hypothesis test for standard deviation

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 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.

Hypothesis Test for Population Standard Deviation for normal population

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

8.6 Hypothesis Tests for a Population Mean with Known Population Standard Deviation

Learning objectives.

• Conduct and interpret hypothesis tests for a population mean with known 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.

Suppose the hypotheses for a hypothesis test are:

[latex]\begin{eqnarray*} H_0: & & \mu=5 \\ H_a: & & \mu \lt 5 \end{eqnarray*}[/latex]

Because the alternative hypothesis is a [latex]\lt[/latex], this is a left-tailed test.  The p -value is the area in the left-tail of the distribution.

[latex]\begin{eqnarray*} H_0: & & \mu=0.5 \\ H_a: & & \mu \neq 0.5  \end{eqnarray*}[/latex]

Because the alternative hypothesis is a [latex]\neq[/latex], this is a two-tailed test.  The p -value is the sum of the areas in the two tails of the distribution.  Each tail contains exactly half of the p -value.

[latex]\begin{eqnarray*} H_0: & & \mu=10 \\ H_a: & & \mu \lt 10  \end{eqnarray*}[/latex]

Steps to Conduct a Hypothesis Test for a Population Mean with Known 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].
• When the population standard deviation is known , we use a normal distribution with [latex]\displaystyle{z=\frac{\overline{x}-\mu}{\frac{\sigma}{\sqrt{n}}}}[/latex] to find the p -value.  The p -value is the area in the corresponding tail of the normal distribution.
• 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 KNOWN 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 known, use the normal distribution to find the p -value.

The p -value is the area in the tail(s) of a normal distribution, so the norm.dist(x,[latex]\mu[/latex],[latex]\sigma[/latex],logic operator) function can be used to calculate the p -value.

• For x , enter the value for [latex]\overline{x}[/latex].
• For [latex]\mu[/latex] , enter the mean of the sample means [latex]\mu[/latex].  Note:  Because the test is run assuming the null hypothesis is true, the value for [latex]\mu[/latex] is the claim from the null hypothesis.
• For [latex]\sigma[/latex] , enter the standard error of the mean [latex]\displaystyle{\frac{\sigma}{\sqrt{n}}}[/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.

Use the appropriate technique with the norm.dist function to find the area in the left-tail or the area in the right-tail.

Jeffrey, as an eight-year old, established a mean time of 16.43 seconds with a standard deviation of 0.8 seconds for swimming the 25-meter freestyle.  His dad, Frank, thought that Jeffrey could swim the 25-meter freestyle faster using goggles.  Frank bought Jeffrey a new pair of goggles and timed Jeffrey swimming the 25-meter freestyle 15 different times.  In the sample of 15 swims, Jeffrey’s mean time was 16 seconds.  Frank thought that the goggles helped Jeffrey swim faster than 16.43 seconds.  At the 5% significance level, did Jeffrey swim faster wearing the goggles?  Assume that the swim times for the 25-meter freestyle are normally distributed.

Hypotheses:

[latex]\begin{eqnarray*} H_0: & & \mu=16.43 \mbox{ seconds} \\ H_a: & & \mu \lt 16.43 \mbox{ seconds} \end{eqnarray*}[/latex]

From the question, we have [latex]n=15[/latex], [latex]\overline{x}=16[/latex], [latex]\sigma=0.8[/latex] and [latex]\alpha=0.05[/latex].

This is a test on a population mean where the population standard deviation is known ([latex]\sigma=0.8[/latex]).  So we use a normal 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.

So the p -value[latex]=0.0187[/latex].

Conclusion:

Because p -value[latex]=0.0187 \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 Jeffrey’s mean swim time with the goggles is less than 16.43 seconds.

• The null hypothesis [latex]\mu=16.43[/latex] is the claim that Jeffrey’s mean swim time with the goggles is 16.43 seconds (the same as it is without the googles).
• The alternative hypothesis [latex]\mu \lt 16.43[/latex] is the claim that Jeffrey’s swim time with the goggles is less than 16.43 seconds.
• The function is norm.dist because we are finding the area in the left tail of a normal distribution.
• Field 1 is the value of [latex]\overline{x}[/latex]
• Field 2 is the value of [latex]\mu[/latex] from the null hypothesis.  Remember, we run the test assuming the null hypothesis is true, so that means we assume [latex]\mu=16.43[/latex].
• Field 3 is the standard deviation for the sample means [latex]\displaystyle{\frac{\sigma}{\sqrt{n}}}[/latex].  Note that we are not using the standard deviation from the population ([latex]\sigma=0.8[/latex]).  This is because the p -value is the area under the curve of the distribution of the sample means, not the distribution of the population.
• The p -value of 0.0187 tells us that under the assumption that Jeffrey’s mean swim time with goggles is 16.43 seconds (the null hypothesis), there is only a 1.87% chance that the mean time for the 15 sample swims is 16 seconds or less.  This is a small probability, 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.
• The Type I error for this problem is to conclude that Jeffrey swims the 25-meter freestyle, on average, in less than 16.43 seconds (the alternative hypothesis) when, in fact, he actually swims the 25-meter freestyle, on average, in 16.43 seconds (the null hypothesis).  That is, reject the null hypothesis when the null hypothesis is actually true.
• The Type II error for this problem is to conclude that Jeffrey swims the 25-meter freestyle, on average, in 16.43 seconds (the null hypothesis) when, in fact, he actually swims the 25-meter freestyle, on average, in less than 16.43 seconds (the alternative hypothesis).  That is, do not reject the null hypothesis when the null hypothesis is actually false.

The mean throwing distance of a football for Marco, a high school freshman quarterback, is 40 yards with a standard deviation of 2 yards.  The team coach tells Marco to adjust his grip to get more distance.  The coach records the distances for 20 throws with the new grip.  For the 20 throws, Marco’s mean distance was 41.5 yards.  The coach thought the different grip helped Marco throw farther than 40 yards.  At the 5% significance level, is Marco’s mean throwing distance higher with the new grip?  Assume the throw distances for footballs are normally distributed.

[latex]\begin{eqnarray*} H_0: & & \mu=40 \mbox{ yards} \\ H_a: & & \mu \gt 40 \mbox{ yards} \end{eqnarray*}[/latex]

From the question, we have [latex]n=20[/latex], [latex]\overline{x}=41.5[/latex], [latex]\sigma=2[/latex] and [latex]\alpha=0.05[/latex].

This is a test on a population mean where the population standard deviation is known ([latex]\sigma=2[/latex]).  So we use a normal 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.

So the p -value[latex]=0.0004[/latex].

Because p -value[latex]=0.0004 \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 Marco’s mean throwing distance is greater than 40 yards with the new grip.

• The null hypothesis [latex]\mu=40[/latex] is the claim that Marco’s mean throwing distance with the new grip is 40 yards (the same as it is without the new grip).
• The alternative hypothesis [latex]\mu \gt 40[/latex] is the claim that Marco’s mean throwing distance with the new grip is greater than 40 yards.
• Field 2 is the value of [latex]\mu[/latex] from the null hypothesis.
• Field 3 is the standard deviation for the sample means [latex]\displaystyle{\frac{\sigma}{\sqrt{n}}}[/latex].
• The p -value of 0.0004 tells us that under the assumption that Marco’s mean throwing distance with the new grip is 40 yards, there is only a 0.047% chance that the mean throwing distance for the 20 sample throws is more than 40 yards.  This is a small probability, 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.

A local college states in its marketing materials that the average age of its first-year students is 18.3 years with a standard deviation of 3.4 years.  But this information is based on old data and does not take into account that more older adults are returning to college.  A researcher at the college believes that the average age of its first-year students has changed.  The researcher takes a sample of 50 first-year students and finds the average age is 19.5 years.  At the 1% significance level, has the average age of the college’s first-year students changed?

[latex]\begin{eqnarray*} H_0: & & \mu=18.3 \mbox{ years} \\ H_a: & & \mu \neq 18.3 \mbox{ years} \end{eqnarray*}[/latex]

From the question, we have [latex]n=50[/latex], [latex]\overline{x}=19.5[/latex], [latex]\sigma=3.4[/latex] and [latex]\alpha=0.01[/latex].

This is a test on a population mean where the population standard deviation is known ([latex]\sigma=3.4[/latex]).  In this case, the sample size is greater than 30.  So we use a normal 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.

Because there is only one sample, we only have information relating to one of the two tails, either the left tail or the right tail.  We need to know if the sample relates to the left tail or right tail because that will determine how we calculate out the area of that tail using the normal distribution.  In this case, the sample mean [latex]\overline{x}=19.5[/latex] is greater than the value of the population mean in the null hypothesis [latex]\mu=18.3[/latex] ([latex]\overline{x}=19.5>18.3=\mu[/latex]), so the sample information relates to the right-tail of the normal distribution.  This means that we will calculate out the area in the right tail using 1-norm.dist .  However, this is a two-tailed test where the p -value is the sum of the area in the two tails and the area in the right-tail is only one half of the p -value.  The area in the left tail equals the area in the right tail and the p -value is the sum of these two areas.

So the area in the right tail is 0.0063 and [latex]\frac{1}{2}[/latex]( p -value)[latex]=0.0063[/latex].  This is also the area in the left tail, so

p -value[latex]=0.0063+0.0063=0.0126[/latex]

Because p -value[latex]=0.0126 \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 the average age of the college’s first-year students has changed.

• The null hypothesis [latex]\mu=18.3[/latex] is the claim that the average age of the first-year students is still 18.3 years.
• The alternative hypothesis [latex]\mu \neq 18.3[/latex] is the claim that the average age of the first-year students has changed from 18.3 years.
• We use norm.dist([latex]\overline{x}[/latex],[latex]\mu[/latex],[latex]\sigma/\mbox{sqrt}(n)[/latex],true) to find the area in the left tail.  The area in the right tail equals the area in the left tail, so we can find the p -value by adding the output from this function to itself.
• We use 1-norm.dist([latex]\overline{x}[/latex],[latex]\mu[/latex],[latex]\sigma/\mbox{sqrt}(n)[/latex],true) to find the area in the right tail.  The area in the left tail equals the area in the right tail, so we can find the p -value by adding the output from this function to itself.
• The p -value of 0.0126  is a large probability compared to the 1% significance level, 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 claim that the average age of first-year students is 18.3 years is most likely correct.

Watch this video: Hypothesis Testing: z -test, right tail by ExcelIsFun [33:47]

Watch this video: Hypothesis Testing: z -test, left tail by ExcelIsFun [10:57]

Watch this video: Hypothesis Testing: z -test, two tail by ExcelIsFun [9:56]

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 known, find the p -value (the area in the corresponding tail) for the test using the normal distribution.
• 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.

• Skip to secondary menu
• Skip to main content
• Skip to primary sidebar

Statistics By Jim

Making statistics intuitive

Z Test: Uses, Formula & Examples

By Jim Frost Leave a Comment

What is a Z Test?

Use a Z test when you need to compare group means. Use the 1-sample analysis to determine whether a population mean is different from a hypothesized value. Or use the 2-sample version to determine whether two population means differ.

A Z test is a form of inferential statistics . It uses samples to draw conclusions about populations.

For example, use Z tests to assess the following:

• One sample : Do students in an honors program have an average IQ score different than a hypothesized value of 100?
• Two sample : Do two IQ boosting programs have different mean scores?

In this post, learn about when to use a Z test vs T test. Then we’ll review the Z test’s hypotheses, assumptions, interpretation, and formula. Finally, we’ll use the formula in a worked example.

Related post : Difference between Descriptive and Inferential Statistics

Z test vs T test

Z tests and t tests are similar. They both assess the means of one or two groups, have similar assumptions, and allow you to draw the same conclusions about population means.

However, there is one critical difference.

Z tests require you to know the population standard deviation, while t tests use a sample estimate of the standard deviation. Learn more about Population Parameters vs. Sample Statistics .

In practice, analysts rarely use Z tests because it’s rare that they’ll know the population standard deviation. It’s even rarer that they’ll know it and yet need to assess an unknown population mean!

A Z test is often the first hypothesis test students learn because its results are easier to calculate by hand and it builds on the standard normal distribution that they probably already understand. Additionally, students don’t need to know about the degrees of freedom .

Z and T test results converge as the sample size approaches infinity. Indeed, for sample sizes greater than 30, the differences between the two analyses become small.

William Sealy Gosset developed the t test specifically to account for the additional uncertainty associated with smaller samples. Conversely, Z tests are too sensitive to mean differences in smaller samples and can produce statistically significant results incorrectly (i.e., false positives).

When to use a T Test vs Z Test

Let’s put a button on it.

When you know the population standard deviation, use a Z test.

When you have a sample estimate of the standard deviation, which will be the vast majority of the time, the best statistical practice is to use a t test regardless of the sample size.

However, the difference between the two analyses becomes trivial when the sample size exceeds 30.

Learn more about a T-Test Overview: How to Use & Examples and How T-Tests Work .

Z Test Hypotheses

This analysis uses sample data to evaluate hypotheses that refer to population means (µ). The hypotheses depend on whether you’re assessing one or two samples.

One-Sample Z Test Hypotheses

• Null hypothesis (H 0 ): The population mean equals a hypothesized value (µ = µ 0 ).
• Alternative hypothesis (H A ): The population mean DOES NOT equal a hypothesized value (µ ≠ µ 0 ).

When the p-value is less or equal to your significance level (e.g., 0.05), reject the null hypothesis. The difference between your sample mean and the hypothesized value is statistically significant. Your sample data support the notion that the population mean does not equal the hypothesized value.

Related posts : Null Hypothesis: Definition, Rejecting & Examples and Understanding Significance Levels

Two-Sample Z Test Hypotheses

• Null hypothesis (H 0 ): Two population means are equal (µ 1 = µ 2 ).
• Alternative hypothesis (H A ): Two population means are not equal (µ 1 ≠ µ 2 ).

Again, when the p-value is less than or equal to your significance level, reject the null hypothesis. The difference between the two means is statistically significant. Your sample data support the idea that the two population means are different.

These hypotheses are for two-sided analyses. You can use one-sided, directional hypotheses instead. Learn more in my post, One-Tailed and Two-Tailed Hypothesis Tests Explained .

Related posts : How to Interpret P Values and Statistical Significance

Z Test Assumptions

For reliable results, your data should satisfy the following assumptions:

You have a random sample

Drawing a random sample from your target population helps ensure that the sample represents the population. Representative samples are crucial for accurately inferring population properties. The Z test results won’t be valid if your data do not reflect the population.

Related posts : Random Sampling and Representative Samples

Continuous data

Z tests require continuous data . Continuous variables can assume any numeric value, and the scale can be divided meaningfully into smaller increments, such as fractional and decimal values. For example, weight, height, and temperature are continuous.

Other analyses can assess additional data types. For more information, read Comparing Hypothesis Tests for Continuous, Binary, and Count Data .

Your sample data follow a normal distribution, or you have a large sample size

All Z tests assume your data follow a normal distribution . However, due to the central limit theorem, you can ignore this assumption when your sample is large enough.

The following sample size guidelines indicate when normality becomes less of a concern:

• One-Sample : 20 or more observations.
• Two-Sample : At least 15 in each group.

Related posts : Central Limit Theorem and Skewed Distributions

Independent samples

For the two-sample analysis, the groups must contain different sets of items. This analysis compares two distinct samples.

Related post : Independent and Dependent Samples

Population standard deviation is known

As I mention in the Z test vs T test section, use a Z test when you know the population standard deviation. However, when n > 30, the difference between the analyses becomes trivial.

Related post : Standard Deviations

Z Test Formula

These Z test formulas allow you to calculate the test statistic. Use the Z statistic to determine statistical significance by comparing it to the appropriate critical values and use it to find p-values.

The correct formula depends on whether you’re performing a one- or two-sample analysis. Both formulas require sample means (x̅) and sample sizes (n) from your sample. Additionally, you specify the population standard deviation (σ) or variance (σ 2 ), which does not come from your sample.

I present a worked example using the Z test formula at the end of this post.

Learn more about Z-Scores and Test Statistics .

One Sample Z Test Formula

The one sample Z test formula is a ratio.

The numerator is the difference between your sample mean and a hypothesized value for the population mean (µ 0 ). This value is often a strawman argument that you hope to disprove.

The denominator is the standard error of the mean. It represents the uncertainty in how well the sample mean estimates the population mean.

Learn more about the Standard Error of the Mean .

Two Sample Z Test Formula

The two sample Z test formula is also a ratio.

The numerator is the difference between your two sample means.

The denominator calculates the pooled standard error of the mean by combining both samples. In this Z test formula, enter the population variances (σ 2 ) for each sample.

Z Test Critical Values

As I mentioned in the Z vs T test section, a Z test does not use degrees of freedom. It evaluates Z-scores in the context of the standard normal distribution. Unlike the t-distribution , the standard normal distribution doesn’t change shape as the sample size changes. Consequently, the critical values don’t change with the sample size.

To find the critical value for a Z test, you need to know the significance level and whether it is one- or two-tailed.

Learn more about Critical Values: Definition, Finding & Calculator .

Z Test Worked Example

Let’s close this post by calculating the results for a Z test by hand!

Suppose we randomly sampled subjects from an honors program. We want to determine whether their mean IQ score differs from the general population. The general population’s IQ scores are defined as having a mean of 100 and a standard deviation of 15.

We’ll determine whether the difference between our sample mean and the hypothesized population mean of 100 is statistically significant.

Specifically, we’ll use a two-tailed analysis with a significance level of 0.05. Looking at the table above, you’ll see that this Z test has critical values of ± 1.960. Our results are statistically significant if our Z statistic is below –1.960 or above +1.960.

The hypotheses are the following:

• Null (H 0 ): µ = 100
• Alternative (H A ): µ ≠ 100

Entering Our Results into the Formula

Here are the values from our study that we need to enter into the Z test formula:

• IQ score sample mean (x̅): 107
• Sample size (n): 25
• Hypothesized population mean (µ 0 ): 100
• Population standard deviation (σ): 15

The Z-score is 2.333. This value is greater than the critical value of 1.960, making the results statistically significant. Below is a graphical representation of our Z test results showing how the Z statistic falls within the critical region.

We can reject the null and conclude that the mean IQ score for the population of honors students does not equal 100. Based on the sample mean of 107, we know their mean IQ score is higher.

Now let’s find the p-value. We could use technology to do that, such as an online calculator. However, let’s go old school and use a Z table.

To find the p-value that corresponds to a Z-score from a two-tailed analysis, we need to find the negative value of our Z-score (even when it’s positive) and double it.

In the truncated Z-table below, I highlight the cell corresponding to a Z-score of -2.33.

The cell value of 0.00990 represents the area or probability to the left of the Z-score -2.33. We need to double it to include the area > +2.33 to obtain the p-value for a two-tailed analysis.

P-value = 0.00990 * 2 = 0.0198

That p-value is an approximation because it uses a Z-score of 2.33 rather than 2.333. Using an online calculator, the p-value for our Z test is a more precise 0.0196. This p-value is less than our significance level of 0.05, which reconfirms the statistically significant results.

See my full Z-table , which explains how to use it to solve other types of problems.

Share this:

Reader Interactions

Comments and questions cancel reply.

• Anatomy & Physiology
• Astrophysics
• Earth Science
• Environmental Science
• Organic Chemistry
• Precalculus
• Trigonometry
• English Grammar
• U.S. History
• World History

... and beyond

• Socratic Meta
• Featured Answers

• Hypothesis Testing for the Standard Deviation
• Why is important to perform a hypothesis test about a standard deviation?
• What are the two conditions that must be met when performing a hypothesis test about the standard deviation?
• What test can be used to determine if two samples have similar variances?
• What causes the null hypothesis to be rejected in an F-test?

Advanced Topics

• Nonlinear Transformations of Data
• Nonparametric Statistics
• Analysis of Variance

Hypothesis Testing

About hypothesis testing.

Watch the video for a brief overview of hypothesis testing:

Can’t see the video? Click here to watch it on YouTube.

Contents (Click to skip to the section):

What is a Hypothesis?

What is hypothesis testing.

• Hypothesis Testing Examples (One Sample Z Test).
• Hypothesis Test on a Mean (TI 83).

Bayesian Hypothesis Testing.

• More Hypothesis Testing Articles
• Hypothesis Tests in One Picture
• Critical Values

What is the Null Hypothesis?

Need help with a homework problem? Check out our tutoring page!

A hypothesis is an educated guess about something in the world around you. It should be testable, either by experiment or observation. For example:

• A new medicine you think might work.
• A way of teaching you think might be better.
• A possible location of new species.
• A fairer way to administer standardized tests.

It can really be anything at all as long as you can put it to the test.

What is a Hypothesis Statement?

If you are going to propose a hypothesis, it’s customary to write a statement. Your statement will look like this: “If I…(do this to an independent variable )….then (this will happen to the dependent variable ).” For example:

• If I (decrease the amount of water given to herbs) then (the herbs will increase in size).
• If I (give patients counseling in addition to medication) then (their overall depression scale will decrease).
• If I (give exams at noon instead of 7) then (student test scores will improve).
• If I (look in this certain location) then (I am more likely to find new species).

A good hypothesis statement should:

• Include an “if” and “then” statement (according to the University of California).
• Include both the independent and dependent variables.
• Be testable by experiment, survey or other scientifically sound technique.
• Be based on information in prior research (either yours or someone else’s).
• Have design criteria (for engineering or programming projects).

Hypothesis testing can be one of the most confusing aspects for students, mostly because before you can even perform a test, you have to know what your null hypothesis is. Often, those tricky word problems that you are faced with can be difficult to decipher. But it’s easier than you think; all you need to do is:

• Figure out your null hypothesis,
• State your null hypothesis,
• Choose what kind of test you need to perform,
• Either support or reject the null hypothesis .

If you trace back the history of science, the null hypothesis is always the accepted fact. Simple examples of null hypotheses that are generally accepted as being true are:

• DNA is shaped like a double helix.
• There are 8 planets in the solar system (excluding Pluto).
• Taking Vioxx can increase your risk of heart problems (a drug now taken off the market).

How do I State the Null Hypothesis?

You won’t be required to actually perform a real experiment or survey in elementary statistics (or even disprove a fact like “Pluto is a planet”!), so you’ll be given word problems from real-life situations. You’ll need to figure out what your hypothesis is from the problem. This can be a little trickier than just figuring out what the accepted fact is. With word problems, you are looking to find a fact that is nullifiable (i.e. something you can reject).

Hypothesis Testing Examples #1: Basic Example

A researcher thinks that if knee surgery patients go to physical therapy twice a week (instead of 3 times), their recovery period will be longer. Average recovery times for knee surgery patients is 8.2 weeks.

The hypothesis statement in this question is that the researcher believes the average recovery time is more than 8.2 weeks. It can be written in mathematical terms as: H 1 : μ > 8.2

Next, you’ll need to state the null hypothesis .  That’s what will happen if the researcher is wrong . In the above example, if the researcher is wrong then the recovery time is less than or equal to 8.2 weeks. In math, that’s: H 0 μ ≤ 8.2

Rejecting the null hypothesis

Ten or so years ago, we believed that there were 9 planets in the solar system. Pluto was demoted as a planet in 2006. The null hypothesis of “Pluto is a planet” was replaced by “Pluto is not a planet.” Of course, rejecting the null hypothesis isn’t always that easy— the hard part is usually figuring out what your null hypothesis is in the first place.

Hypothesis Testing Examples (One Sample Z Test)

The one sample z test isn’t used very often (because we rarely know the actual population standard deviation ). However, it’s a good idea to understand how it works as it’s one of the simplest tests you can perform in hypothesis testing. In English class you got to learn the basics (like grammar and spelling) before you could write a story; think of one sample z tests as the foundation for understanding more complex hypothesis testing. This page contains two hypothesis testing examples for one sample z-tests .

One Sample Hypothesis Testing Example: One Tailed Z Test

Watch the video for an example:

A principal at a certain school claims that the students in his school are above average intelligence. A random sample of thirty students IQ scores have a mean score of 112.5. Is there sufficient evidence to support the principal’s claim? The mean population IQ is 100 with a standard deviation of 15.

Step 1: State the Null hypothesis . The accepted fact is that the population mean is 100, so: H 0 : μ = 100.

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 4: State the alpha level . If you aren’t given an alpha level , use 5% (0.05).

Step 5: Find the rejection region area (given by your alpha level above) from the z-table . An area of .05 is equal to a z-score of 1.645.

Step 6: If Step 6 is greater than Step 5, reject the null hypothesis. If it’s less than Step 5, you cannot reject the null hypothesis. In this case, it is more (4.56 > 1.645), so you can reject the null.

One Sample Hypothesis Testing Examples: #3

Watch the video for an example of a two-tailed z-test:

Blood glucose levels for obese patients have a mean of 100 with a standard deviation of 15. A researcher thinks that a diet high in raw cornstarch will have a positive or negative effect on blood glucose levels. A sample of 30 patients who have tried the raw cornstarch diet have a mean glucose level of 140. Test the hypothesis that the raw cornstarch had an effect.

• State the null hypothesis : H 0 :μ=100
• State the alternate hypothesis : H 1 :≠100
• State your alpha level. We’ll use 0.05 for this example. As this is a two-tailed test, split the alpha into two. 0.05/2=0.025
• Find the z-score associated with your alpha level . You’re looking for the area in one tail only . A z-score for 0.75(1-0.025=0.975) is 1.96. As this is a two-tailed test, you would also be considering the left tail (z = 1.96)
•   If Step 5 is less than -1.96 or greater than 1.96 (Step 3), reject the null hypothesis . In this case, it is greater, so you can reject the null.

*This process is made much easier if you use a TI-83 or Excel to calculate the z-score (the “critical value”). See:

• Critical z value TI 83
• Z Score in Excel

Hypothesis Testing Examples: Mean (Using TI 83)

You can use the TI 83 calculator for hypothesis testing, but the calculator won’t figure out the null and alternate hypotheses; that’s up to you to read the question and input it into the calculator.

Example problem : A sample of 200 people has a mean age of 21 with a population standard deviation (σ) of 5. Test the hypothesis that the population mean is 18.9 at α = 0.05.

Step 1: State the null hypothesis. In this case, the null hypothesis is that the population mean is 18.9, so we write: H 0 : μ = 18.9

Step 2: State the alternative hypothesis. We want to know if our sample, which has a mean of 21 instead of 18.9, really is different from the population, therefore our alternate hypothesis: H 1 : μ ≠ 18.9

Step 3: Press Stat then press the right arrow twice to select TESTS.

Step 4: Press 1 to select 1:Z-Test… . Press ENTER.

Step 5: Use the right arrow to select Stats .

Step 6: Enter the data from the problem: μ 0 : 18.9 σ: 5 x : 21 n: 200 μ: ≠μ 0

Step 7: Arrow down to Calculate and press ENTER. The calculator shows the p-value: p = 2.87 × 10 -9

This is smaller than our alpha value of .05. That means we should reject the null hypothesis .

Bayesian Hypothesis Testing: What is it?

Bayesian hypothesis testing helps to answer the question: Can the results from a test or survey be repeated? Why do we care if a test can be repeated? Let’s say twenty people in the same village came down with leukemia. A group of researchers find that cell-phone towers are to blame. However, a second study found that cell-phone towers had nothing to do with the cancer cluster in the village. In fact, they found that the cancers were completely random. If that sounds impossible, it actually can happen! Clusters of cancer can happen simply by chance . There could be many reasons why the first study was faulty. One of the main reasons could be that they just didn’t take into account that sometimes things happen randomly and we just don’t know why.

It’s good science to let people know if your study results are solid, or if they could have happened by chance. The usual way of doing this is to test your results with a p-value . A p value is a number that you get by running a hypothesis test on your data. A P value of 0.05 (5%) or less is usually enough to claim that your results are repeatable. However, there’s another way to test the validity of your results: Bayesian Hypothesis testing. This type of testing gives you another way to test the strength of your results.

Traditional testing (the type you probably came across in elementary stats or AP stats) is called Non-Bayesian. It is how often an outcome happens over repeated runs of the experiment. It’s an objective view of whether an experiment is repeatable. Bayesian hypothesis testing is a subjective view of the same thing. It takes into account how much faith you have in your results. In other words, would you wager money on the outcome of your experiment?

Differences Between Traditional and Bayesian Hypothesis Testing.

Traditional testing (Non Bayesian) requires you to repeat sampling over and over, while Bayesian testing does not. The main different between the two is in the first step of testing: stating a probability model. In Bayesian testing you add prior knowledge to this step. It also requires use of a posterior probability , which is the conditional probability given to a random event after all the evidence is considered.

Arguments for Bayesian Testing.

Many researchers think that it is a better alternative to traditional testing, because it:

• Includes prior knowledge about the data.
• Takes into account personal beliefs about the results.

Arguments against.

• Including prior data or knowledge isn’t justifiable.
• It is difficult to calculate compared to non-Bayesian testing.

Back to top

Hypothesis Testing Articles

• What is Ad Hoc Testing?
• Composite Hypothesis Test
• What is a Rejection Region?
• What is a Two Tailed Test?
• How to Decide if a Hypothesis Test is a One Tailed Test or a Two Tailed Test.
• How to Decide if a Hypothesis is a Left Tailed Test or a Right-Tailed Test.
• How to State the Null Hypothesis in Statistics.
• How to Find a Critical Value .
• How to Support or Reject a Null Hypothesis.

Specific Tests:

• Brunner Munzel Test (Generalized Wilcoxon Test).
• Chi Square Test for Normality.
• Cochran-Mantel-Haenszel Test.
• Granger Causality Test .
• Hotelling’s T-Squared.
• KPSS Test .
• What is a Likelihood-Ratio Test?
• Log rank test .
• MANCOVA Assumptions.
• MANCOVA Sample Size.
• Marascuilo Procedure
• Rao’s Spacing Test
• Rayleigh test of uniformity.
• Sequential Probability Ratio Test.
• How to Run a Sign Test.
• T Test: one sample.
• T-Test: Two sample .
• Welch’s ANOVA .
• Welch’s Test for Unequal Variances .
• Z-Test: one sample .
• Z Test: Two Proportion.
• Wald Test .

Related Articles:

• What is an Acceptance Region?
• How to Calculate Chebyshev’s Theorem.
• Contrast Analysis
• Decision Rule.
• Degrees of Freedom .
• Directional Test
• False Discovery Rate
• How to calculate the Least Significant Difference.
• Levels in Statistics.
• How to Calculate Margin of Error.
• Mean Difference (Difference in Means)
• The Multiple Testing Problem .
• What is the Neyman-Pearson Lemma?
• What is an Omnibus Test?
• One Sample Median Test .
• How to Find a Sample Size (General Instructions).
• Sig 2(Tailed) meaning in results
• What is a Standardized Test Statistic?
• How to Find Standard Error
• Standardized values: Example.
• How to Calculate a T-Score.
• T-Score Vs. a Z.Score.
• Testing a Single Mean.
• Unequal Sample Sizes.
• Uniformly Most Powerful Tests.
• How to Calculate a Z-Score.

Teach yourself statistics

Hypothesis Test: Difference Between Means

This lesson explains how to conduct a hypothesis test for the difference between two means. The test procedure, called the two-sample t-test , is appropriate when the following conditions are met:

• The sampling method for each sample is simple random sampling .
• The samples are independent .
• Each population is at least 20 times larger than its respective sample .
• The population distribution is normal.
• The population data are symmetric , unimodal , without outliers , and the sample size is 15 or less.
• The population data are slightly skewed , unimodal, without outliers, and the sample size is 16 to 40.
• The sample size is greater than 40, without outliers.

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

State the Hypotheses

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

The table below shows three sets of null and alternative hypotheses. Each makes a statement about the difference d between the mean of one population μ 1 and the mean of another population μ 2 . (In the table, the symbol ≠ means " not equal to ".)

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

When the null hypothesis states that there is no difference between the two population means (i.e., d = 0), the null and alternative hypothesis are often stated in the following form.

H o : μ 1 = μ 2

H a : μ 1 ≠ μ 2

Formulate an Analysis Plan

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

• Significance level. Often, researchers choose significance levels equal to 0.01, 0.05, or 0.10; but any value between 0 and 1 can be used.
• Test method. Use the two-sample t-test to determine whether the difference between means found in the sample is significantly different from the hypothesized difference between means.

Analyze Sample Data

Using sample data, find the standard error, degrees of freedom, test statistic, and the P-value associated with the test statistic.

SE = sqrt[ (s 1 2 /n 1 ) + (s 2 2 /n 2 ) ]

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) ] }

t = [ ( x 1 - x 2 ) - d ] / SE

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

Interpret Results

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

Test Your Understanding

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

Problem 1: Two-Tailed Test

Within a school district, students were randomly assigned to one of two Math teachers - Mrs. Smith and Mrs. Jones. After the assignment, Mrs. Smith had 30 students, and Mrs. Jones had 25 students.

At the end of the year, each class took the same standardized test. Mrs. Smith's students had an average test score of 78, with a standard deviation of 10; and Mrs. Jones' students had an average test score of 85, with a standard deviation of 15.

Test the hypothesis that Mrs. Smith and Mrs. Jones are equally effective teachers. Use a 0.10 level of significance. (Assume that student performance is approximately normal.)

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

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: μ 1 - μ 2 = 0

Alternative hypothesis: μ 1 - μ 2 ≠ 0

• Formulate an analysis plan . For this analysis, the significance level is 0.10. Using sample data, we will conduct a two-sample t-test of the null hypothesis.

SE = sqrt[(s 1 2 /n 1 ) + (s 2 2 /n 2 )]

SE = sqrt[(10 2 /30) + (15 2 /25] = sqrt(3.33 + 9)

SE = sqrt(12.33) = 3.51

DF = (10 2 /30 + 15 2 /25) 2 / { [ (10 2 / 30) 2 / (29) ] + [ (15 2 / 25) 2 / (24) ] }

DF = (3.33 + 9) 2 / { [ (3.33) 2 / (29) ] + [ (9) 2 / (24) ] } = 152.03 / (0.382 + 3.375) = 152.03/3.757 = 40.47

t = [ ( x 1 - x 2 ) - d ] / SE = [ (78 - 85) - 0 ] / 3.51 = -7/3.51 = -1.99

where s 1 is the standard deviation of sample 1, s 2 is the standard deviation of sample 2, n 1 is the size of sample 1, n 2 is the size of sample 2, x 1 is the mean of sample 1, x 2 is the mean of sample 2, d is the hypothesized difference between the population means, and SE is the standard error.

Since we have a two-tailed test , the P-value is the probability that a t statistic having 40 degrees of freedom is more extreme than -1.99; that is, less than -1.99 or greater than 1.99.

We use the t Distribution Calculator to find P(t < -1.99) is about 0.027.

• If you enter 1.99 as the sample mean in the t Distribution Calculator, you will find the that the P(t ≤ 1.99) is about 0.973. Therefore, P(t > 1.99) is 1 minus 0.973 or 0.027. Thus, the P-value = 0.027 + 0.027 = 0.054.
• Interpret results . Since the P-value (0.054) is less than the significance level (0.10), we cannot accept the null hypothesis.

Note: If you use this approach on an exam, you may also want to mention why this approach is appropriate. Specifically, the approach is appropriate because the sampling method was simple random sampling, the samples were independent, the sample size was much smaller than the population size, and the samples were drawn from a normal population.

Problem 2: One-Tailed Test

The Acme Company has developed a new battery. The engineer in charge claims that the new battery will operate continuously for at least 7 minutes longer than the old battery.

To test the claim, the company selects a simple random sample of 100 new batteries and 100 old batteries. The old batteries run continuously for 190 minutes with a standard deviation of 20 minutes; the new batteries, 200 minutes with a standard deviation of 40 minutes.

Test the engineer's claim that the new batteries run at least 7 minutes longer than the old. Use a 0.05 level of significance. (Assume that there are no outliers in either sample.)

Null hypothesis: μ 1 - μ 2 <= 7

Alternative hypothesis: μ 1 - μ 2 > 7

where μ 1 is battery life for the new battery, and μ 2 is battery life for the old battery.

• Formulate an analysis plan . For this analysis, the significance level is 0.05. Using sample data, we will conduct a two-sample t-test of the null hypothesis.

SE = sqrt[(40 2 /100) + (20 2 /100]

SE = sqrt(16 + 4) = 4.472

DF = (40 2 /100 + 20 2 /100) 2 / { [ (40 2 / 100) 2 / (99) ] + [ (20 2 / 100) 2 / (99) ] }

DF = (20) 2 / { [ (16) 2 / (99) ] + [ (2) 2 / (99) ] } = 400 / (2.586 + 0.162) = 145.56

t = [ ( x 1 - x 2 ) - d ] / SE = [(200 - 190) - 7] / 4.472 = 3/4.472 = 0.67

where s 1 is the standard deviation of sample 1, s 2 is the standard deviation of sample 2, n 1 is the size of sample 1, n 2 is the size of sample 2, x 1 is the mean of sample 1, x 2 is the mean of sample 2, d is the hypothesized difference between population means, and SE is the standard error.

Here is the logic of the analysis: Given the alternative hypothesis (μ 1 - μ 2 > 7), we want to know whether the observed difference in sample means is big enough (i.e., sufficiently greater than 7) to cause us to reject the null hypothesis.

Interpret results . Suppose we replicated this study many times with different samples. If the true difference in population means were actually 7, we would expect the observed difference in sample means to be 10 or less in 75% of our samples. And we would expect to find an observed difference to be more than 10 in 25% of our samples Therefore, the P-value in this analysis is 0.25.

Pardon Our Interruption

As you were browsing something about your browser made us think you were a bot. There are a few reasons this might happen:

• You've disabled JavaScript in your web browser.
• You're a power user moving through this website with super-human speed.
• You've disabled cookies in your web browser.
• A third-party browser plugin, such as Ghostery or NoScript, is preventing JavaScript from running. Additional information is available in this support article .

To regain access, please make sure that cookies and JavaScript are enabled before reloading the page.