AP Statistics t Intervals and t Tests for a Mean Worksheet Solutions (PDF)

AP Statistics math worksheet solutions: t Intervals and t Tests for a Mean – Maths PDF

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This inference one mean answers for AP Statistics covers t intervals and t tests for a mean (Common Core AP STAT 7.1, AP STAT 7.2, AP STAT 7.3, AP STAT 7.4, AP STAT 7.5, HSS-IC.B.4), with a free PDF to print.

Each solution states the conditions, the degrees of freedom, the critical value or t statistic, and a conclusion in context. Values are rounded to three decimals for intervals and four for p-values.

AP Statistics · Common Core State Standards: AP STAT 7.1, AP STAT 7.2, AP STAT 7.3, AP STAT 7.4, AP STAT 7.5, HSS-IC.B.4

Inference one mean answers: exercises 1 to 10

Exercise 1 – t versus normal

The t distribution has heavier tails than the normal; it approaches the normal as degrees of freedom grow.

The t curve is shorter in the middle and has more area in the tails.

Heavier tails and a lower peak; the t distribution depends on df and approaches the normal as df increases.

Exercise 2 – Degrees of freedom and t*

n = 21: df = 20, $t^{*}=2.086$. n = 31: df = 30, $t^{*}=2.750$.

df 20, t* = 2.086; df 30, t* = 2.750.

Exercise 3 – Standard error

$SE=\frac{1.4}{\sqrt{25}}=0.28$.

0.28 hours.

Exercise 4 – Compute an interval

$df=24,\ t^{*}=2.064$. $ME=2.064(0.28)=0.578$.

(6.622, 7.778) hours.

Exercise 5 – Interpret the interval

We name the parameter: the true mean battery life.

We are 95% confident that the interval from 38.96 to 41.75 hours captures the true mean life of all batteries of this type.

Exercise 6 – Check the Normal condition

The sample is small and has a strong outlier at 30.

No: with n = 10 and an outlier, the Normal condition is not satisfied, so a t interval is not appropriate.

Exercise 7 – Write hypotheses

Let $\mu$ be the true mean delivery time.

$H_0:\mu=30$, $H_a:\mu>30$.

Exercise 8 – Compute t

$SE=\frac{6.5}{\sqrt{40}}=1.028$. $t=\frac{33.2-30}{1.028}=3.11$.

t = 3.11.

Exercise 9 – p-value

Right-tail p-value.

$P(t_{39} > 3.11)=0.0017$.

0.0017.

Exercise 10 – Conclusion

0.0017 < 0.05.

We reject H0. There is convincing evidence that the mean delivery time is greater than 30 minutes.

Exercises 11 and up

Exercise 11 – Two-sided t test

$H_0:\mu=98.6,\ H_a:\mu\neq 98.6$. $SE=0.1833,\ t=-2.18,\ df=35$, p-value $=0.0361$.

Reject H0: convincing evidence that the true mean differs from 98.6 degrees F.

Exercise 12 – Interval and test

40 lies outside the interval.

Reject H0: there is convincing evidence that mu differs from 40.

Exercise 13 – Matched pairs differences

Differences: 4, 5, 2, 8, 2, 4, 5, 2.

Mean $\overline{x}_d=4.000$; $s_d=2.070$.

Differences (before minus after) in mg/dL.

Mean difference 4.000; standard deviation 2.070.

Exercise 14 – Matched pairs interval

$SE=\frac{2.070}{\sqrt{8}}=0.732$. $df=7,\ t^{*}=2.365$. $ME=1.731$.

(2.269, 5.731) mg/dL. The interval is positive, which suggests the diet lowers cholesterol.

Exercise 15 – Matched pairs test

$t=\frac{4.000}{0.732}=5.47$, df = 7, p-value $=0.0005$.

Reject H0: convincing evidence that the diet lowers mean cholesterol.

Exercise 16 – Effect of n

n = 16: $2.131(10/4)=5.329$. n = 64: $1.998(10/8)=2.498$.

5.33 compared with 2.50: the larger sample has about half the margin of error.

Exercise 17 – Read output

$SE=\frac{6.8}{\sqrt{20}}=1.52$. $t^{*}=1.729$, $ME=2.628$.

(49.672, 54.928) minutes.

Exercise 18 – Outlier warning

The extreme value is an outlier and the sample is small.

The Normal condition fails: the outlier inflates the mean and standard deviation, so the t interval would be misleading.

Exercise 19 – Left-tailed test

$H_0:\mu=25,\ H_a:\mu < 25$. $SE=0.552,\ t=-1.63,\ df=49$, p-value $=0.0548$.

Fail to reject H0: not convincing evidence that the mean is below 25 hours.

Exercise 20 – Multi-step

$df=29,\ t^{*}=2.756,\ SE=32.86,\ ME=90.58$.

($1,259.42, $1,440.58). $1,300 is inside the interval, so it is a plausible value for the mean rent.

Key facts behind this inference one mean answers

Key facts

t distribution: bell-shaped, centered at 0, heavier tails; df = n – 1.
Interval: x-bar +/- t* s/sqrt(n); test: t = (x-bar – mu0)/(s/sqrt(n)).
Conditions: random, 10% condition, and Normal/Large Sample (population normal, n >= 30, or no strong skew or outliers).
As df grows, t* approaches z*.
Matched pairs: take differences and do a one-sample t procedure.

How to use this inference one mean answers

Cover each solution, solve the exercise again, then uncover it and compare every line. If your answer differs, find the first step that changed and redo the exercise from there.

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