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Two-sample t hypothesis tests · Tutorial 740 of 1000

Writing the Full Two-Sample t Test Solution

Practice the four parts of a two-sample t test solution and see how to communicate each part clearly and accurately.

Intermediate 12 min read

What You'll Learn

  • Define both population means and state hypotheses that match the research question.
  • Check the randomization or sampling, independence, 10% condition, and Nearly Normal condition.
  • Calculate the unpooled standard error and test statistic for a difference in means.
  • Use Welch degrees of freedom and the correct tail to find a p-value.
  • Write a conclusion that connects the decision to the evidence and the study design.

A Complete Test Is More Than a Calculator Result

A two-sample t test evaluates whether the data provide evidence of a difference between two population means. A complete response explains what those means represent, why the procedure is appropriate, how the statistic and p-value were obtained, and what the result means in context.

In “Writing a Complete Paired t Test Solution,” you practiced a four-part response for paired data. For two independent samples, the same overall structure applies, but the parameter, conditions, and calculation are different. As covered in “Two-Sample Versus Paired Test on the Same Numbers,” the study design determines whether groups are independent; matching sample sizes or arranging observations in rows does not create pairs.

Key takeaway: A full two-sample t test response has four parts: State the population means and hypotheses; Plan the procedure and check conditions; Do the calculation and report the p-value; Conclude in context.

The Four Parts of the Response

1
State.
Define \(\mu_1\) and \(\mu_2\) as the population means for the same quantitative variable in the two specified groups. State \(H_0:\mu_1-\mu_2=0\) and an alternative hypothesis that matches the research question.
2
Plan and check conditions.
Name the unpooled two-sample t test. Explain the random design and why the observations and groups are independent. Check the 10% condition when sampling without replacement, and assess the Nearly Normal condition for each group.
3
Do.
Calculate the standard error and test statistic, use Welch degrees of freedom, and find the p-value in the direction specified by \(H_a\). Keep the sample order consistent with the hypotheses.
4
Conclude in context.
Compare the p-value with the stated significance level, make the decision to reject or fail to reject \(H_0\), and describe the evidence about the population means in the setting of the question.

The hypotheses are about population parameters, not sample statistics. The two-sample test statistic for a null difference of zero uses the difference in sample means divided by its estimated standard error. The standard error keeps the two groups’ variance contributions separate.

$$ t=\frac{(\bar{x}_1-\bar{x}_2)-0} {\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}} $$

As in “Computing Two-Sample Degrees of Freedom With the Welch Formula,” use the calculator’s Welch degrees of freedom for the unpooled test, or the conservative method if required. The alternative hypothesis determines the tail area used for the p-value; it is not chosen after seeing the sign of the statistic.

Worked Example: No Observed Difference in Filter Flow

In a fictional quality-control study, engineers randomly sample 12 cartridges of Filter A and 15 cartridges of Filter B from their respective production runs. They measure water flow rate in liters per minute. The sample mean is 25.0 liters per minute for each filter type; the sample standard deviations are 3.0 and 3.6 liters per minute, respectively. The samples are less than 10% of their production runs. Assume plots of each group show no strong skewness or outliers. Test for a difference in mean flow rate at \(\alpha=0.05\).

1
State.
Let \(\mu_1\) be the true mean flow rate of all Filter A cartridges from the production run, and let \(\mu_2\) be the true mean flow rate of all Filter B cartridges from its production run. The hypotheses are \(H_0:\mu_1-\mu_2=0\) and \(H_a:\mu_1-\mu_2\ne0\).
2
Plan and check conditions.
Use an unpooled two-sample t test because the samples contain different cartridges, with no matching between them. The cartridges were randomly sampled, supporting inference to the production runs. Within each group the observations are independent, and the two groups are independent because no cartridge is in both. The 10% condition holds for both samples. The stated plots show no strong skewness or outliers, so the Nearly Normal condition is reasonable.
3
Do.
The observed difference in sample means is \(25.0-25.0=0\) liters per minute. The standard error uses each sample variance and size separately.
$$ SE=\sqrt{\frac{3.0^2}{12}+\frac{3.6^2}{15}} =\sqrt{0.750+0.864} =\sqrt{1.614} \approx1.27\text{ liters per minute}, \qquad t=\frac{0}{1.27}=0. $$

The calculator gives Welch degrees of freedom of approximately 24.94. For the two-sided alternative, a statistic of zero is not farther from zero in either direction, so the p-value is \(1.0000\).

4
Conclude in context.
Because \(1.0000\) is greater than \(0.05\), fail to reject \(H_0\). These data do not provide convincing evidence of a difference in the true mean flow rates of Filter A and Filter B cartridges from these production runs.

Failing to reject does not establish that the population means are equal. It says that this sample, with these standard deviations and sample sizes, provides no convincing evidence of a difference. The conclusion refers to the populations identified in the State step, not just to the particular cartridges measured.

Worked Example: One-Sided Test of Seedling Growth

A fictional greenhouse study randomly selects 30 seedlings from a seedling population of more than 300 seedlings and randomly assigns 16 to Nutrient A and 14 to a standard nutrient solution. After three weeks, the mean growth is 9.8 cm for Nutrient A and 9.0 cm for the standard solution. The sample standard deviations are 1.2 cm and 1.0 cm. Each treatment group is less than 10% of the source population. Plots show no strong skewness or outliers. Researchers ask whether the mean growth with Nutrient A is greater. Use \(\alpha=0.05\).

1
State.
Let \(\mu_1\) be the true mean three-week growth of seedlings receiving Nutrient A, and let \(\mu_2\) be the true mean three-week growth of seedlings receiving the standard solution, for the target seedling population. The hypotheses are \(H_0:\mu_1-\mu_2=0\) and \(H_a:\mu_1-\mu_2>0\).
2
Plan and check conditions.
Use an unpooled two-sample t test. The seedlings were randomly selected and randomly assigned to treatments. Each seedling received only one treatment, so the groups are independent, and the individual seedlings are treated as independent units. The total sample of 30 is less than 10% of the source population, which has more than 300 seedlings. The group plots show no strong skewness or outliers, supporting the Nearly Normal condition.
3
Do.
The observed difference is \(9.8-9.0=0.8\) cm. Calculate the standard error from the separate sample variances, then divide the observed difference by that standard error.
$$ SE=\sqrt{\frac{1.2^2}{16}+\frac{1.0^2}{14}} =\sqrt{0.090000+0.071429} \approx0.4018\text{ cm}, \qquad t=\frac{0.8}{0.4018}\approx1.9911. $$

Welch degrees of freedom are approximately 27.95. Because the alternative is \(H_a:\mu_1-\mu_2>0\), the p-value is the area to the right of \(t=1.9911\), not a two-sided area. A calculator gives \(p\approx0.02816\), rounded.

4
Conclude in context.
Because \(0.02816\) is less than \(0.05\), reject \(H_0\). The data provide convincing evidence that the true mean three-week growth is greater for seedlings receiving Nutrient A than for seedlings receiving the standard solution. Because treatments were randomly assigned, the result supports a cause-and-effect conclusion for these treatments under the study conditions.

This example also shows why the subtraction order matters. The first group in the hypotheses is Nutrient A, so the calculation uses \(\bar{x}_1-\bar{x}_2\). That difference is positive, consistent with the right-tailed alternative. Had the groups been reversed, both the statistic’s sign and the direction of the alternative would need to change.

Worked Example: A Difference That Is Not Convincing

A fictional transit analyst randomly samples six weekday commuters on Route 1 and six different weekday commuters on Route 2, recording commute time in minutes. The sample means are 10.1547 and 9.0000 minutes, and each sample standard deviation is 2.0 minutes. The groups are independent, each sample is less than 10% of its route’s weekday commuters, and plots show no strong skewness or outliers. Test whether the population mean commute times differ, using \(\alpha=0.05\).

1
State.
Let \(\mu_1\) be the true mean weekday commute time for commuters on Route 1, and \(\mu_2\) the true mean weekday commute time for commuters on Route 2. The hypotheses are \(H_0:\mu_1-\mu_2=0\) and \(H_a:\mu_1-\mu_2\ne0\).
2
Plan and check conditions.
Use an unpooled two-sample t test. The commuters were randomly sampled; different people make up the two samples, so there is no pairing. The observations within each route sample and the two groups are treated as independent. The 10% condition holds for both samples. The plots show no strong skewness or outliers, which is important with only six observations per group.
3
Do.
The observed difference in means is \(10.1547-9.0000=1.1547\) minutes. The standard error and statistic are:
$$ SE=\sqrt{\frac{2.0^2}{6}+\frac{2.0^2}{6}} =\sqrt{\frac{4}{3}} \approx1.1547\text{ minutes}, \qquad t=\frac{1.1547}{1.1547}=1.000. $$

The Welch degrees of freedom are 10. The two-sided p-value for \(t=1.000\) with 10 degrees of freedom is approximately \(0.3409\), rounded.

4
Conclude in context.
Because \(0.3409\) is greater than \(0.05\), fail to reject \(H_0\). The data do not provide convincing evidence that the true mean weekday commute times differ between commuters on Route 1 and Route 2.

The observed sample means are not identical, but a test asks whether their difference is large relative to the estimated sampling variability. Here the difference is only one estimated standard error from zero. Do not turn a failure to reject into a claim that the routes have exactly equal population means.

Common Mistakes and AP Exam Tips

  • Writing hypotheses about sample means: The hypotheses describe \(\mu_1\) and \(\mu_2\), not \(\bar{x}_1\) and \(\bar{x}_2\). Sample means are statistics used to assess the population claim.
  • Leaving the populations vague: Define both means using the same quantitative variable, name each group, and identify the target setting or time period. “The mean for group 1” is not enough if the groups are unclear.
  • Skipping a condition or merely naming it: Explain what in the design supports randomization and independence, address the 10% condition when applicable, and describe the shape evidence for each sample. A calculator cannot verify these conditions.
  • Using the wrong standard error: For the unpooled test, calculate \(\sqrt{s_1^2/n_1+s_2^2/n_2}\). Do not pool the variances or use the standard deviation of paired differences unless the design actually consists of pairs.
  • Choosing the tail after seeing the result: The wording of the research question determines \(H_a\). A claim about either direction calls for a two-sided alternative; a specified direction calls for a one-sided alternative.
  • Reporting a decision without a contextual conclusion: “Reject” or “fail to reject” alone is incomplete. Name the evidence about the population means and keep the conclusion consistent with the study design.
  • Claiming that failure to reject proves equality: A large p-value means the data do not provide convincing evidence against the null at the chosen level. It does not prove the null hypothesis true.

For a full-credit response, make each part easy to find. State the parameter definitions and hypotheses; name the test and justify its conditions; show the standard error, statistic, degrees of freedom, and p-value; then compare the p-value with \(\alpha\) and conclude in context. If the study is observational, do not claim that a difference was caused by group membership. Random assignment, as opposed to random sampling alone, is what supports a cause-and-effect interpretation.

Key takeaway: A complete two-sample t test connects the research question to defined population means, checks the design and data conditions, shows the unpooled calculation and the correct p-value, and states what the evidence means in context.

Check Your Understanding

Use the four-part structure to assess each response or plan.

  1. A question asks whether two independent populations have different mean delivery times. Write the alternative hypothesis using \(\mu_1\) and \(\mu_2\).
  2. Why should a response explain how the two samples were collected or assigned instead of simply saying “the samples are independent”?
  3. A two-sample test has \(t=1.7\), but the stated alternative is \(H_a:\mu_1-\mu_2<0\). Which tail area is needed, and why might the observed statistic be inconsistent with the direction of the claim?
  4. A test has p-value \(0.12\) and significance level \(0.05\). Write a conclusion in context that avoids claiming the population means are equal.
  5. What is the difference between random sampling and random assignment when explaining the scope of a conclusion?

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