For x = 52, _0 = 50, s = 8, n = 16, find t and the degrees of freedom. [2 points]

College Board AP 2022 (style)-style practice: Section II (free response). A bakery claims its loaves have a mean mass of 800 grams. A random sample of n = 16 loaves has x = 788 grams and s = 20 grams; a dotplot is roughly symmetric. Test at α = 0.05 whether the true mean mass is less than 800 grams. State hypotheses, check conditions, compute the t statistic and P-value (use df = 15), and conclude in context (justify in context).

A 4-point complete one-sample t-test. (1) (1 point) Let μ be the true mean mass. H_0: μ = 800 versus H_a: μ < 800. (2) (1 point) Random sample stated; n = 16 < 30 but the dotplot is roughly symmetric with no outliers; 10\% condition reasonable. One-sample t-test appropriate. (3) (1 point) t = 788 - 80020/sqrt(16) = -1220/4 = -125 = -2.40, df = 15. P-value = P(t_15 < -2.40) ≈ 0.015. (4) (1 point) Since P-value ≈ 0.015 < 0.05, reject H_0. There is convincing evidence that the true mean mass of the loaves is less than 800 grams. Markers reward the t statistic with df = 15, the lower-tail P-value, and a contextual reject conclusion.

United StatesStatisticsSyllabus dot point

How do you compute the t test statistic and P-value and conclude a test about a population mean?

Topic 7.5 Carrying Out a Test for a Population Mean: compute the t test statistic with n minus 1 degrees of freedom, find the P-value, compare to the significance level, and state a conclusion in context.

A focused answer to AP Statistics Topic 7.5, on computing the one-sample t statistic with n minus 1 degrees of freedom, finding the P-value, comparing to alpha, and stating a conclusion in context, with a full worked t-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 t test statistic
From t to the P-value and decision
Test and interval agree
Try this

What this topic is asking

The College Board (Topic 7.5) wants you to carry out and conclude a one-sample mean test: compute the $t$ statistic with $n - 1$ degrees of freedom, find the P-value from the $t$ -distribution, compare to $\alpha$ , and state a conclusion in context, completing the test set up in Topic 7.4.

The t test statistic

The numerator is the gap between the observed mean and the claimed mean; the denominator is the standard error $s/\sqrt{n}$ , which uses the sample standard deviation $s$ because $\sigma$ is unknown. That substitution is why the statistic follows a $t$ -distribution with $n - 1$ degrees of freedom, not the normal. Using the right $df$ is essential, because it sets how heavy the tails are and hence the P-value.

From t to the P-value and decision

Match the tail to $H_a$ and double for a two-sided test. The P-value is the area beyond $t$ in the appropriate tail of the $t_{df}$ curve, read from a calculator or table. The conclusion sentence states the decision, ties it to the $P$ -versus- $\alpha$ comparison, and translates it into context ("there is convincing evidence that the mean ... is less than ..."). Never write "accept $H_0$ "; the choices are "reject" or "fail to reject."

Test and interval agree

A two-sided t-test at level $\alpha$ and a $C = 1 - \alpha$ t-interval reach the same verdict: $H_0: \mu = \mu_0$ is rejected exactly when $\mu_0$ falls outside the interval. Because both procedures use the same standard error $s/\sqrt{n}$ (unlike the proportion case, where the test and interval used different standard deviations), this agreement is exact. So a confidence interval doubles as a two-sided test, a frequently examined connection and a useful self-check.

Complete one-sample t-test

A supplier guarantees a mean tensile strength of $\mu_0 = 500$ N. A random sample of $n = 25$ specimens has $\bar{x} = 492$ N and $s = 15$ N; a graph of the data is roughly symmetric with no outliers. Test at $\alpha = 0.05$ whether the true mean strength is below $500$ N.

step 1 Hypotheses

Let $\mu$ be the true mean tensile strength.

H_0: \mu = 500 \qquad H_a: \mu < 500.

step 2 Conditions

Random: stated random sample. Normal/large: $n = 25 < 30$ , but the data are roughly symmetric with no outliers, so the t-procedure is appropriate. $10\%$ : $25$ specimens is plausibly under $10\%$ of production. One-sample t-test appropriate.

step 3 Test statistic

t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} = \frac{492 - 500}{15/\sqrt{25}} = \frac{-8}{15/5} = \frac{-8}{3} \approx -2.67, \qquad df = 24.

step 4 P-value

Lower-tailed: $\text{P-value} = P(t_{24} < -2.67) \approx 0.0067$ .

step 5 Conclusion in context

Since P-value $0.0067 < 0.05 = \alpha$ , reject $H_0$ . There is convincing evidence that the true mean tensile strength is below $500$ N, so the data contradict the supplier's guarantee.

Try this

Q1. For $\bar{x} = 52$ , $\mu_0 = 50$ , $s = 8$ , $n = 16$ , find $t$ and the degrees of freedom. [2 points]

Cue. $t = \dfrac{52 - 50}{8/\sqrt{16}} = \dfrac{2}{2} = 1.00$ ; $df = 15$ .

Q2. A one-sided lower test gives $t = -1.80$ with $df = 20$ . Roughly, is the P-value above or below $0.05$ ? [1 point]

Cue. $P(t_{20} < -1.80)$ is about $0.043$ , just below $0.05$ , so it would reject at $\alpha = 0.05$ .

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). A test of

H_0: \mu = 100

uses

n = 25

\bar{x} = 104

s = 10

. The t statistic is (A)

0.40

(B)

2.00

(C)

4.00

(D)

0.04

Show worked answer →

The correct answer is (B).

$t = \dfrac{\bar{x} - \mu_0}{s/\sqrt{n}} = \dfrac{104 - 100}{10/\sqrt{25}} = \dfrac{4}{10/5} = \dfrac{4}{2} = 2.00$ , with $df = 24$ .

(A) divides by $s$ instead of the standard error. (C) forgets to divide by the standard error term properly. (D) misplaces a factor. The statistic is $2.00$ .

AP 2022 (style)4 marksSection II (free response). A bakery claims its loaves have a mean mass of

800

grams. A random sample of

n = 16

loaves has

\bar{x} = 788

grams and

s = 20

grams; a dotplot is roughly symmetric. Test at

\alpha = 0.05

whether the true mean mass is less than

800

grams. State hypotheses, check conditions, compute the t statistic and P-value (use

df = 15

), and conclude in context (justify in context).

Show worked answer →

A 4-point complete one-sample t-test.

(1) (1 point) Let $\mu$ be the true mean mass. $H_0: \mu = 800$ versus $H_a: \mu < 800$ .
(2) (1 point) Random sample stated; $n = 16 < 30$ but the dotplot is roughly symmetric with no outliers; $10\%$ condition reasonable. One-sample t-test appropriate.
(3) (1 point) $t = \dfrac{788 - 800}{20/\sqrt{16}} = \dfrac{-12}{20/4} = \dfrac{-12}{5} = -2.40$ , $df = 15$ . P-value $= P(t_{15} < -2.40) \approx 0.015$ .
(4) (1 point) Since P-value $\approx 0.015 < 0.05$ , reject $H_0$ . There is convincing evidence that the true mean mass of the loaves is less than $800$ grams.

Markers reward the t statistic with $df = 15$ , the lower-tail P-value, and a contextual reject conclusion.

Related dot points

Sources & how we know this

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