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Hypothesis Testing Calculator

Type II Error

The first step in hypothesis testing is to calculate the test statistic. The formula for the test statistic depends on whether the population standard deviation (σ) is known or unknown. If σ is known, our hypothesis test is known as a z test and we use the z distribution. If σ is unknown, our hypothesis test is known as a t test and we use the t distribution. Use of the t distribution relies on the degrees of freedom, which is equal to the sample size minus one. Furthermore, if the population standard deviation σ is unknown, the sample standard deviation s is used instead. To switch from σ known to σ unknown, click on $\boxed{\sigma}$ and select $\boxed{s}$ in the Hypothesis Testing Calculator.

Next, the test statistic is used to conduct the test using either the p-value approach or critical value approach. The particular steps taken in each approach largely depend on the form of the hypothesis test: lower tail, upper tail or two-tailed. The form can easily be identified by looking at the alternative hypothesis (H a ). If there is a less than sign in the alternative hypothesis then it is a lower tail test, greater than sign is an upper tail test and inequality is a two-tailed test. To switch from a lower tail test to an upper tail or two-tailed test, click on $\boxed{\geq}$ and select $\boxed{\leq}$ or $\boxed{=}$, respectively.

In the p-value approach, the test statistic is used to calculate a p-value. If the test is a lower tail test, the p-value is the probability of getting a value for the test statistic at least as small as the value from the sample. If the test is an upper tail test, the p-value is the probability of getting a value for the test statistic at least as large as the value from the sample. In a two-tailed test, the p-value is the probability of getting a value for the test statistic at least as unlikely as the value from the sample.

To test the hypothesis in the p-value approach, compare the p-value to the level of significance. If the p-value is less than or equal to the level of signifance, reject the null hypothesis. If the p-value is greater than the level of significance, do not reject the null hypothesis. This method remains unchanged regardless of whether it's a lower tail, upper tail or two-tailed test. To change the level of significance, click on $\boxed{.05}$. Note that if the test statistic is given, you can calculate the p-value from the test statistic by clicking on the switch symbol twice.

In the critical value approach, the level of significance ($\alpha$) is used to calculate the critical value. In a lower tail test, the critical value is the value of the test statistic providing an area of $\alpha$ in the lower tail of the sampling distribution of the test statistic. In an upper tail test, the critical value is the value of the test statistic providing an area of $\alpha$ in the upper tail of the sampling distribution of the test statistic. In a two-tailed test, the critical values are the values of the test statistic providing areas of $\alpha / 2$ in the lower and upper tail of the sampling distribution of the test statistic.

To test the hypothesis in the critical value approach, compare the critical value to the test statistic. Unlike the p-value approach, the method we use to decide whether to reject the null hypothesis depends on the form of the hypothesis test. In a lower tail test, if the test statistic is less than or equal to the critical value, reject the null hypothesis. In an upper tail test, if the test statistic is greater than or equal to the critical value, reject the null hypothesis. In a two-tailed test, if the test statistic is less than or equal the lower critical value or greater than or equal to the upper critical value, reject the null hypothesis.

When conducting a hypothesis test, there is always a chance that you come to the wrong conclusion. There are two types of errors you can make: Type I Error and Type II Error. A Type I Error is committed if you reject the null hypothesis when the null hypothesis is true. Ideally, we'd like to accept the null hypothesis when the null hypothesis is true. A Type II Error is committed if you accept the null hypothesis when the alternative hypothesis is true. Ideally, we'd like to reject the null hypothesis when the alternative hypothesis is true.

Hypothesis testing is closely related to the statistical area of confidence intervals. If the hypothesized value of the population mean is outside of the confidence interval, we can reject the null hypothesis. Confidence intervals can be found using the Confidence Interval Calculator . The calculator on this page does hypothesis tests for one population mean. Sometimes we're interest in hypothesis tests about two population means. These can be solved using the Two Population Calculator . The probability of a Type II Error can be calculated by clicking on the link at the bottom of the page.

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Here you will find everything about hypothesis testing: One sample t-test , Unpaired t-test , Paired t-test and Chi-square test . You will also find tutorials for non-parametric statistical procedures such as the Mann-Whitney u-Test and Wilcoxon-Test . mann-whitney-u-test and the Wilcoxon test

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Hypothesis Testing Calculator Online

Hypothesis testing is a foundational method used in statistics to infer the validity of a hypothesis about a population parameter. The Hypothesis Testing Calculator facilitates this process by automating the computations necessary for the t-test , a method used to compare sample means against a hypothesized mean or against each other. Let’s delve into the formulas this calculator uses to execute one-sample and two-sample t-tests.

One-Sample t-Test

This test is used to determine if the mean (x̄) of your sample is statistically different from a hypothesized population mean (μ₀).

Two-Sample t-Test

Equal variances:, unequal variances (welch’s t-test):.

t = (x̄₁ – x̄₂) / (√((s₁² / n₁) + (s₂² / n₂)))

Table of Critical t-Values

These values are crucial in hypothesis testing as they help define the threshold for significance, assisting users of the calculator in interpreting their results accurately.

Using the one-sample t-test formula:

t = (74 – 70) / (8 / √36) = (4 / 1.333) = 3.00

Most Common FAQs

The p-value represents the probability of obtaining test results at least as extreme as the results observed, under the assumption that the null hypothesis is correct. A low p-value (typically below 0.05) indicates strong evidence against the null hypothesis, hence it is usually rejected.

Yes, while the t-test is specifically design for means, the principles of hypothesis testing apply to other parameters such as proportions and variances. Which can also be tested using appropriate versions of hypothesis tests such as the z-test and F-test.

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Test Statistic Calculator

Choose the method, enter the values into the test statistic calculator, and click on the “Calculate” button to calculate the statistical value for hypothesis evaluation.

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This test statistic calculator helps to find the static value for hypothesis testing. The calculated test value shows if there’s enough evidence to reject a null hypothesis. Also, this calculator performs calculations of either for one population mean, comparing two means, single population proportion, and two population proportions.

Our tool is highly useful in various fields like research, experimentation, quality control, and data analysis.

What is Test Statistics?

A test statistic is a numerical value obtained from the sample data set. It summarizes the differences between what you observe within your sample and what would be expected if a hypothesis were true. 

The t-test statistic also shows how closely your data matches the predicted distribution among the sample tests you perform. 

How to Calculate Test Statistics Value?

• Collect the data from the populations
• Use the data to find the standard deviation of the population
• Calculate the mean (μ) of the population using this data
• Determine the z-value or sample size 
• Use the suitable test statistic formula and get the results

Test Statistic For One Population Mean:

Test statistics for a single population mean is calculated when a variable is numeric and involves one population or a group. 

x̄ - µ 0 σ / √n

• x̄ = Mean of your sample data
• µ 0 = Hypothesized population mean that you are comparing to your sample mean
• σ = Population standard deviation
• n = number of observations (sample size) in your data set

Suppose we want to test if the average height of adult males in a city is 70 inches. We take a sample of 25 adult males and find the sample mean height to be 71 inches with a sample standard deviation of 3 inches. We use a significance level of 0.05.

t = 70 - 71 3√25

Test Statistic Comparing Two Population Means:

This test is applied when the numeric value is compared across the various populations or groups. To compute the resulting t statistic, two distinct random samples must be chosen, one from each population.

\(\frac{√x̄ - √ȳ}{\sqrt{\frac{\sigma_{1}^{2}}{n_{1}}+\frac{\sigma_{2}^{2}}{n_{2}}}}\)

• ȳ = means of hypothesized population

Suppose we want to test if there is a difference in average test scores between two schools. We take a sample of 30 students from school A with an average score of 85 and a standard deviation of 5, and a sample of 35 students from school B with an average score of 82 and a standard deviation of 6.

t = 85 - 82 √5 2 / 30 + 6 2 / 35

t = 3 √ 25/30 + 36/35

t = 3 √0.833 + 1.029

t = 3 √1.862

Test Statistic For a Single Population Proportion:

This test is used to determine if a single population's proportion differs from a specified standard. The t statistic calculator works for a population proportion when dealing with data by having a limit of P₀ because proportions represent parts of a whole and cannot logically exceed the total or be negative.

\(\frac{\hat{p}-p_{0}}{\sqrt{\frac{p_{0}(1-p_{0})}{n}}}\)

• P̂ = Sample proportion
• P 0 = Population proportion

Suppose we want to test if the proportion of left-handed people in a population is 10%. We take a sample of 100 people and find that 8 are left-handed. We use a significance level of 0.05.

= P̂ - P₀ √0.10 (1 - 0.10)/100

= 0.08 - 0.10 √0.10 (1 - 0.10)/100

= -0.02 √0.10 (0.9)/100

= -0.02 √0.009

= -0.02 0.03

= −0.67

Test Statistic For Two Population Proportion:

This test identifies the difference in proportions between two independent groups to assess their significance.

\(\frac{\hat{p}_{1}-\hat{p}_{2}}{\sqrt{\hat{p}(1-\hat{p})(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}\)

• P̂ 1 and P̂ 2 = Sample proportions for two groups

Suppose we want to test if the proportion of smokers is different between two cities. We take a sample of 150 people from City A and find that 30 are smokers, and a sample of 200 people from City B and find that 50 are smokers.

• P̂ 1 = 30 / 150 = 0.20
• P̂ 2 = 50 / 200 = 0.25
• P̂ = 30 + 50 / 150 + 200 = 0.229

Calculation:

= 0.20 - 0.25 √0.229 (1 - 0.229) (1 / 150 + 1/200)

= -0.05 √0.229 (0.771) (1 / 150 + 1 / 200)

= -0.05 √0.176 (1/150 + 1/200)

= -0.05 √0.176 (0.0113)

= -0.05 √0.002

= -0.05 0.045

= −1.11

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Testing a Population  Standard Deviation (data)

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Sample standard deviation, standardized sample score, sample variance.

Z-test Calculator

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This Z-test calculator is a tool that helps you perform a one-sample Z-test on the population's mean . Two forms of this test - a two-tailed Z-test and a one-tailed Z-tests - exist, and can be used depending on your needs. You can also choose whether the calculator should determine the p-value from Z-test or you'd rather use the critical value approach!

Read on to learn more about Z-test in statistics, and, in particular, when to use Z-tests, what is the Z-test formula, and whether to use Z-test vs. t-test. As a bonus, we give some step-by-step examples of how to perform Z-tests!

Or you may also check our t-statistic calculator , where you can learn the concept of another essential statistic. If you are also interested in F-test, check our F-statistic calculator .

What is a Z-test?

A one sample Z-test is one of the most popular location tests. The null hypothesis is that the population mean value is equal to a given number, μ 0 \mu_0 μ 0 ​ :

We perform a two-tailed Z-test if we want to test whether the population mean is not μ 0 \mu_0 μ 0 ​ :

and a one-tailed Z-test if we want to test whether the population mean is less/greater than μ 0 \mu_0 μ 0 ​ :

Let us now discuss the assumptions of a one-sample Z-test.

When do I use Z-tests?

You may use a Z-test if your sample consists of independent data points and:

the data is normally distributed , and you know the population variance ;

the sample is large , and data follows a distribution which has a finite mean and variance. You don't need to know the population variance.

The reason these two possibilities exist is that we want the test statistics that follow the standard normal distribution N ( 0 , 1 ) \mathrm N(0, 1) N ( 0 , 1 ) . In the former case, it is an exact standard normal distribution, while in the latter, it is approximately so, thanks to the central limit theorem.

The question remains, "When is my sample considered large?" Well, there's no universal criterion. In general, the more data points you have, the better the approximation works. Statistics textbooks recommend having no fewer than 50 data points, while 30 is considered the bare minimum.

Z-test formula

Let x 1 , . . . , x n x_1, ..., x_n x 1 ​ , ... , x n ​ be an independent sample following the normal distribution N ( μ , σ 2 ) \mathrm N(\mu, \sigma^2) N ( μ , σ 2 ) , i.e., with a mean equal to μ \mu μ , and variance equal to σ 2 \sigma ^2 σ 2 .

We pose the null hypothesis, H 0  ⁣  ⁣ :  ⁣  ⁣   μ = μ 0 \mathrm H_0 \!\!:\!\! \mu = \mu_0 H 0 ​ :   μ = μ 0 ​ .

We define the test statistic, Z , as:

x ˉ \bar x x ˉ is the sample mean, i.e., x ˉ = ( x 1 + . . . + x n ) / n \bar x = (x_1 + ... + x_n) / n x ˉ = ( x 1 ​ + ... + x n ​ ) / n ;

μ 0 \mu_0 μ 0 ​ is the mean postulated in H 0 \mathrm H_0 H 0 ​ ;

n n n is sample size; and

σ \sigma σ is the population standard deviation.

In what follows, the uppercase Z Z Z stands for the test statistic (treated as a random variable), while the lowercase z z z will denote an actual value of Z Z Z , computed for a given sample drawn from N(μ,σ²).

If H 0 \mathrm H_0 H 0 ​ holds, then the sum S n = x 1 + . . . + x n S_n = x_1 + ... + x_n S n ​ = x 1 ​ + ... + x n ​ follows the normal distribution, with mean n μ 0 n \mu_0 n μ 0 ​ and variance n 2 σ n^2 \sigma n 2 σ . As Z Z Z is the standardization (z-score) of S n / n S_n/n S n ​ / n , we can conclude that the test statistic Z Z Z follows the standard normal distribution N ( 0 , 1 ) \mathrm N(0, 1) N ( 0 , 1 ) , provided that H 0 \mathrm H_0 H 0 ​ is true. By the way, we have the z-score calculator if you want to focus on this value alone.

If our data does not follow a normal distribution, or if the population standard deviation is unknown (and thus in the formula for Z Z Z we substitute the population standard deviation σ \sigma σ with sample standard deviation), then the test statistics Z Z Z is not necessarily normal. However, if the sample is sufficiently large, then the central limit theorem guarantees that Z Z Z is approximately N ( 0 , 1 ) \mathrm N(0, 1) N ( 0 , 1 ) .

In the sections below, we will explain to you how to use the value of the test statistic, z z z , to make a decision , whether or not you should reject the null hypothesis . Two approaches can be used in order to arrive at that decision: the p-value approach, and critical value approach - and we cover both of them! Which one should you use? In the past, the critical value approach was more popular because it was difficult to calculate p-value from Z-test. However, with help of modern computers, we can do it fairly easily, and with decent precision. In general, you are strongly advised to report the p-value of your tests!

p-value from Z-test

Formally, the p-value is the smallest level of significance at which the null hypothesis could be rejected. More intuitively, p-value answers the questions: provided that I live in a world where the null hypothesis holds, how probable is it that the value of the test statistic will be at least as extreme as the z z z - value I've got for my sample? Hence, a small p-value means that your result is very improbable under the null hypothesis, and so there is strong evidence against the null hypothesis - the smaller the p-value, the stronger the evidence.

To find the p-value, you have to calculate the probability that the test statistic, Z Z Z , is at least as extreme as the value we've actually observed, z z z , provided that the null hypothesis is true. (The probability of an event calculated under the assumption that H 0 \mathrm H_0 H 0 ​ is true will be denoted as P r ( event ∣ H 0 ) \small \mathrm{Pr}(\text{event} | \mathrm{H_0}) Pr ( event ∣ H 0 ​ ) .) It is the alternative hypothesis which determines what more extreme means :

• Two-tailed Z-test: extreme values are those whose absolute value exceeds ∣ z ∣ |z| ∣ z ∣ , so those smaller than − ∣ z ∣ -|z| − ∣ z ∣ or greater than ∣ z ∣ |z| ∣ z ∣ . Therefore, we have:

The symmetry of the normal distribution gives:

• Left-tailed Z-test: extreme values are those smaller than z z z , so
• Right-tailed Z-test: extreme values are those greater than z z z , so

To compute these probabilities, we can use the cumulative distribution function, (cdf) of N ( 0 , 1 ) \mathrm N(0, 1) N ( 0 , 1 ) , which for a real number, x x x , is defined as:

Also, p-values can be nicely depicted as the area under the probability density function (pdf) of N ( 0 , 1 ) \mathrm N(0, 1) N ( 0 , 1 ) , due to:

Two-tailed Z-test and one-tailed Z-test

With all the knowledge you've got from the previous section, you're ready to learn about Z-tests.

• Two-tailed Z-test:

From the fact that Φ ( − z ) = 1 − Φ ( z ) \Phi(-z) = 1 - \Phi(z) Φ ( − z ) = 1 − Φ ( z ) , we deduce that

The p-value is the area under the probability distribution function (pdf) both to the left of − ∣ z ∣ -|z| − ∣ z ∣ , and to the right of ∣ z ∣ |z| ∣ z ∣ :

• Left-tailed Z-test:

The p-value is the area under the pdf to the left of our z z z :

• Right-tailed Z-test:

The p-value is the area under the pdf to the right of z z z :

The decision as to whether or not you should reject the null hypothesis can be now made at any significance level, α \alpha α , you desire!

if the p-value is less than, or equal to, α \alpha α , the null hypothesis is rejected at this significance level; and

if the p-value is greater than α \alpha α , then there is not enough evidence to reject the null hypothesis at this significance level.

Z-test critical values & critical regions

The critical value approach involves comparing the value of the test statistic obtained for our sample, z z z , to the so-called critical values . These values constitute the boundaries of regions where the test statistic is highly improbable to lie . Those regions are often referred to as the critical regions , or rejection regions . The decision of whether or not you should reject the null hypothesis is then based on whether or not our z z z belongs to the critical region.

The critical regions depend on a significance level, α \alpha α , of the test, and on the alternative hypothesis. The choice of α \alpha α is arbitrary; in practice, the values of 0.1, 0.05, or 0.01 are most commonly used as α \alpha α .

Once we agree on the value of α \alpha α , we can easily determine the critical regions of the Z-test:

To decide the fate of H 0 \mathrm H_0 H 0 ​ , check whether or not your z z z falls in the critical region:

If yes, then reject H 0 \mathrm H_0 H 0 ​ and accept H 1 \mathrm H_1 H 1 ​ ; and

If no, then there is not enough evidence to reject H 0 \mathrm H_0 H 0 ​ .

As you see, the formulae for the critical values of Z-tests involve the inverse, Φ − 1 \Phi^{-1} Φ − 1 , of the cumulative distribution function (cdf) of N ( 0 , 1 ) \mathrm N(0, 1) N ( 0 , 1 ) .

How to use the one-sample Z-test calculator?

Our calculator reduces all the complicated steps:

Choose the alternative hypothesis: two-tailed or left/right-tailed.

In our Z-test calculator, you can decide whether to use the p-value or critical regions approach. In the latter case, set the significance level, α \alpha α .

Enter the value of the test statistic, z z z . If you don't know it, then you can enter some data that will allow us to calculate your z z z for you:

• sample mean x ˉ \bar x x ˉ (If you have raw data, go to the average calculator to determine the mean);
• tested mean μ 0 \mu_0 μ 0 ​ ;
• sample size n n n ; and
• population standard deviation σ \sigma σ (or sample standard deviation if your sample is large).

Results appear immediately below the calculator.

If you want to find z z z based on p-value , please remember that in the case of two-tailed tests there are two possible values of z z z : one positive and one negative, and they are opposite numbers. This Z-test calculator returns the positive value in such a case. In order to find the other possible value of z z z for a given p-value, just take the number opposite to the value of z z z displayed by the calculator.

Z-test examples

To make sure that you've fully understood the essence of Z-test, let's go through some examples:

• A bottle filling machine follows a normal distribution. Its standard deviation, as declared by the manufacturer, is equal to 30 ml. A juice seller claims that the volume poured in each bottle is, on average, one liter, i.e., 1000 ml, but we suspect that in fact the average volume is smaller than that...

Formally, the hypotheses that we set are the following:

H 0  ⁣ :   μ = 1000  ml \mathrm H_0 \! : \mu = 1000 \text{ ml} H 0 ​ :   μ = 1000  ml

H 1  ⁣ :   μ < 1000  ml \mathrm H_1 \! : \mu \lt 1000 \text{ ml} H 1 ​ :   μ < 1000  ml

We went to a shop and bought a sample of 9 bottles. After carefully measuring the volume of juice in each bottle, we've obtained the following sample (in milliliters):

1020 , 970 , 1000 , 980 , 1010 , 930 , 950 , 980 , 980 \small 1020, 970, 1000, 980, 1010, 930, 950, 980, 980 1020 , 970 , 1000 , 980 , 1010 , 930 , 950 , 980 , 980 .

Sample size: n = 9 n = 9 n = 9 ;

Sample mean: x ˉ = 980   m l \bar x = 980 \ \mathrm{ml} x ˉ = 980   ml ;

Population standard deviation: σ = 30   m l \sigma = 30 \ \mathrm{ml} σ = 30   ml ;

And, therefore, p-value = Φ ( − 2 ) ≈ 0.0228 \text{p-value} = \Phi(-2) \approx 0.0228 p-value = Φ ( − 2 ) ≈ 0.0228 .

As 0.0228 < 0.05 0.0228 \lt 0.05 0.0228 < 0.05 , we conclude that our suspicions aren't groundless; at the most common significance level, 0.05, we would reject the producer's claim, H 0 \mathrm H_0 H 0 ​ , and accept the alternative hypothesis, H 1 \mathrm H_1 H 1 ​ .

We tossed a coin 50 times. We got 20 tails and 30 heads. Is there sufficient evidence to claim that the coin is biased?

Clearly, our data follows Bernoulli distribution, with some success probability p p p and variance σ 2 = p ( 1 − p ) \sigma^2 = p (1-p) σ 2 = p ( 1 − p ) . However, the sample is large, so we can safely perform a Z-test. We adopt the convention that getting tails is a success.

Let us state the null and alternative hypotheses:

H 0  ⁣ :   p = 0.5 \mathrm H_0 \! : p = 0.5 H 0 ​ :   p = 0.5 (the coin is fair - the probability of tails is 0.5 0.5 0.5 )

H 1  ⁣ :   p ≠ 0.5 \mathrm H_1 \! : p \ne 0.5 H 1 ​ :   p  = 0.5 (the coin is biased - the probability of tails differs from 0.5 0.5 0.5 )

In our sample we have 20 successes (denoted by ones) and 30 failures (denoted by zeros), so:

Sample size n = 50 n = 50 n = 50 ;

Sample mean x ˉ = 20 / 50 = 0.4 \bar x = 20/50 = 0.4 x ˉ = 20/50 = 0.4 ;

Population standard deviation is given by σ = 0.5 × 0.5 \sigma = \sqrt{0.5 \times 0.5} σ = 0.5 × 0.5 ​ (because 0.5 0.5 0.5 is the proportion p p p hypothesized in H 0 \mathrm H_0 H 0 ​ ). Hence, σ = 0.5 \sigma = 0.5 σ = 0.5 ;

• And, therefore

Since 0.1573 > 0.1 0.1573 \gt 0.1 0.1573 > 0.1 we don't have enough evidence to reject the claim that the coin is fair , even at such a large significance level as 0.1 0.1 0.1 . In that case, you may safely toss it to your Witcher or use the coin flip probability calculator to find your chances of getting, e.g., 10 heads in a row (which are extremely low!).

What is the difference between Z-test vs t-test?

We use a t-test for testing the population mean of a normally distributed dataset which had an unknown population standard deviation . We get this by replacing the population standard deviation in the Z-test statistic formula by the sample standard deviation, which means that this new test statistic follows (provided that H₀ holds) the t-Student distribution with n-1 degrees of freedom instead of N(0,1) .

When should I use t-test over the Z-test?

For large samples, the t-Student distribution with n degrees of freedom approaches the N(0,1). Hence, as long as there are a sufficient number of data points (at least 30), it does not really matter whether you use the Z-test or the t-test, since the results will be almost identical. However, for small samples with unknown variance, remember to use the t-test instead of Z-test .

How do I calculate the Z test statistic?

To calculate the Z test statistic:

• Compute the arithmetic mean of your sample .
• From this mean subtract the mean postulated in null hypothesis .
• Multiply by the square root of size sample .
• Divide by the population standard deviation .
• That's it, you've just computed the Z test statistic!

Here, we perform a Z-test for population mean μ. Null hypothesis H₀: μ = μ₀.

Alternative hypothesis H₁

Significance level α

The probability that we reject the true hypothesis H₀ (type I error).

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8.7 Hypothesis Tests for a Population Mean with Unknown Population Standard Deviation

Learning objectives.

• Conduct and interpret hypothesis tests for a population mean with unknown population standard deviation.

Some notes about conducting a hypothesis test:

• The null hypothesis [latex]H_0[/latex] is always an “equal to.”  The null hypothesis is the original claim about the population parameter.
• The alternative hypothesis [latex]H_a[/latex] is a “less than,” “greater than,” or “not equal to.”  The form of the alternative hypothesis depends on the context of the question.
• If the alternative hypothesis is a “less than”,  then the test is left-tail.  The p -value is the area in the left-tail of the distribution.
• If the alternative hypothesis is a “greater than”, then the test is right-tail.  The p -value is the area in the right-tail of the distribution.
• If the alternative hypothesis is a “not equal to”, then the test is two-tail.  The p -value is the sum of the area in the two-tails of the distribution.  Each tail represents exactly half of the p -value.
• Think about the meaning of the p -value.  A data analyst (and anyone else) should have more confidence that they made the correct decision to reject the null hypothesis with a smaller p -value (for example, 0.001 as opposed to 0.04) even if using a significance level of  0.05.  Similarly, for a large p -value such as 0.4, as opposed to a p -value of 0.056 (a significance level of 0.05 is less than either number), a data analyst should have more confidence that they made the correct decision in not rejecting the null hypothesis.  This makes the data analyst use judgment rather than mindlessly applying rules.
• The significance level must be identified before collecting the sample data and conducting the test.  Generally, the significance level will be included in the question.  If no significance level is given, a common standard is to use a significance level of 5%.
• An alternative approach for hypothesis testing is to use what is called the critical value approach .  In this book, we will only use the p -value approach.  Some of the videos below may mention the critical value approach, but this approach will not be used in this book.

Steps to Conduct a Hypothesis Test for a Population Mean with Unknown Population Standard Deviation

• Write down the null and alternative hypotheses in terms of the population mean [latex]\mu[/latex].  Include appropriate units with the values of the mean.
• Use the form of the alternative hypothesis to determine if the test is left-tailed, right-tailed, or two-tailed.
• Collect the sample information for the test and identify the significance level [latex]\alpha[/latex].

[latex]\begin{eqnarray*} t & = & \frac{\overline{x}-\mu}{\frac{s}{\sqrt{n}}} \\ \\ df & = & n-1 \\ \\ \end{eqnarray*}[/latex]

• The results of the sample data are significant. There is sufficient evidence to conclude that the null hypothesis [latex]H_0[/latex] is an incorrect belief and that the alternative hypothesis [latex]H_a[/latex] is most likely correct.
• The results of the sample data are not significant. There is not sufficient evidence to conclude that the alternative hypothesis [latex]H_a[/latex] may be correct.
• Write down a concluding sentence specific to the context of the question.

USING EXCEL TO CALCULE THE P -VALUE FOR A HYPOTHESIS TEST ON A POPULATION MEAN WITH UNKNOWN POPULATION STANDARD DEVIATION

The p -value for a hypothesis test on a population mean is the area in the tail(s) of the distribution of the sample mean.  When the population standard deviation is unknown, use the [latex]t[/latex]-distribution to find the p -value.

If the p -value is the area in the left-tail:

• For t-score , enter the value of [latex]t[/latex] calculated from [latex]\displaystyle{t=\frac{\overline{x}-\mu}{\frac{s}{\sqrt{n}}}}[/latex].
• For degrees of freedom , enter the degrees of freedom for the [latex]t[/latex]-distribution [latex]n-1[/latex].
• For the logic operator , enter true .  Note:  Because we are calculating the area under the curve, we always enter true for the logic operator.
• The output from the t.dist function is the area under the [latex]t[/latex]-distribution to the left of the entered [latex]t[/latex]-score.
• Visit the Microsoft page for more information about the t.dist function.

If the p -value is the area in the right-tail:

• The output from the t.dist.rt function is the area under the [latex]t[/latex]-distribution to the right of the entered [latex]t[/latex]-score.
• Visit the Microsoft page for more information about the t.dist.rt function.

If the p -value is the sum of area in the tails:

• For t-score , enter the absolute value of [latex]t[/latex] calculated from [latex]\displaystyle{t=\frac{\overline{x}-\mu}{\frac{s}{\sqrt{n}}}}[/latex].  Note:  In the t.dist.2t function, the value of the [latex]t[/latex]-score must be a positive number.  If the [latex]t[/latex]-score is negative, enter the absolute value of the [latex]t[/latex]-score into the t.dist.2t function.
• The output from the t.dist.2t function is the sum of areas in the tails under the [latex]t[/latex]-distribution.
• Visit the Microsoft page for more information about the t.dist.2t function.

Statistics students believe that the mean score on the first statistics test is 65.  A statistics instructor thinks the mean score is higher than 65.  He samples ten statistics students and obtains the following scores:

The instructor performs a hypothesis test using a 1% level of significance. The test scores are assumed to be from a normal distribution.

Hypotheses:

[latex]\begin{eqnarray*} H_0: & & \mu=65  \\ H_a: & & \mu \gt 65  \end{eqnarray*}[/latex]

From the question, we have [latex]n=10[/latex], [latex]\overline{x}=67[/latex], [latex]s=3.1972...[/latex] and [latex]\alpha=0.01[/latex].

This is a test on a population mean where the population standard deviation is unknown (we only know the sample standard deviation [latex]s=3.1972...[/latex]).  So we use a [latex]t[/latex]-distribution to calculate the p -value.  Because the alternative hypothesis is a [latex]\gt[/latex], the p -value is the area in the right-tail of the distribution.

To use the t.dist.rt function, we need to calculate out the [latex]t[/latex]-score:

[latex]\begin{eqnarray*} t & = & \frac{\overline{x}-\mu}{\frac{s}{\sqrt{n}}} \\ & = & \frac{67-65}{\frac{3.1972...}{\sqrt{10}}} \\ & = & 1.9781... \end{eqnarray*}[/latex]

The degrees of freedom for the [latex]t[/latex]-distribution is [latex]n-1=10-1=9[/latex].

So the p -value[latex]=0.0396[/latex].

Conclusion:

Because p -value[latex]=0.0396 \gt 0.01=\alpha[/latex], we do not reject the null hypothesis.  At the 1% significance level there is not enough evidence to suggest that mean score on the test is greater than 65.

• The null hypothesis [latex]\mu=65[/latex] is the claim that the mean test score is 65.
• The alternative hypothesis [latex]\mu \gt 65[/latex] is the claim that the mean test score is greater than 65.
• Keep all of the decimals throughout the calculation (i.e. in the sample standard deviation, the [latex]t[/latex]-score, etc.) to avoid any round-off error in the calculation of the p -value.  This ensures that we get the most accurate value for the p -value.
• The p -value is the area in the right-tail of the [latex]t[/latex]-distribution, to the right of [latex]t=1.9781...[/latex].
• The p -value of 0.0396 tells us that under the assumption that the mean test score is 65 (the null hypothesis), there is a 3.96% chance that the mean test score is 65 or more.  Compared to the 1% significance level, this is a large probability, and so is likely to happen assuming the null hypothesis is true.  This suggests that the assumption that the null hypothesis is true is most likely correct, and so the conclusion of the test is to not reject the null hypothesis.

A company claims that the average change in the value of their stock is $3.50 per week.  An investor believes this average is too high. The investor records the changes in the company’s stock price over 30 weeks and finds the average change in the stock price is $2.60 with a standard deviation of $1.80.  At the 5% significance level, is the average change in the company’s stock price lower than the company claims?

[latex]\begin{eqnarray*} H_0: & & \mu=$3.50  \\ H_a: & & \mu \lt $3.50  \end{eqnarray*}[/latex]

From the question, we have [latex]n=30[/latex], [latex]\overline{x}=2.6[/latex], [latex]s=1.8[/latex] and [latex]\alpha=0.05[/latex].

This is a test on a population mean where the population standard deviation is unknown (we only know the sample standard deviation [latex]s=1.8.[/latex]).  So we use a [latex]t[/latex]-distribution to calculate the p -value.  Because the alternative hypothesis is a [latex]\lt[/latex], the p -value is the area in the left-tail of the distribution.

To use the t.dist function, we need to calculate out the [latex]t[/latex]-score:

[latex]\begin{eqnarray*} t & = & \frac{\overline{x}-\mu}{\frac{s}{\sqrt{n}}} \\ & = & \frac{2.6-3.5}{\frac{1.8}{\sqrt{30}}} \\ & = & -1.5699... \end{eqnarray*}[/latex]

The degrees of freedom for the [latex]t[/latex]-distribution is [latex]n-1=30-1=29[/latex].

So the p -value[latex]=0.0636[/latex].

Because p -value[latex]=0.0636 \gt 0.05=\alpha[/latex], we do not reject the null hypothesis.  At the 5% significance level there is not enough evidence to suggest that average change in the stock price is lower than $3.50.

• The null hypothesis [latex]\mu=$3.50[/latex] is the claim that the average change in the company’s stock is $3.50 per week.
• The alternative hypothesis [latex]\mu \lt $3.50[/latex] is the claim that the average change in the company’s stock is less than $3.50 per week.
• The p -value is the area in the left-tail of the [latex]t[/latex]-distribution, to the left of [latex]t=-1.5699...[/latex].
• The p -value of 0.0636 tells us that under the assumption that the average change in the stock is $3.50 (the null hypothesis), there is a 6.36% chance that the average change is $3.50 or less.  Compared to the 5% significance level, this is a large probability, and so is likely to happen assuming the null hypothesis is true.  This suggests that the assumption that the null hypothesis is true is most likely correct, and so the conclusion of the test is to not reject the null hypothesis.  In other words, the company’s claim that the average change in their stock price is $3.50 per week is most likely correct.

A paint manufacturer has their production line set-up so that the average volume of paint in a can is 3.78 liters.  The quality control manager at the plant believes that something has happened with the production and the average volume of paint in the cans has changed.  The quality control department takes a sample of 100 cans and finds the average volume is 3.62 liters with a standard deviation of 0.7 liters.  At the 5% significance level, has the volume of paint in a can changed?

[latex]\begin{eqnarray*} H_0: & & \mu=3.78 \mbox{ liters}  \\ H_a: & & \mu \neq 3.78 \mbox{ liters}  \end{eqnarray*}[/latex]

From the question, we have [latex]n=100[/latex], [latex]\overline{x}=3.62[/latex], [latex]s=0.7[/latex] and [latex]\alpha=0.05[/latex].

This is a test on a population mean where the population standard deviation is unknown (we only know the sample standard deviation [latex]s=0.7[/latex]).  So we use a [latex]t[/latex]-distribution to calculate the p -value.  Because the alternative hypothesis is a [latex]\neq[/latex], the p -value is the sum of area in the tails of the distribution.

To use the t.dist.2t function, we need to calculate out the [latex]t[/latex]-score:

[latex]\begin{eqnarray*} t & = & \frac{\overline{x}-\mu}{\frac{s}{\sqrt{n}}} \\ & = & \frac{3.62-3.78}{\frac{0.07}{\sqrt{100}}} \\ & = & -2.2857... \end{eqnarray*}[/latex]

The degrees of freedom for the [latex]t[/latex]-distribution is [latex]n-1=100-1=99[/latex].

So the p -value[latex]=0.0244[/latex].

Because p -value[latex]=0.0244 \lt 0.05=\alpha[/latex], we reject the null hypothesis in favour of the alternative hypothesis.  At the 5% significance level there is enough evidence to suggest that average volume of paint in the cans has changed.

• The null hypothesis [latex]\mu=3.78[/latex] is the claim that the average volume of paint in the cans is 3.78.
• The alternative hypothesis [latex]\mu \neq 3.78[/latex] is the claim that the average volume of paint in the cans is not 3.78.
• Keep all of the decimals throughout the calculation (i.e. in the [latex]t[/latex]-score) to avoid any round-off error in the calculation of the p -value.  This ensures that we get the most accurate value for the p -value.
• The p -value is the sum of the area in the two tails.  The output from the t.dist.2t function is exactly the sum of the area in the two tails, and so is the p -value required for the test.  No additional calculations are required.
• The t.dist.2t function requires that the value entered for the [latex]t[/latex]-score is positive .  A negative [latex]t[/latex]-score entered into the t.dist.2t function generates an error in Excel.  In this case, the value of the [latex]t[/latex]-score is negative, so we must enter the absolute value of this [latex]t[/latex]-score into field 1.
• The p -value of 0.0244 is a small probability compared to the significance level, and so is unlikely to happen assuming the null hypothesis is true.  This suggests that the assumption that the null hypothesis is true is most likely incorrect, and so the conclusion of the test is to reject the null hypothesis in favour of the alternative hypothesis.  In other words, the average volume of paint in the cans has most likely changed from 3.78 liters.

Watch this video: Hypothesis Testing: t -test, right tail by ExcelIsFun [11:02]

Watch this video: Hypothesis Testing: t -test, left tail by ExcelIsFun [7:48]

Watch this video: Hypothesis Testing: t -test, two tail by ExcelIsFun [8:54]

Concept Review

The hypothesis test for a population mean is a well established process:

• Collect the sample information for the test and identify the significance level.
• When the population standard deviation is unknown, find the p -value (the area in the corresponding tail) for the test using the [latex]t[/latex]-distribution with [latex]\displaystyle{t=\frac{\overline{x}-\mu}{\frac{s}{\sqrt{n}}}}[/latex] and [latex]df=n-1[/latex].
• Compare the p -value to the significance level and state the outcome of the test.

Attribution

“ 9.6   Hypothesis Testing of a Single Mean and Single Proportion “ in Introductory Statistics by OpenStax  is licensed under a  Creative Commons Attribution 4.0 International License.

Introduction to Statistics Copyright © 2022 by Valerie Watts is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License , except where otherwise noted.

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• [email protected]

Hypothesis testing for the mean Calculator

State the null and alternative hypothesis:

Calculate our test statistic z:.

Determine rejection region:

What is the answer, how does the hypothesis testing for the mean calculator work, what 1 formula is used for the hypothesis testing for the mean calculator, what 7 concepts are covered in the hypothesis testing for the mean calculator.

• alternative hypothesis
• hypothesis testing for the mean
• null hypothesis
• test statistic

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Section 10.4: Hypothesis Tests for a Population Standard Deviation

• 10.1 The Language of Hypothesis Testing
• 10.2 Hypothesis Tests for a Population Proportion
• 10.3 Hypothesis Tests for a Population Mean
• 10.4 Hypothesis Tests for a Population Standard Deviation
• 10.5 Putting It Together: Which Method Do I Use?

By the end of this lesson, you will be able to...

• test hypotheses about a population standard deviation

For a quick overview of this section, watch this short video summary:

Before we begin this section, we need a quick refresher of the Χ 2 distribution.

The Chi-Square ( Χ 2 ) distribution

Reminder: "chi-square" is pronounced "kai" as in sky, not "chai" like the tea .

If a random sample size n is obtained from a normally distributed population with mean μ and standard deviation σ , then

has a chi-square distribution with n-1 degrees of freedom.

Properties of the Χ 2 distribution

• It is not symmetric.
• The shape depends on the degrees of freedom.
• As the number of degrees of freedom increases, the distribution becomes more symmetric.
• Χ 2 ≥0

Finding Probabilities Using StatCrunch

We again have some conditions that need to be true in order to perform the test 

• the sample was randomly selected, and
• the population from which the sample is drawn is normally distributed

Note that in the second requirement, the population must be normally distributed. The steps in performing the hypothesis test should be familiar by now.

Performing a Hypothesis Test Regarding σ

Step 1 : State the null and alternative hypotheses.

Step 2 : Decide on a level of significance, α .

Step 4 : Determine the P -value.

Step 5 : Reject the null hypothesis if the P -value is less than the level of significance, α.

Step 6 : State the conclusion.

In Example 2 , in Section 10.2, we assumed that the standard deviation for the resting heart rates of ECC students was 12 bpm. Later, in Example 2 in Section 10.3, we considered the actual sample data below.

( Click here to view the data in a format more easily copied.)

Based on this sample, is there enough evidence to say that the standard deviation of the resting heart rates for students in this class is different from 12 bpm?

Note: Be sure to check that the conditions for performing the hypothesis test are met.

[ reveal answer ]

From the earlier examples, we know that the resting heart rates could come from a normally distributed population and there are no outliers.

Step 1 : H 0 : σ = 12 H 1 : σ ≠ 12

Step 2 : α = 0.05

Step 4 : P -value = 2P( Χ 2 > 15.89) ≈ 0.2159

Step 5 : Since P -value > α , we do not reject H 0 .

Step 6 : There is not enough evidence at the 5% level of significance to support the claim that the standard deviation of the resting heart rates for students in this class is different from 12 bpm.

Hypothesis Testing Regarding σ Using StatCrunch

Let's look at Example 1 again, and try the hypothesis test with technology.

Using DDXL:

Using StatCrunch:

<< previous section | next section >>

Hypothesis Tests for One or Two Variances or Standard Deviations

Chi-Square-tests and F-tests for variance or standard deviation both require that the original population be normally distributed.

Testing a Claim about a Variance or Standard Deviation

To test a claim about the value of the variance or the standard deviation of a population, then the test statistic will follow a chi-square distribution with $n-1$ dgrees of freedom, and is given by the following formula.

The television habits of 30 children were observed. The sample mean was found to be 48.2 hours per week, with a standard deviation of 12.4 hours per week. Test the claim that the standard deviation was at least 16 hours per week.

• The hypotheses are: $H_0: \sigma = 16$ $H_a: \sigma < 16$
• We shall choose   $\alpha = 0.05$.
• The test statistic is   $\chi^2 = \dfrac{(n-1)s^2}{\sigma_0^2} = \dfrac{(30-1)12.4^2}{16^2} = 17.418$.
• The p-value is   $p = \chi^2\text{cdf}(0,17.418,29) = 0.0447$.
• The variation in television watching was less than 16 hours per week.

Testing a the Difference of Two Variances or Two Standard Deviations

Two equal variances would satisfy the equation   $\sigma_1^2 = \sigma_2^2$,   which is equivalent to   $\dfrac{ \sigma_1^2}{\sigma_2^2} = 1$.   Since sample variances are related to chi-square distributions, and the ratio of chi-square distributions is an F-distribution, we can use the F-distribution to test against a null hypothesis of equal variances. Note that this approach does not allow us to test for a particular magnitude of difference between variances or standard deviations.

Given sample sizes of $n_1$ and $n_2$, the test statistic will have   $n_1-1$   and   $n_2-1$   degrees of freedom, and is given by the following formula.

If the larger variance (or standard deviation) is present in the first sample, then the test is right-tailed. Otherwise, the test is left-tailed. Most tables of the F-distribution assume right-tailed tests, but that requirement may not be necessary when using technology.

Samples from two makers of ball bearings are collected, and their diameters (in inches) are measured, with the following results:

• Acme: $n_1 = 80$, $s_1 = 0.0395$
• Bigelow: $n_2 = 120$, $s_2 = 0.0428$
• The hypotheses are: $H_0: \sigma_1 = \sigma_2$ $H_a: \sigma_1 \neq \sigma_2$
• The test statistic is   $F = \dfrac{s_1^2}{s_2^2} = \dfrac{0.0395^2}{0.0428^2} = 0.8517$.
• Since the first sample had the smaller standard deviation, this is a left-tailed test. The p-value is   $p = \operatorname{Fcdf}(0,0.8517,79,119) = 0.2232$.
• There is insufficient evidence to conclude that the diameters of the ball bearings in the two companies have different standard deviations.

If the two samples had been reversed in our computations, we would have obtained the test statistic   $F = 1.1741$,   and performing a right-tailed test, found the p-value   $p = \operatorname{Fcdf}(1.1741,\infty,119,79) = 0.2232$.   Of course, the answer is the same.

T-test for two Means – Unknown Population Standard Deviations

Instructions : Use this T-Test Calculator for two Independent Means calculator to conduct a t-test for two population means (\(\mu_1\) and \(\mu_2\)), with unknown population standard deviations. This test apply when you have two-independent samples, and the population standard deviations \(\sigma_1\) and \(\sigma_2\) and not known. Please select the null and alternative hypotheses, type the significance level, the sample means, the sample standard deviations, the sample sizes, and the results of the t-test for two independent samples will be displayed for you:

The T-test for Two Independent Samples

More about the t-test for two means so you can better interpret the output presented above: A t-test for two means with unknown population variances and two independent samples is a hypothesis test that attempts to make a claim about the population means (\(\mu_1\) and \(\mu_2\)).

More specifically, a t-test uses sample information to assess how plausible it is for the population means \(\mu_1\) and \(\mu_2\) to be equal. The test has two non-overlapping hypotheses, the null and the alternative hypothesis.

The null hypothesis is a statement about the population means, specifically the assumption of no effect, and the alternative hypothesis is the complementary hypothesis to the null hypothesis.

Properties of the two sample t-test

The main properties of a two sample t-test for two population means are:

• Depending on our knowledge about the "no effect" situation, the t-test can be two-tailed, left-tailed or right-tailed
• The main principle of hypothesis testing is that the null hypothesis is rejected if the test statistic obtained is sufficiently unlikely under the assumption that the null hypothesis is true
• The p-value is the probability of obtaining sample results as extreme or more extreme than the sample results obtained, under the assumption that the null hypothesis is true
• In a hypothesis tests there are two types of errors. Type I error occurs when we reject a true null hypothesis, and the Type II error occurs when we fail to reject a false null hypothesis

How do you compute the t-statistic for the t test for two independent samples?

The formula for a t-statistic for two population means (with two independent samples), with unknown population variances shows us how to calculate t-test with mean and standard deviation and it depends on whether the population variances are assumed to be equal or not. If the population variances are assumed to be unequal, then the formula is:

On the other hand, if the population variances are assumed to be equal, then the formula is:

Normally, the way of knowing whether the population variances must be assumed to be equal or unequal is by using an F-test for equality of variances.

With the above t-statistic, we can compute the corresponding p-value, which allows us to assess whether or not there is a statistically significant difference between two means.

Why is it called t-test for independent samples?

This is because the samples are not related with each other, in a way that the outcomes from one sample are unrelated from the other sample. If the samples are related (for example, you are comparing the answers of husbands and wives, or identical twins), you should use a t-test for paired samples instead .

What if the population standard deviations are known?

The main purpose of this calculator is for comparing two population mean when sigma is unknown for both populations. In case that the population standard deviations are known, then you should use instead this z-test for two means .

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Hypothesis Test for Population Standard Deviation for normal population

Difference in Means Hypothesis Test Calculator

Use the calculator below to analyze the results of a difference in sample means hypothesis test. Enter your sample means, sample standard deviations, sample sizes, hypothesized difference in means, test type, and significance level to calculate your results.

You will find a description of how to conduct a two sample t-test below the calculator.

Define the Two Sample t-test

The Difference Between the Sample Means Under the Null Distribution

Conducting a hypothesis test for the difference in means.

When two populations are related, you can compare them by analyzing the difference between their means.

A hypothesis test for the difference in samples means can help you make inferences about the relationships between two population means.

Testing for a Difference in Means

For the results of a hypothesis test to be valid, you should follow these steps:

Check Your Conditions

State your hypothesis, determine your analysis plan, analyze your sample, interpret your results.

To use the testing procedure described below, you should check the following conditions:

• Independence of Samples - Your samples should be collected independently of one another.
• Simple Random Sampling - You should collect your samples with simple random sampling. This type of sampling requires that every occurrence of a value in a population has an equal chance of being selected when taking a sample.
• Normality of Sample Distributions - The sampling distributions for both samples should follow the Normal or a nearly Normal distribution. A sampling distribution will be nearly Normal when the samples are collected independently and when the population distribution is nearly Normal. Generally, the larger the sample size, the more normally distributed the sampling distribution. Additionally, outlier data points can make a distribution less Normal, so if your data contains many outliers, exercise caution when verifying this condition.

You must state a null hypothesis and an alternative hypothesis to conduct an hypothesis test of the difference in means.

The null hypothesis is a skeptical claim that you would like to test.

The alternative hypothesis represents the alternative claim to the null hypothesis.

Your null hypothesis and alternative hypothesis should be stated in one of three mutually exclusive ways listed in the table below.

D is the hypothesized difference between the populations' means that you would like to test.

Before conducting a hypothesis test, you must determine a reasonable significance level, α, or the probability of rejecting the null hypothesis assuming it is true. The lower your significance level, the more confident you can be of the conclusion of your hypothesis test. Common significance levels are 10%, 5%, and 1%.

To evaluate your hypothesis test at the significance level that you set, consider if you are conducting a one or two tail test:

• Two-tail tests divide the rejection region, or critical region, evenly above and below the null distribution, i.e. to the tails of the null sampling distribution. For example, in a two-tail test with a 5% significance level, your rejection region would be the upper and lower 2.5% of the null distribution. An alternative hypothesis of μ 1 - μ 2 ≠ D requires a two tail test.
• One-tail tests place the rejection region entirely on one side of the distribution i.e. to the right or left tail of the null distribution. For example, in a one-tail test evaluating if the actual difference in means, D, is above the null distribution with a 5% significance level, your rejection region would be the upper 5% of the null distribution. μ 1 - μ 2 > D and μ 1 - μ 2 < D alternative hypotheses require one-tail tests.

The graphical results section of the calculator above shades rejection regions blue.

After checking your conditions, stating your hypothesis, determining your significance level, and collecting your sample, you are ready to analyze your hypothesis.

Sample means follow the Normal distribution with the following parameters:

• The Difference in the Population Means, D - The true difference in the population means is unknown, but we use the hypothesized difference in the means, D, from the null hypothesis in the calculations.
• The Standard Error, SE - The standard error of the difference in the sample means can be computed as follows:      SE = (s 1 2 /n 1 + s 2 2 /n 2 ) (1/2) with s 1 being the standard deviation of sample one, n 1 being the sample size of sample one, s 2 being the standard deviation of sample one, and n 2 being the sample size of sample two. The standard error defines how differences in sample means are expected to vary around the null difference in means sampling distribution given the sample sizes and under the assumption that the null hypothesis is true.
• The Degrees of Freedom, DF - The degrees of freedom calculation can be estimated as the smaller of n 1 - 1 or n 2 - 1. For more accurate results, use the following formula for the degrees of freedom (DF):      DF = (s 1 2 /n 1 + s 2 2 /n 2 ) 2 / ((s 1 2 /n 1 ) 2 / (n 1 - 1) + (s 2 2 /n 2 ) 2 / (n 2 - 1))

In a difference in means hypothesis test, we calculate the probability that we would observe the difference in sample means (x̄ 1 - x̄ 2 ), assuming the null hypothesis is true, also known as the p-value . If the p-value is less than the significance level, then we can reject the null hypothesis.

You can determine a precise p-value using the calculator above, but we can find an estimate of the p-value manually by calculating the t-score, or t-statistic, as follows: t = (x̄ 1 - x̄ 2 - D) / SE

The t-score is a test statistic that tells you how far our observation is from the null hypothesis's difference in means under the null distribution. Using any t-score table, you can look up the probability of observing the results under the null distribution. You will need to look up the t-score for the type of test you are conducting, i.e. one or two tail. A hypothesis test for the difference in means is sometimes known as a two sample mean t-test because of the use of a t-score in analyzing results.

The conclusion of a hypothesis test for the difference in means is always either:

• Reject the null hypothesis
• Do not reject the null hypothesis

If you reject the null hypothesis, you cannot say that your sample difference in means is the true difference between the means. If you do not reject the null hypothesis, you cannot say that the hypothesized difference in means is true.

A hypothesis test is simply a way to look at evidence and conclude if it provides sufficient evidence to reject the null hypothesis.

Example: Hypothesis Test for the Difference in Two Means

Let’s say you are a manager at a company that designs batteries for smartphones. One of your engineers believes that she has developed a battery that will last more than two hours longer than your standard battery.

Before you can consider if you should replace your standard battery with the new one, you need to test the engineer’s claim. So, you decided to run a difference in means hypothesis test to see if her claim that the new battery will last two hours longer than the standard one is reasonable.

You direct your team to run a study. They will take a sample of 100 of the new batteries and compare their performance to 1,000 of the old standard batteries.

• Check the conditions - Your test consists of independent samples . Your team collects your samples using simple random sampling , and you have reason to believe that all your batteries' performances are always close to normally distributed . So, the conditions are met to conduct a two sample t-test.
• State Your Hypothesis - Your null hypothesis is that the charge of the new battery lasts at most two hours longer than your standard battery (i.e. μ 1 - μ 2 ≤ 2). Your alternative hypothesis is that the new battery lasts more than two hours longer than the standard battery (i.e. μ 1 - μ 2 > 2).
• Determine Your Analysis Plan - You believe that a 1% significance level is reasonable. As your test is a one-tail test, you will evaluate if the difference in mean charge between the samples would occur at the upper 1% of the null distribution.
• Analyze Your Sample - After collecting your samples (which you do after steps 1-3), you find the new battery sample had a mean charge of 10.4 hours, x̄ 1 , with a 0.8 hour standard deviation, s 1 . Your standard battery sample had a mean charge of 8.2 hours, x̄ 2 , with a standard deviation of 0.2 hours, s 2 . Using the calculator above, you find that a difference in sample means of 2.2 hours [2 = 10.4 – 8.2] would results in a t-score of 2.49 under the null distribution, which translates to a p-value of 0.72%.
• Interpret Your Results - Since your p-value of 0.72% is less than the significance level of 1%, you have sufficient evidence to reject the null hypothesis.

In this example, you found that you can reject your null hypothesis that the new battery design does not result in more than 2 hours of extra battery life. The test does not guarantee that your engineer’s new battery lasts two hours longer than your standard battery, but it does give you strong reason to believe her claim.

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Standard deviation is found in statistics which represents the scattering of data about its mean. The measuring of standard deviation also tells the value of dispersion or variation of a given data set around its mean. Taking the square root of the variance of the given data to evaluate the Standard deviation.

The standard deviation is commonly used in various statistical analyses such as hypothesis testing and confidence interval calculations. Also used in the field of finance to measure the profit or volatility of a stock or investment forum.

In this article, we will discuss the definition of standard deviation, steps to find the standard deviation, its application in different fields, and understanding of the concept of the solution of the standard deviation solve example with detailed steps.

What is Standard deviation?

Standard deviation is statistically measured that shows the dispersion and variation amount in the given data value. It measured how the data spread around the mean. The given data points are shown to be more closely separated around the mean if the value is small and are more widely spread if the standard deviation value is very high. It can be represented by the letter “ S.D ”.

The decimal/numerical value of the standard deviation is evaluated by taking the square root of the variance. The variance can be measured by the average of the squared deviations of each data point from the mean. For this use the specific formula of the standard deviation.

Formula of Standard deviation

The mathematical formula of standard deviation is different according to the size of the data set or the selection of the data points from the randomly distributed data. The mathematical formula for sample/population is stated as:

For sample data :

S = S.D= √ [∑ (x j – x̄) 2 / (N-1)]

For population data :

σ = S.D= √ [∑ (x j – x̄) 2 / (N)]

• σ = standard deviation for population data,
• S = standard deviation for sample data. 
• Σ = sum of the values for all “j”,
• x j = “j th ” value in the given data,
• N = total number,
• x̄ = mean value of the dataset.

Steps to find the Standard Deviation

Follow the below steps to evaluate the standard deviation value for any sample/population data set: 

Step 1: Evaluate the Mean: First, find the mean (average) of the data set by summing up all the data points and dividing the total by the number of data points.

Step 2: Find the Deviations: Subtract the mean from each data point to find the deviation of the data set and square each deviation value. 

Step 3: Evaluate the Variance : For the variance of the dataset now find the average of the squared deviations by summing them up and dividing by the one less number of data points for the sample data set (while for the population dividing it by the number of population). 

Step 4: Find Standard Deviation: To find the standard deviation value take the square root of the variance value. 

For a better understanding of the above steps follow the below example.

Example section:

In this section, we will solve the example that helps to understand the method to find standard deviation with the formula.

Example: If the statistical data is 12, 5, 4, 10, and 15 then evaluate the standard deviation of the given sample data set.

Step 1: Write the above data in the set form carefully.

Dataset = {5, 4, 10, 12, 15}, N = 5, S.D =?

To find the standard deviation first find the variance of the given data. For variance, find the mean and square of deviation.

Step 2: Now, Calculate the mean of the given data points by dividing the sum of data points by the number of data set elements.

x̄ = (∑ x j )/N

x̄ = ∑ (5 + 4 + 10 + 12 +15) /5

Step 3:  Now evaluate the squared differences from the mean to evaluate the variance of the data set.

 (5 – 9.2) 2 = (-4.2) 2 = 17.64

(4 – 9.2) 2 = (-5.2) 2 = 27.04

(10 – 9.2) 2 = (0.8) 2 = 0.64

(12 – 9.2) 2 = (2.8) 2 = 7.64

(15 – 9.2) 2 = (5.8) 2 = 33.64

Step 4: Now, Find the Variance by taking the sum of all the above values and dividing by the difference of the total number by one.

V = Variance = (17.64 + 27.04 + 0.64 + 7.64 + 33.64) / (5 – 1) 

          = (86.8) / 4

      V = 21.7

Step 5: Now, take the square root of the variance to find the value of the “S.D.”.

S = √21.7 = 4.66

To overcome this long calculation process, use the Standard Deviation Calculator to make your calculation faster and save your time. This tool provides the solution of standard deviation for both data sets (sample/population) with detailed steps in a single click, quickly and accurately. 

Physical Interpretations of Standard Deviation

The physical interpretation of standard deviation depends on the context of the data being analyzed. 

• Measurement Precision : In experimental sciences, the standard deviation represents the measurement precision or the degree of uncertainty associated with the measurements. The small value of the standard deviation indicates the measurements are more precise and consistent and the larger standard deviation shows that there are chances of greater variability and potential errors in the measurements.
• Statistical Models : In Statistical models , Standard deviation plays a crucial role that helps assess the accuracy or goodness-of-fit of a model to the observed data. In regression analysis, the standard deviation of the residuals (the differences between observed and predicted values) provides insights into the variability of the model’s predictions.

Application of Standard Deviation:

• Finance: It is commonly used to measure the risk of a particular stock or portfolio. It helps the investors to understand the potential risk and percentage of return of their investments.
• Quality control: In a manufacturing company, the standard deviation is used to measure the consistency/quality of a product. A lower standard deviation determines that the product is good and on the other hand a higher value of standard deviation shows that the product is not suitable for use.
• Statistical research: Standard deviation is used to measure the dispersion or variation of data in experimental research . It helps researchers to understand whether finding results are statistically significant or not.

Related Blog: Top 5 Calculators Every College Student Needs for Academic Success

In this article, we explained the basic definition and formula of standard deviation for sample and population data sets. Also discussed the applications of standard deviation in different fields of science and statistics. For a better understanding of the calculation of standard deviation, I solved a detailed example using the variance with the detailed steps. 

The example helps to understand the calculation method and find its value easily. I hope by the proofreading of this article everyone can solve the related problem easily.

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