AP Statistics Inference for the Difference of Two Means Worksheet Solutions (PDF)

AP Statistics math worksheet solutions: Inference for the Difference of Two Means – Maths PDF

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This two means answers for AP Statistics covers inference for the difference of two means (Common Core AP STAT 7.6, AP STAT 7.7, AP STAT 7.8, AP STAT 7.9, HSS-IC.B.5), with a free PDF to print.

Each solution gives the conditions, the standard error, the degrees of freedom from a calculator, the interval or t statistic, and a conclusion in context. Probabilities and critical values use the exact (non-integer) degrees of freedom.

AP Statistics · Common Core State Standards: AP STAT 7.6, AP STAT 7.7, AP STAT 7.8, AP STAT 7.9, HSS-IC.B.5

Two means answers: exercises 1 to 10

Exercise 1 – Parameters and hypotheses

$\mu_A$ = true mean score of all students taught with Method A; $\mu_B$ likewise for Method B.

$H_0:\mu_A-\mu_B=0$, $H_a:\mu_A-\mu_B > 0$.

Exercise 2 – Difference and standard error

Difference: $78.4-72.1=6.3$.

$SE=\sqrt{\frac{9.2^{2}}{35}+\frac{11.5^{2}}{40}}=\sqrt{2.4183+3.3062}=2.393$.

6.3 points; SE = 2.393.

Exercise 3 – Degrees of freedom

Conservative: $\min(35-1,\ 40-1)=34$. Calculator: $df\approx 72.5$.

34 compared with 72.5; either gives a similar critical value.

Exercise 4 – Confidence interval

95% interval for mu_A – mu_B: (1.53, 11.07); 0 is not included.

$t^{*}=1.993$, $ME=4.769$. Interval: $6.3\pm 4.769$.

(1.531, 11.069) points.

Exercise 5 – Interpret the interval

State the parameter, the groups, and the direction.

We are 95% confident that the true mean score with Method A is between 1.531 and 11.069 points higher than with Method B. Because 0 is not in the interval, there is convincing evidence of a difference.

Exercise 6 – Check conditions

Boxplots of both classes: no outliers and roughly symmetric.

Random and independent: given. 10%: each sample is far less than 10% of its population. Normal/Large: both sample sizes are at least 30, and the boxplots show no outliers.

All conditions are met.

Exercise 7 – Compute t

$t=\frac{6.3}{2.393}=2.63$.

t = 2.63.

Exercise 8 – p-value

Two-sided p-value for t = 2.63: 0.0104.

$p=2P(t_{72.5} > 2.63)=0.0104$.

0.0104.

Exercise 9 – Conclusion

0.0104 < 0.05.

Reject H0. There is convincing evidence of a difference in the true mean scores of students taught with Method A and Method B (Method A higher).

Exercise 10 – Experiment with caffeine

$H_0:\mu_c-\mu_p=0,\ H_a:\mu_c-\mu_p > 0$. Random assignment; small samples, assume approximately normal.

$SE=1.963,\ df\approx 23.8,\ t=1.68$, p-value $=0.0530$.

Fail to reject H0: not convincing evidence that caffeine increases the mean score (the p-value 0.0530 exceeds 0.05).

Exercises 11 and up

Exercise 11 – From raw data

Fertilizer 1: mean 59.000, s 4.472. Fertilizer 2: mean 52.000, s 5.071.

$SE=2.390,\ df\approx 13.8,\ t^{*}=2.148,\ ME=5.135$.

(1.865, 12.135) cm.

Exercise 12 – Paired or independent?

(a) The same runners are measured twice: paired. (b) Different people in two groups: two-sample. (c) Twin pairs are matched: paired.

(a) paired; (b) two-sample; (c) paired.

Exercise 13 – Wrong procedure

The scores are dependent because they come from the same students.

Use a matched-pairs t procedure on the differences (after minus before) instead.

Exercise 14 – Interval and test

0 is inside the interval.

Fail to reject H0: there is not convincing evidence that the means differ.

Exercise 15 – Homework hours

$SE=0.260,\ df\approx 83.4,\ t=1.92$, p-value $=0.0583$.

Fail to reject H0: p-value 0.0583 exceeds 0.05, so there is not convincing evidence that the mean weekly homework time differs between boys and girls.

Exercise 16 – Effect of sample size

n = 25: $\sqrt{\frac{100}{25}+\frac{100}{25}}=\sqrt{8}=2.828$. n = 100: $\sqrt{\frac{100}{100}+\frac{100}{100}}=\sqrt{2}=1.414$.

The larger samples halve the standard error, giving a narrower interval and a larger |t| for the same difference.

Exercise 17 – Outlier problem

The outlier inflates the mean and standard deviation of group 2, and the sample is small.

The Normal condition fails (small n with an outlier), so the t interval may be misleading; investigate the outlier first.

Exercise 18 – 99 percent interval

$SE=1.072,\ df\approx 113.0,\ t^{*}=2.620,\ ME=2.810$.

(0.690, 6.310) minutes.

Exercise 19 – Multi-step

$t=2.33,\ df\approx 55.8$, p-value $=0.0117$.

90% interval: (0.652, 3.948).

Reject H0 (p < 0.05); the 90% interval is entirely positive, which agrees.

Exercise 20 – Describe the difference

0 is in the interval.

We are 95% confident that the true mean reaction time for men is between 8 ms less and 21 ms more than for women. Since 0 is plausible, there is not convincing evidence of a difference.

Key facts behind this two means answers

Key facts

Parameter: mu1 – mu2; statistic: x-bar1 – x-bar2.
SE = sqrt(s1^2/n1 + s2^2/n2); df from the calculator, or min(n1 – 1, n2 – 1) conservatively.
Interval: (x-bar1 – x-bar2) +/- t* SE; test: t = (x-bar1 – x-bar2)/SE with H0: mu1 – mu2 = 0.
Conditions: independent random samples or random assignment; 10%; each group normal or n >= 30.
Same subjects measured twice or matched pairs: use a one-sample t on the differences.

How to use this two means 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.

Download the two means answers as a PDF with the button below and keep it with your notes.

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