Paired data give x_d = 3, s_d = 6, n = 16. Find the paired t statistic. [1 point]

College Board AP 2022 (style)-style practice: Section II (free response). Independent random samples compare a supplement to a placebo on reaction time (ms). Treatment: n_1 = 36, x_1 = 245, s_1 = 30. Placebo: n_2 = 36, x_2 = 260, s_2 = 28. Test at α = 0.05 whether the supplement lowers mean reaction time. Compute the test statistic and P-value, and conclude in context (justify in context). Use a conservative df = 35.

A 4-point two-sample t-test. (1) (1 point) Let _1, _2 be the true mean reaction times for supplement and placebo. H_0: _1 = _2 versus H_a: _1 < _2. Random/independent; both n 30, so normal/large holds. (2) (1 point) SE = 30^236 + 28^236 = sqrt(25 + 21.78) = sqrt(46.78) ≈ 6.84. (3) (1 point) t = 245 - 2606.84 = -156.84 ≈ -2.19, conservative df = 35. P-value = P(t_35 < -2.19) ≈ 0.017. (4) (1 point) Since 0.017 < 0.05, reject H_0. There is convincing evidence that the supplement lowers the true mean reaction time relative to placebo. Markers reward the unpooled standard error, the t statistic, the lower-tail P-value, and a contextual conclusion.

United StatesStatisticsSyllabus dot point

How do you compute the t test statistic and P-value and conclude a test comparing two means?

Topic 7.9 Carrying Out a Test for the Difference of Two Population Means: compute the two-sample (or paired) t test statistic, find the P-value, compare to the significance level, and state a conclusion in context.

A focused answer to AP Statistics Topic 7.9, on computing the two-sample t statistic with the unpooled standard error (or the paired one-sample t statistic on differences), finding the P-value, and concluding in context, with a full worked test.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-04

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this topic is asking
The two-sample t statistic
The paired t statistic
P-value, decision, and conclusion
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What this topic is asking

The College Board (Topic 7.9) wants you to carry out and conclude a test comparing two means: compute the two-sample t statistic with the unpooled standard error (or the paired one-sample t statistic on the differences), find the P-value, compare to $\alpha$ , and state a conclusion in context.

The two-sample t statistic

The numerator is the observed difference of means; the denominator is the same unpooled standard error as the two-sample interval (variances added). Unlike the proportion case, the two-sample mean test and interval use the same standard error, because there is no pooled-proportion analogue required on the AP exam by default. The degrees of freedom are messy, so report the calculator value or use the conservative smaller- $df$ choice; AP grading accepts either.

The paired t statistic

Pairing turns a two-sample problem into a one-sample problem on a single list of differences. This is why the procedure, statistic, and degrees of freedom all match Topic 7.5 applied to the differences. The payoff is that pairing removes between-subject variability, often giving a smaller standard error and more power than an independent design with the same number of measurements.

P-value, decision, and conclusion

Find the P-value from the appropriate $t$ -distribution in the direction of $H_a$ : upper tail for $>$ , lower for $<$ , both (doubled) for $\ne$ . Then compare to $\alpha$ : P-value $\le \alpha$ rejects $H_0$ (convincing evidence of a difference in the stated direction); P-value $> \alpha$ fails to reject (not convincing evidence). The conclusion sentence states the decision, ties it to the $P$ -versus- $\alpha$ comparison, and interprets in context ("there is convincing evidence the supplement lowers the mean reaction time"). As ever, never write "accept $H_0$ ," and for a two-sided two-sample test, the two-sided interval at $C = 1 - \alpha$ gives the same verdict via the zero-check.

Complete two-sample t-test

A teacher compares two review methods. Independent random samples: method 1, $n_1 = 25$ , $\bar{x}_1 = 82$ , $s_1 = 9$ ; method 2, $n_2 = 28$ , $\bar{x}_2 = 77$ , $s_2 = 10$ . Both samples are roughly symmetric. Test at $\alpha = 0.05$ whether the mean scores differ. Use a conservative $df = 24$ .

step 1 Hypotheses and conditions

Let $\mu_1, \mu_2$ be the true mean scores. $H_0: \mu_1 = \mu_2$ versus $H_a: \mu_1 \ne \mu_2$ . Random and independent: stated. Normal/large: both samples roughly symmetric (and reasonably sized); $10\%$ condition reasonable. Two-sample t-test appropriate.

step 2 Unpooled standard error

SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} = \sqrt{\frac{81}{25} + \frac{100}{28}} = \sqrt{3.24 + 3.571} = \sqrt{6.811} \approx 2.610.

step 3 Test statistic

t = \frac{\bar{x}_1 - \bar{x}_2}{SE} = \frac{82 - 77}{2.610} = \frac{5}{2.610} \approx 1.92, \qquad df = 24 \text{ (conservative)}.

step 4 P-value

Two-sided: $\text{P-value} = 2 \cdot P(t_{24} > 1.92) \approx 2(0.0334) = 0.0668$ .

step 5 Conclusion in context

Since P-value $0.0668 > 0.05 = \alpha$ , fail to reject $H_0$ . There is not convincing evidence that the mean scores for the two review methods differ. (A calculator $df$ , about $51$ , would give a slightly smaller P-value but the same conclusion at $\alpha = 0.05$ .)

Try this

Q1. Write the two-sample t statistic and name the standard error type. [2 points]

Cue. $t = \dfrac{\bar{x}_1 - \bar{x}_2}{\sqrt{s_1^2/n_1 + s_2^2/n_2}}$ ; it uses the unpooled standard error.

Q2. Paired data give $\bar{x}_d = 3$ , $s_d = 6$ , $n = 16$ . Find the paired t statistic. [1 point]

Cue. $t = \dfrac{3}{6/\sqrt{16}} = \dfrac{3}{1.5} = 2.00$ , $df = 15$ .

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2018 (style)1 marksSection I (multiple choice). For a two-sample t-test, the test statistic is (A)

\dfrac{\bar{x}_1 - \bar{x}_2}{\sqrt{s_1^2/n_1 + s_2^2/n_2}}

(B)

\dfrac{\bar{x}_1 - \bar{x}_2}{s_p\sqrt{1/n_1 + 1/n_2}}

only (C)

\dfrac{\bar{x}_1 - \bar{x}_2}{\sqrt{n_1 + n_2}}

(D)

\dfrac{\bar{x}_d}{s_d/\sqrt{n}}

Show worked answer →

The correct answer is (A).

The standard two-sample t-test (unpooled) uses $t = \dfrac{\bar{x}_1 - \bar{x}_2}{\sqrt{s_1^2/n_1 + s_2^2/n_2}}$ , the difference over the unpooled standard error.

(B) is a pooled-variance version not required on the AP exam as the default. (C) is not a valid standard error. (D) is the paired statistic, used only for matched data.

AP 2022 (style)4 marksSection II (free response). Independent random samples compare a supplement to a placebo on reaction time (ms). Treatment:

n_1 = 36

\bar{x}_1 = 245

s_1 = 30

. Placebo:

n_2 = 36

\bar{x}_2 = 260

s_2 = 28

. Test at

\alpha = 0.05

whether the supplement lowers mean reaction time. Compute the test statistic and P-value, and conclude in context (justify in context). Use a conservative

df = 35

Show worked answer →

A 4-point two-sample t-test.

(1) (1 point) Let $\mu_1, \mu_2$ be the true mean reaction times for supplement and placebo. $H_0: \mu_1 = \mu_2$ versus $H_a: \mu_1 < \mu_2$ . Random/independent; both $n \ge 30$ , so normal/large holds.
(2) (1 point) $SE = \sqrt{\dfrac{30^2}{36} + \dfrac{28^2}{36}} = \sqrt{25 + 21.78} = \sqrt{46.78} \approx 6.84$ .
(3) (1 point) $t = \dfrac{245 - 260}{6.84} = \dfrac{-15}{6.84} \approx -2.19$ , conservative $df = 35$ . P-value $= P(t_{35} < -2.19) \approx 0.017$ .
(4) (1 point) Since $0.017 < 0.05$ , reject $H_0$ . There is convincing evidence that the supplement lowers the true mean reaction time relative to placebo.

Markers reward the unpooled standard error, the t statistic, the lower-tail P-value, and a contextual conclusion.

Related dot points

Sources & how we know this

AP Statistics Course and Exam Description — College Board (2020)