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Last updated: October 22, 2026 · Written by the Math Made Easy math teachers
Standards: AP Statistics Unit 4 (2026-27) · Level: Grade 12 AP Statistics
Problem 2 Finding a Critical Value
(a) df = 11, t* = 2.201. (b) df = 19, t* = 2.861. (c) df = 7, t* = 1.895. (calculator: invT(0.975, 11), invT(0.995, 19), invT(0.95, 7)).
(a) 2.201; (b) 2.861; (c) 1.895
Problem 3 A One-Sample t Interval
df = 15, t* = 2.131. SE = 0.85/√16 = 0.2125. ME = 2.131 × 0.2125 = 0.4529. Interval: 4.62 ± 0.453.
(4.17, 5.07) pounds
Problem 4 Checking Conditions
The sample is small (n = 10), so the t procedure needs the data to show no strong skew or outliers. The dotplot has one value far above the rest (15 minutes), which is an outlier, so the normal/large sample condition is not met.
Not appropriate: strong outlier with n = 10
Problem 5 Interpreting an Interval
We are 95% confident that the mean weekly screen time of all students at the school is between 18.47 and 24.33 hours. The confidence level refers to the method: about 95% of intervals constructed this way from random samples capture the true mean.
95% confident the true mean is in (18.47, 24.33)
Problem 6 Writing Hypotheses for Means
(a) H0: μ = 450, Ha: μ < 450 (μ is the true mean loaf weight). (b) H0: μ = 8, Ha: μ ≠ 8 (μ is the true mean sleep time on school nights). (c) H0: μ = 12, Ha: μ > 12 (μ is the true mean battery life in hours).
(a) μ < 450; (b) μ ≠ 8; (c) μ > 12
Problem 7 Test Statistic and p-Value
SE = 4.1/√22 = 0.8741. t = (448.2 − 450)/0.8741 = −2.059, df = 21. p-value = tcdf(−∞, −2.059, 21) = 0.0260.
t ≈ −2.06, p-value ≈ 0.0260
Problem 8 Two-Sided t Test
SE = 1.3/√30 = 0.2373. t = (7.4 − 8)/0.2373 = −2.528, df = 29. Two-sided p-value = 0.0172. Since 0.0172 < 0.05, reject H0: there is convincing evidence that the mean sleep time differs from 8 hours.
t ≈ −2.53, p-value ≈ 0.0172; reject H0
Problem 9 A Claim from an Interval
df = 23, t* = 2.069, ME = 0.633, interval (5.47, 6.73). Since 7 is outside the interval, a two-sided test at the 5% level rejects μ = 7: there is convincing evidence that the mean is not 7 hours.
Yes; 7 is outside (5.47, 6.73)
Problem 10 Sample Size and Interval Width
n = 9: t* = 2.306, ME = 2.306 × 12/3 = 9.22. n = 36: t* = 2.030, ME = 2.030 × 12/6 = 4.06. The small sample is wider because the standard error is larger and the critical value is larger.
ME ≈ 9.22 vs 4.06
Problem 11 Paired or Two-Sample
(a) The same runners are measured twice: paired. (b) Two unrelated groups: two-sample. (c) Twins form natural pairs and treatments are randomized within each pair: paired.
(a) paired; (b) two-sample; (c) paired
Problem 12 Paired t Interval
The dotplot is roughly symmetric with no outliers, so the conditions are reasonable. d̄ = 3.750, sd = 2.493. df = 7, t* = 2.365, ME = 2.365 × 2.493/√8 = 2.084.
(1.67, 5.83) points
Problem 13 Paired t Test
H0: μd = 0; Ha: μd > 0. d̄ = 10.0, sd = 11.106. t = 10.0/(11.106/√10) = 2.847, df = 9, p-value = 0.0096. Since 0.0096 < 0.05, reject H0: there is convincing evidence that texting increases the mean braking reaction time.
t ≈ 2.85, p-value ≈ 0.0096; reject H0
Problem 14 Two-Sample t Interval
SE = √(5.22/28 + 6.12/31) = √(0.9657 + 1.2003) = 1.4717. Technology gives df ≈ 56.8, t* = 2.003. ME = 2.947. Difference = 3.5.
(0.55, 6.45) grams
Problem 15 Two-Sample t Test
The boxplots are roughly symmetric with no outliers, and both samples are around 30. H0: μ1 − μ2 = 0; Ha: μ1 − μ2 ≠ 0. SE = 2.7481, t = 5.3/2.7481 = 1.929, df ≈ 53.4, p-value = 0.0591. Since 0.0591 > 0.05, fail to reject H0: not convincing evidence that the means differ.
t ≈ 1.93, p-value ≈ 0.059; fail to reject H0
Problem 16 Interval and Test Agree
t* = 2.005, ME = 5.511. Interval: 5.3 ± 5.51 gives (−0.21, 10.81). The interval contains 0, which agrees with failing to reject H0 at α = 0.05.
(−0.21, 10.81); contains 0
Problem 17 Selecting a Procedure
(a) Two-sample z interval or test for a difference of proportions. (b) One-sample t interval for a mean. (c) Paired t procedure (one-sample t on the differences). (d) Chi-square test for homogeneity.
(a) two-proportion z; (b) one-sample t; (c) paired t; (d) chi-square homogeneity
Problem 18 A Full Two-Sample Test
State: H0: μA = μB; Ha: μA ≠ μB. Plan: two-sample t test; random independent samples; the boxplots below show no outliers and no strong skew. Do: x̄A = 73.11, sA = 5.925; x̄B = 65.00, sB = 4.346; SE = 2.406; t = 3.371; df ≈ 14.58; p-value = 0.0043. Conclude: 0.0043 < 0.05, so there is convincing evidence that the mean scores differ.
t ≈ 3.37, df ≈ 14.6, p-value ≈ 0.0043; reject H0
Problem 19 Confidence Level and Two-Sided Tests
(a) 50 is outside the interval, so reject H0 at α = 0.01. (b) 45 is inside, so fail to reject: 45 is a plausible value. (c) A lower confidence level uses a smaller t*, so the interval is narrower.
(a) reject; (b) fail to reject; (c) narrower
Problem 20 Why Pairing Matters
(a) Differences (before minus after): 4, 6, 1, 8, 8, 3, 9, 5; d̄ = 5.500, sd = 2.777; t = 5.601, df = 7, p-value = 0.00082. (b) Treating the groups as independent: t = 1.564, p-value = 0.1429. (c) The paired analysis removes the person-to-person variation in pulse and gives a much stronger result. The independent analysis is wrong here because the two measurements are linked.
Paired: t ≈ 5.60, p ≈ 0.0008; independent: t ≈ 1.56, p ≈ 0.14
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