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

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

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This two proportions answers for AP Statistics covers inference for the difference of two proportions (Common Core AP STAT 6.8, AP STAT 6.9, AP STAT 6.10, AP STAT 6.11, HSS-IC.B.5), with a free PDF to print.

Each solution lists the conditions, the sample proportions, the standard error, the interval or test statistic, and a conclusion in context. Values are rounded to four decimals for standard errors and p-values.

AP Statistics · Common Core State Standards: AP STAT 6.8, AP STAT 6.9, AP STAT 6.10, AP STAT 6.11, HSS-IC.B.5

Two proportions answers: exercises 1 to 10

Exercise 1 – State the parameters

\(p_A\) = true proportion of all students at School A who support the proposal; \(p_B\) likewise at School B.

\(H_0: p_A=p_B\); \(H_a: p_A\neq p_B\).

Exercise 2 – Check conditions

Random assignment: given. Counts: new app \(52,\ 28\); old app \(45,\ 45\), all at least 10. (The 10% condition is not needed with random assignment.)

Conditions are met.

Exercise 3 – Difference and standard error

\(\hat{p}_1=\frac{52}{80}=0.650\), \(\hat{p}_2=\frac{45}{90}=0.500\). Difference \(0.150\).

\(SE=\sqrt{\frac{0.650(1-0.650)}{80}+\frac{0.5(0.5)}{90}}=0.0750\).

Difference 0.150; SE 0.0750.

Exercise 4 – Confidence interval

95% interval.

\(ME=1.96(0.0750)=0.1470\). Interval: \(0.150\pm 0.147\).

(0.003, 0.297).

Exercise 5 – Interpret the interval

The interval does not contain 0 and is entirely positive.

We are 95% confident that the interval from 0.003 to 0.297 captures the true difference (new minus old) in the proportions of volunteers who improve. Since 0 is not in the interval, there is convincing evidence that the new app is better.

Exercise 6 – 90 percent interval

\(\hat{p}_1=0.45,\ \hat{p}_2=0.375,\ SE=0.0347\). \(ME=1.645(0.0347)=0.0571\).

(0.018, 0.132).

Exercise 7 – Pooled proportion

\(\hat{p}_c=\frac{126+90}{300+250}=\frac{216}{550}=0.3927\).

\(SE=\sqrt{0.3927(1-0.3927)\left(\frac{1}{300}+\frac{1}{250}\right)}=0.0418\).

0.3927 and 0.0418.

Exercise 8 – z statistic and p-value

Two-sided p-value for z = 1.43: 0.1527.

\(z=\frac{0.42-0.36}{0.0418}=1.43\). \(p\text{-value}=2P(Z > 1.43)=0.1527\).

z = 1.43; p-value = 0.1527.

Exercise 9 – Conclusion

0.1527 > 0.05.

We fail to reject H0. There is not convincing evidence that the proportions of students who support the later start time differ between School A and School B.

Exercise 10 – One-sided test

Right-tailed p-value.

\(H_0: p_{new}=p_{old},\ H_a: p_{new} > p_{old}\). \(\hat{p}_c=\frac{97}{170}=0.5706\). \(SE=0.0761,\ z=1.97\), p-value \(=0.0244\).

Reject H0: convincing evidence that the new app helps a greater proportion.

Exercises 11 and up

Exercise 11 – Interval and test agree

0 is not in the interval.

Reject H0; there is convincing evidence that p1 and p2 differ (p1 appears larger).

Exercise 12 – Two-way table

Placebo \(\frac{45}{300}=0.15\), vaccine \(\frac{18}{300}=0.06\).

\(SE=\sqrt{\frac{0.15(0.85)}{300}+\frac{0.06(0.94)}{300}}=0.0248\).

Interval (0.041, 0.139), all positive: the vaccine seems to lower the flu rate.

Exercise 13 – Significance in the flu study

\(P(Z < -3.60)=0.000159\).

The p-value is tiny, so reject H0. Because the volunteers were randomly assigned, there is convincing evidence that the vaccine reduces the chance of getting the flu.

Exercise 14 – Large counts fails

Successes: 4 and 9; failures: 56 and 41.

The large counts condition fails because 4 and 9 are less than 10.

Exercise 15 – Dependent samples

The same students are measured twice, so the samples are not independent.

The two groups are dependent (paired data), so the independence condition fails.

Exercise 16 – Margin of error comparison

Larger samples reduce the standard error and narrow the interval.

With larger samples, the same difference of 0.06 could give an interval that excludes 0 and a significant test result.

Exercise 17 – Sample proportions

\(\hat{p}_{men}=\frac{90}{200}=0.45\), \(\hat{p}_{women}=\frac{130}{250}=0.52\).

Difference 0.07 (women minus men).

Exercise 18 – Interval for the electric car survey

\(SE=\sqrt{\frac{0.52(0.48)}{250}+\frac{0.45(0.55)}{200}}=0.0473\). \(ME=0.0927\).

(-0.023, 0.163). 0 is inside, so there is not convincing evidence that the proportions differ.

Exercise 19 – Test for the electric car survey

\(\hat{p}_c=\frac{220}{450}=0.4889\). \(SE=0.0474,\ z=1.48\), p-value \(=0.1389\).

Fail to reject H0: not convincing evidence that the proportions differ.

Exercise 20 – Multi-step

\(SE=0.0559,\ ME=2.576(0.0559)=0.1441\).

(-0.039, 0.249). 0 is in the interval, so at 99% confidence there is no convincing evidence of a difference.

Key facts behind this two proportions answers

Key facts

Parameter: p1 – p2; statistic: p-hat1 – p-hat2.
Interval: (p-hat1 – p-hat2) +/- z* sqrt(p-hat1(1 – p-hat1)/n1 + p-hat2(1 – p-hat2)/n2).
Test: pooled p-hat_c = (x1 + x2)/(n1 + n2); z = (p-hat1 – p-hat2)/sqrt(p-hat_c(1 – p-hat_c)(1/n1 + 1/n2)).
Conditions: independent random samples (or random assignment); 10% condition; at least 10 successes and 10 failures in each group.
If the interval contains 0, a difference of 0 is plausible; if it does not, there is evidence of a difference.

How to use this two proportions 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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