is pronounced H-naught.

The Alternate Hypothesis is the logical opposite of the null hypothesis. The acceptance of the alternative hypothesis follows the rejection of the null hypothesis. H1 is the symbol for it.

Let's understand this with an example.

A sanitizer manufacturer claims that its product kills 95 percent of germs on average.

To put this company's claim to the test, create a null and alternate hypothesis.

H0 (Null Hypothesis): Average = 95%.

Alternative Hypothesis (H1): The average is less than 95%.

Another straightforward example to understand this concept is determining whether or not a coin is fair and balanced. The null hypothesis states that the probability of a show of heads is equal to the likelihood of a show of tails. In contrast, the alternate theory states that the probability of a show of heads and tails would be very different.

## Become a Data Scientist with Hands-on Training!

## Hypothesis Testing Calculation With Examples

Let's consider a hypothesis test for the average height of women in the United States. Suppose our null hypothesis is that the average height is 5'4". We gather a sample of 100 women and determine that their average height is 5'5". The standard deviation of population is 2.

To calculate the z-score, we would use the following formula:

z = ( x̅ – μ0 ) / (σ /√n)

z = (5'5" - 5'4") / (2" / √100)

z = 0.5 / (0.045)

We will reject the null hypothesis as the z-score of 11.11 is very large and conclude that there is evidence to suggest that the average height of women in the US is greater than 5'4".

## Steps of Hypothesis Testing

Step 1: specify your null and alternate hypotheses.

It is critical to rephrase your original research hypothesis (the prediction that you wish to study) as a null (Ho) and alternative (Ha) hypothesis so that you can test it quantitatively. Your first hypothesis, which predicts a link between variables, is generally your alternate hypothesis. The null hypothesis predicts no link between the variables of interest.

## Step 2: Gather Data

For a statistical test to be legitimate, sampling and data collection must be done in a way that is meant to test your hypothesis. You cannot draw statistical conclusions about the population you are interested in if your data is not representative.

## Step 3: Conduct a Statistical Test

Other statistical tests are available, but they all compare within-group variance (how to spread out the data inside a category) against between-group variance (how different the categories are from one another). If the between-group variation is big enough that there is little or no overlap between groups, your statistical test will display a low p-value to represent this. This suggests that the disparities between these groups are unlikely to have occurred by accident. Alternatively, if there is a large within-group variance and a low between-group variance, your statistical test will show a high p-value. Any difference you find across groups is most likely attributable to chance. The variety of variables and the level of measurement of your obtained data will influence your statistical test selection.

## Step 4: Determine Rejection Of Your Null Hypothesis

Your statistical test results must determine whether your null hypothesis should be rejected or not. In most circumstances, you will base your judgment on the p-value provided by the statistical test. In most circumstances, your preset level of significance for rejecting the null hypothesis will be 0.05 - that is, when there is less than a 5% likelihood that these data would be seen if the null hypothesis were true. In other circumstances, researchers use a lower level of significance, such as 0.01 (1%). This reduces the possibility of wrongly rejecting the null hypothesis.

## Step 5: Present Your Results

The findings of hypothesis testing will be discussed in the results and discussion portions of your research paper, dissertation, or thesis. You should include a concise overview of the data and a summary of the findings of your statistical test in the results section. You can talk about whether your results confirmed your initial hypothesis or not in the conversation. Rejecting or failing to reject the null hypothesis is a formal term used in hypothesis testing. This is likely a must for your statistics assignments.

## Types of Hypothesis Testing

To determine whether a discovery or relationship is statistically significant, hypothesis testing uses a z-test. It usually checks to see if two means are the same (the null hypothesis). Only when the population standard deviation is known and the sample size is 30 data points or more, can a z-test be applied.

A statistical test called a t-test is employed to compare the means of two groups. To determine whether two groups differ or if a procedure or treatment affects the population of interest, it is frequently used in hypothesis testing.

## Chi-Square

You utilize a Chi-square test for hypothesis testing concerning whether your data is as predicted. To determine if the expected and observed results are well-fitted, the Chi-square test analyzes the differences between categorical variables from a random sample. The test's fundamental premise is that the observed values in your data should be compared to the predicted values that would be present if the null hypothesis were true.

## Hypothesis Testing and Confidence Intervals

Both confidence intervals and hypothesis tests are inferential techniques that depend on approximating the sample distribution. Data from a sample is used to estimate a population parameter using confidence intervals. Data from a sample is used in hypothesis testing to examine a given hypothesis. We must have a postulated parameter to conduct hypothesis testing.

Bootstrap distributions and randomization distributions are created using comparable simulation techniques. The observed sample statistic is the focal point of a bootstrap distribution, whereas the null hypothesis value is the focal point of a randomization distribution.

A variety of feasible population parameter estimates are included in confidence ranges. In this lesson, we created just two-tailed confidence intervals. There is a direct connection between these two-tail confidence intervals and these two-tail hypothesis tests. The results of a two-tailed hypothesis test and two-tailed confidence intervals typically provide the same results. In other words, a hypothesis test at the 0.05 level will virtually always fail to reject the null hypothesis if the 95% confidence interval contains the predicted value. A hypothesis test at the 0.05 level will nearly certainly reject the null hypothesis if the 95% confidence interval does not include the hypothesized parameter.

## Simple and Composite Hypothesis Testing

Depending on the population distribution, you can classify the statistical hypothesis into two types.

Simple Hypothesis: A simple hypothesis specifies an exact value for the parameter.

Composite Hypothesis: A composite hypothesis specifies a range of values.

A company is claiming that their average sales for this quarter are 1000 units. This is an example of a simple hypothesis.

Suppose the company claims that the sales are in the range of 900 to 1000 units. Then this is a case of a composite hypothesis.

## One-Tailed and Two-Tailed Hypothesis Testing

The One-Tailed test, also called a directional test, considers a critical region of data that would result in the null hypothesis being rejected if the test sample falls into it, inevitably meaning the acceptance of the alternate hypothesis.

In a one-tailed test, the critical distribution area is one-sided, meaning the test sample is either greater or lesser than a specific value.

In two tails, the test sample is checked to be greater or less than a range of values in a Two-Tailed test, implying that the critical distribution area is two-sided.

If the sample falls within this range, the alternate hypothesis will be accepted, and the null hypothesis will be rejected.

## Become a Data Scientist With Real-World Experience

## Right Tailed Hypothesis Testing

If the larger than (>) sign appears in your hypothesis statement, you are using a right-tailed test, also known as an upper test. Or, to put it another way, the disparity is to the right. For instance, you can contrast the battery life before and after a change in production. Your hypothesis statements can be the following if you want to know if the battery life is longer than the original (let's say 90 hours):

- The null hypothesis is (H0 <= 90) or less change.
- A possibility is that battery life has risen (H1) > 90.

The crucial point in this situation is that the alternate hypothesis (H1), not the null hypothesis, decides whether you get a right-tailed test.

## Left Tailed Hypothesis Testing

Alternative hypotheses that assert the true value of a parameter is lower than the null hypothesis are tested with a left-tailed test; they are indicated by the asterisk "<".

Suppose H0: mean = 50 and H1: mean not equal to 50

According to the H1, the mean can be greater than or less than 50. This is an example of a Two-tailed test.

In a similar manner, if H0: mean >=50, then H1: mean <50

Here the mean is less than 50. It is called a One-tailed test.

## Type 1 and Type 2 Error

A hypothesis test can result in two types of errors.

Type 1 Error: A Type-I error occurs when sample results reject the null hypothesis despite being true.

Type 2 Error: A Type-II error occurs when the null hypothesis is not rejected when it is false, unlike a Type-I error.

Suppose a teacher evaluates the examination paper to decide whether a student passes or fails.

H0: Student has passed

H1: Student has failed

Type I error will be the teacher failing the student [rejects H0] although the student scored the passing marks [H0 was true].

Type II error will be the case where the teacher passes the student [do not reject H0] although the student did not score the passing marks [H1 is true].

## Level of Significance

The alpha value is a criterion for determining whether a test statistic is statistically significant. In a statistical test, Alpha represents an acceptable probability of a Type I error. Because alpha is a probability, it can be anywhere between 0 and 1. In practice, the most commonly used alpha values are 0.01, 0.05, and 0.1, which represent a 1%, 5%, and 10% chance of a Type I error, respectively (i.e. rejecting the null hypothesis when it is in fact correct).

## Future-Proof Your AI/ML Career: Top Dos and Don'ts

A p-value is a metric that expresses the likelihood that an observed difference could have occurred by chance. As the p-value decreases the statistical significance of the observed difference increases. If the p-value is too low, you reject the null hypothesis.

Here you have taken an example in which you are trying to test whether the new advertising campaign has increased the product's sales. The p-value is the likelihood that the null hypothesis, which states that there is no change in the sales due to the new advertising campaign, is true. If the p-value is .30, then there is a 30% chance that there is no increase or decrease in the product's sales. If the p-value is 0.03, then there is a 3% probability that there is no increase or decrease in the sales value due to the new advertising campaign. As you can see, the lower the p-value, the chances of the alternate hypothesis being true increases, which means that the new advertising campaign causes an increase or decrease in sales.

## Why is Hypothesis Testing Important in Research Methodology?

Hypothesis testing is crucial in research methodology for several reasons:

- Provides evidence-based conclusions: It allows researchers to make objective conclusions based on empirical data, providing evidence to support or refute their research hypotheses.
- Supports decision-making: It helps make informed decisions, such as accepting or rejecting a new treatment, implementing policy changes, or adopting new practices.
- Adds rigor and validity: It adds scientific rigor to research using statistical methods to analyze data, ensuring that conclusions are based on sound statistical evidence.
- Contributes to the advancement of knowledge: By testing hypotheses, researchers contribute to the growth of knowledge in their respective fields by confirming existing theories or discovering new patterns and relationships.

## Limitations of Hypothesis Testing

Hypothesis testing has some limitations that researchers should be aware of:

- It cannot prove or establish the truth: Hypothesis testing provides evidence to support or reject a hypothesis, but it cannot confirm the absolute truth of the research question.
- Results are sample-specific: Hypothesis testing is based on analyzing a sample from a population, and the conclusions drawn are specific to that particular sample.
- Possible errors: During hypothesis testing, there is a chance of committing type I error (rejecting a true null hypothesis) or type II error (failing to reject a false null hypothesis).
- Assumptions and requirements: Different tests have specific assumptions and requirements that must be met to accurately interpret results.

After reading this tutorial, you would have a much better understanding of hypothesis testing, one of the most important concepts in the field of Data Science . The majority of hypotheses are based on speculation about observed behavior, natural phenomena, or established theories.

If you are interested in statistics of data science and skills needed for such a career, you ought to explore Simplilearn’s Post Graduate Program in Data Science.

If you have any questions regarding this ‘Hypothesis Testing In Statistics’ tutorial, do share them in the comment section. Our subject matter expert will respond to your queries. Happy learning!

## 1. What is hypothesis testing in statistics with example?

Hypothesis testing is a statistical method used to determine if there is enough evidence in a sample data to draw conclusions about a population. It involves formulating two competing hypotheses, the null hypothesis (H0) and the alternative hypothesis (Ha), and then collecting data to assess the evidence. An example: testing if a new drug improves patient recovery (Ha) compared to the standard treatment (H0) based on collected patient data.

## 2. What is hypothesis testing and its types?

Hypothesis testing is a statistical method used to make inferences about a population based on sample data. It involves formulating two hypotheses: the null hypothesis (H0), which represents the default assumption, and the alternative hypothesis (Ha), which contradicts H0. The goal is to assess the evidence and determine whether there is enough statistical significance to reject the null hypothesis in favor of the alternative hypothesis.

Types of hypothesis testing:

- One-sample test: Used to compare a sample to a known value or a hypothesized value.
- Two-sample test: Compares two independent samples to assess if there is a significant difference between their means or distributions.
- Paired-sample test: Compares two related samples, such as pre-test and post-test data, to evaluate changes within the same subjects over time or under different conditions.
- Chi-square test: Used to analyze categorical data and determine if there is a significant association between variables.
- ANOVA (Analysis of Variance): Compares means across multiple groups to check if there is a significant difference between them.

## 3. What are the steps of hypothesis testing?

The steps of hypothesis testing are as follows:

- Formulate the hypotheses: State the null hypothesis (H0) and the alternative hypothesis (Ha) based on the research question.
- Set the significance level: Determine the acceptable level of error (alpha) for making a decision.
- Collect and analyze data: Gather and process the sample data.
- Compute test statistic: Calculate the appropriate statistical test to assess the evidence.
- Make a decision: Compare the test statistic with critical values or p-values and determine whether to reject H0 in favor of Ha or not.
- Draw conclusions: Interpret the results and communicate the findings in the context of the research question.

## 4. What are the 2 types of hypothesis testing?

- One-tailed (or one-sided) test: Tests for the significance of an effect in only one direction, either positive or negative.
- Two-tailed (or two-sided) test: Tests for the significance of an effect in both directions, allowing for the possibility of a positive or negative effect.

The choice between one-tailed and two-tailed tests depends on the specific research question and the directionality of the expected effect.

## 5. What are the 3 major types of hypothesis?

The three major types of hypotheses are:

- Null Hypothesis (H0): Represents the default assumption, stating that there is no significant effect or relationship in the data.
- Alternative Hypothesis (Ha): Contradicts the null hypothesis and proposes a specific effect or relationship that researchers want to investigate.
- Nondirectional Hypothesis: An alternative hypothesis that doesn't specify the direction of the effect, leaving it open for both positive and negative possibilities.

## Find our Data Analyst Online Bootcamp in top cities:

About the author.

Avijeet is a Senior Research Analyst at Simplilearn. Passionate about Data Analytics, Machine Learning, and Deep Learning, Avijeet is also interested in politics, cricket, and football.

## Recommended Resources

Free eBook: Top Programming Languages For A Data Scientist

Normality Test in Minitab: Minitab with Statistics

Machine Learning Career Guide: A Playbook to Becoming a Machine Learning Engineer

- PMP, PMI, PMBOK, CAPM, PgMP, PfMP, ACP, PBA, RMP, SP, and OPM3 are registered marks of the Project Management Institute, Inc.

## Lab 7 – Hypothesis Testing

Introduction.

This lab is intended to serve as an introduction to hypothesis testing. Where available, we will use built-in functions in R to find and report p-values. In particular, we will investigate the finding of p-values for proportions, differences in proportions, odds ratios, means, and differences of means

## Hypothesis Testing

Testing proportions.

While we are able to use our normality assumptions in constructing confidence intervals and p-values for single proportions, we are also able to compute them exactly using the theoretical properties of a true distribution. That is, while counting the number of heads in successive coin flips will become approximately normal as the number of flips increases, we can also utilize the fact that coin flipping follows a binomial distribution . As such, we can use an exact binomial to compute our confidence intervals and p-values rather than relying a normal approximations. This is done with the R function binom.test() . This function takes several arugments which you can learn more about with ?binom.test .

For now, let’s consider again the Johns Hopkins study that found 31 of 39 infants born preterm survived to 6 months. Here, we are wanting to check the hypothesis \(H_0: p_0 = 0.7\) . To do this, we will use binom.test() with the arguments:

- x – the number of “successes” we observed
- n – the total number of observations
- p – our hypothesized proportion

The function will print out a litany of useful information

First, observed the bottom line, “probability of success” which is simply our observed test statistic, \(\hat{p} = 31/39\) . Note, however, that the \(p\) -value found here is not the same as what we found in the slides. This is a consequence of the fact that here we are utilizing assumptions from the binomial distribution rather than our normal approximation.

Just as the \(p\) -values change, so too do our confidence intervals. For example, in our slides Wednesday we found a 95% confidence interval to be (0.668, 0.922), while here they are slightly different, being (0.635, 0.907).

It is worth noting that we can also use binom.test() for the construction of confidence intervals. By default, the interval printed out is 95%, but we could just as well find an 80% confidence interval using the argument conf.level

Question 1: We see above that changing our confidence interval to 80% from 95% changed the interval, but not the resulting p-value. Why is this the case?

Question 2: You might also notice that the confidence interval we found using binom.test() is no longer symmetric around the estimate \(\hat{p}\) . Why is this the case?

Question 3: There is an intimate relationship between p-values and confidence intervals in that the exact same attributes of our sample data are used to construct each. Presented below are two scenarios:

- We flipped a coin 5 times and it landed on head 4 times
- We flipped a coin 50 times and it landed on head 40 times

Using the binom.test() function, create 95% confidence intervals and construct a \(p\) -value for the null hypothesis \(p_0 = 0.5\) . Then answer the following:

- What is our estimate of \(\hat{p}\) for each of these scenarios?
- What changed in our \(p\) -values and confidence intervals?
- Which of these scenarios provides more evidence against the null hypothesis? Why?

## Testing means

Testing odds ratios, testing things beyond our wildest dreams.

Help | Advanced Search

## Statistics > Methodology

Title: a bayes factor framework for unified parameter estimation and hypothesis testing.

Abstract: The Bayes factor, the data-based updating factor of the prior to posterior odds of two hypotheses, is a natural measure of statistical evidence for one hypothesis over the other. We show how Bayes factors can also be used for parameter estimation. The key idea is to consider the Bayes factor as a function of the parameter value under the null hypothesis. This 'Bayes factor function' is inverted to obtain point estimates ('maximum evidence estimates') and interval estimates ('support intervals'), similar to how P-value functions are inverted to obtain point estimates and confidence intervals. This provides data analysts with a unified inference framework as Bayes factors (for any tested parameter value), support intervals (at any level), and point estimates can be easily read off from a plot of the Bayes factor function. This approach shares similarities but is also distinct from conventional Bayesian and frequentist approaches: It uses the Bayesian evidence calculus, but without synthesizing data and prior, and it defines statistical evidence in terms of (integrated) likelihood ratios, but also includes a natural way for dealing with nuisance parameters. Applications to real-world examples illustrate how our framework is of practical value for making make quantitative inferences.

## Submission history

Access paper:.

- Download PDF
- HTML (experimental)
- Other Formats

## References & Citations

- Google Scholar
- Semantic Scholar

## BibTeX formatted citation

## Bibliographic and Citation Tools

Code, data and media associated with this article, recommenders and search tools.

- Institution

## arXivLabs: experimental projects with community collaborators

arXivLabs is a framework that allows collaborators to develop and share new arXiv features directly on our website.

Both individuals and organizations that work with arXivLabs have embraced and accepted our values of openness, community, excellence, and user data privacy. arXiv is committed to these values and only works with partners that adhere to them.

Have an idea for a project that will add value for arXiv's community? Learn more about arXivLabs .

## IMAGES

## VIDEO

## COMMENTS

A hypothesis test is carried out at the 5% level of significance to test if a normal coin is fair or not. (i) Describe what the population parameter could be for the hypothesis test. (ii) State whether the hypothesis test should be a one-tailed test or a two-tailed test, give a reason for your answer. (iii)

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.

We choose a region known as the critical region. If the result of our test lies in this region, then we reject the null hypothesis in favour of the alternative. Hypothesis testing A-Level Maths Statistics revision looking at Hypothesis testing. Topics include null hypothesis, alternative hypothesis, testing and critical regions.

A hypothesis test is the means by which we generate a test statistic that directs us to either reject or not reject the null hypothesis. The test statistic is a "summary" of the collected data, and should have a sampling distribution specified by the null hypothesis. A Level AQA Edexcel OCR.

The actual significance 5.level of a hypothesis test is the probability of incorrectly rejecting the null hypothesis. Example 2: A single observation is taken from the binomial distribution ) B(6,$). The observation is used to test H!:$ = 0.35 against H ":$ > 0.35 a. Using a 5% significance level, find the critical region for this test. Assume H

A teacher carried out a hypothesis test at the 10% significance level to test if her students perform better in exams after using a new revision technique. Under the null hypothesis she calculates the probability that a value will be at least as extreme as the observed value to be 0.09142. Write a conclusion for her hypothesis test.

Now we carry out the above steps in order to come to a conclusion. Step 1: We state the null hypothesis and the alternate hypothesis: and. Step 2: We select the level of significance which is stated in the problem as 5% or α = 0.05. Step 3: Compute the test statistics. We first identify the test to be used.

Hypothesis testing involves using the null hypothesis and the alternative hypothesis in order to assess whether an assumption is true. This video takes a loo...

AS Level Mechanics and Statistics - Hypothesis Testing. Maths revision videos and notes on the topics of hypothesis testing, correlation hypothesis testing, mean of normal distribution hypothesis testing and non linear regression.

Step 2: State the Alternate Hypothesis. The claim is that the students have above average IQ scores, so: H 1: μ > 100. The fact that we are looking for scores "greater than" a certain point means that this is a one-tailed test. Step 3: Draw a picture to help you visualize the problem. Step 4: State the alpha level.

Statistical Hypothesis Testing Topics for A-Level Maths. This module will teach you the following: Hypothesis testing in a binomial distribution. Hypothesis testing in a normal distribution. Hypothesis test using Pearson's correlation coefficient. Download the Sample Chapters →.

How is the critical value found in a hypothesis test for the mean of a normal distribution? The critical value(s) will be the boundary of the critical region. The probability of the observed value being within the critical region, given a true null hypothesis will be the same as the significance level; For an % significance level: In a one-tailed test the critical region will consist of % in ...

Revision notes on 3.1.2 Type I & Type II Errors for the CIE A Level Maths: Probability & Statistics 2 syllabus, written by the Maths experts at Save My Exams. ... Any hypothesis test will only provide evidence about whether a parameter has changed or not. A conclusion can not claim with certainty whether to accept or reject the null hypothesis ...

b. State the actual significance level of this test. P(reject null hypothesis)= P(" ≥ 5) = 0.0223 = 2.23% Two-tailed Test A two-tailed test is used to test if the probability is changed in either direction. The critical region is split at either end of distribution. The significance level at each end is halved. For two-tailed tests, H ":$ ≠ ⋯

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 ...

A hypothesis test consists of five steps: 1. State the hypotheses. State the null and alternative hypotheses. These two hypotheses need to be mutually exclusive, so if one is true then the other must be false. 2. Determine a significance level to use for the hypothesis. Decide on a significance level.

Probably the toughest chapter for the 1st year material of Edexcel's A Level Maths and Statistics material.This chapter takes a look at Hypothesis Testing fo...

A test of the null hypothesis is carried out for the random variable . The observed value of the test statistic is . You are given the following probabilities: Determine the outcome of the test, with reasons, when the alternative hypothesis is: with a 1% level of significance.

Conduct a hypothesis test. At a significance level of \(a = 0.05\), what is the correct conclusion? ... Use a significance level of 0.05 and using appropriate statistical evidence, conduct a hypothesis test and state the conclusions. Q 9.6.35. A recent survey in the N.Y. Times Almanac indicated that 48.8% of families own stock. A broker wanted ...

2-tailed test crit under Ho Allow use Of t method AWFW 0.99 to 1.00 (allow 1) Or z = 1.96 On their z and critical value Or t Accept Ho at 5% level Of significance. Sufficient evidence at the 5% level Of significance to support the manufacturer's belief. Total 6(a) 471 = 94.2 - 6.058 4 1 -tailed test --2.132 Ho Hi : p < 100 94.2-100 --2.14 6.058

Hypothesis testing is a statistical method used to determine if there is enough evidence in a sample data to draw conclusions about a population. It involves formulating two competing hypotheses, the null hypothesis (H0) and the alternative hypothesis (Ha), and then collecting data to assess the evidence.

Here, we are wanting to check the hypothesis H0: p0 = 0.7 H 0: p 0 = 0.7. To do this, we will use binom.test () with the arguments: x - the number of "successes" we observed. n - the total number of observations. p - our hypothesized proportion. The function will print out a litany of useful information.

Calculate the interquartile range for the LSAT scores b. Test the hypothesis that the mean is greater than 1500. Use the 1% level of significance. c. Would your conclusion change at the 10% level of significance? Show all steps for the hypothesis test. Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018. 18th Edition.

In this case, we need to plot. against to obtain a linear model. Example 1: The heights, h cm, and masses, kg, of a sample of Galapagos penguins are recorded. The data are coded using = and = h and it is found that a linear relationship exists between and . The equation of the regression line of on is = 0.0023 + 1.8 .

The Bayes factor, the data-based updating factor of the prior to posterior odds of two hypotheses, is a natural measure of statistical evidence for one hypothesis over the other. We show how Bayes factors can also be used for parameter estimation. The key idea is to consider the Bayes factor as a function of the parameter value under the null hypothesis. This 'Bayes factor function' is ...

Nationally 44% of A Level mathematics students identify as female. The headteacher of a particular school claims that the proportion of A Level mathematics students in the school who identify as female is higher than the national average. (i) State a suitable null hypothesis to test the headteacher's claim. (ii)