ConceptConceptDocsDocuments

AP Statistics 3.10 Two-Proportion Confidence Intervals

Practise choosing, checking, and calculating two-sample z-intervals for differences between population proportions, with group order kept explicit.

Syllabus
Effective Fall 2026
Course
AP Statistics

Exam points

  • Choose a two-sample z-interval for independent groups and define the parameter as p1 minus p2.
  • Verify independent random groups, the 10% condition, and at least 10 successes and failures per group.
  • Calculate p-hat1 minus p-hat2 plus or minus z-star times the unpooled standard error.

3.10 Constructing a Confidence Interval for the Difference Between Two Population Proportions question 1

[Maximum number: 1]

The campaign manager for a candidate for governor wants to estimate the difference between the percentage of voters who viewed an advertisement and favor the candidate and the percentage of voters who had not viewed the advertisement and favor the candidate. Two independent random samples of 500 voters were selected. One random sample consisted of voters who viewed the advertisement and the other random sample consisted of voters who had not viewed the advertisement. The table shows the frequencies of responses for those surveyed.

Table for Question 3.10 Constructing a Confidence Interval for the Difference Between Two Population Proportions question 1 — AP Statistics

Which of the following is the most appropriate method for the campaign manager to use to estimate the difference between the percentages of voters who favor the candidate?

A

A one-sample z-interval for estimating a population proportion

B

A t-interval for estimating the slope of the least-squares regression line

C

A matched-pairs t-interval for estimating a mean difference

D

A two-sample z-interval for estimating a difference between sample proportions

E

A two-sample z-interval for estimating a difference between population proportions

3.10 Constructing a Confidence Interval for the Difference Between Two Population Proportions question 2

[Maximum number: 1]

35. In a random sample of 1,500 college students, a pollster found 45\% prefer a female president and 42%42 \% prefer a male president. To calculate a 95 percent confidence interval for the difference in the proportion of college students who prefer a female president over a male president, he uses (0.450.42)±1.96(0.45)(0.55)1,500+(0.42)(0.58)1,500(0.45-0.42) \pm 1.96 \sqrt{\frac{(0.45)(0.55)}{1,500}+\frac{(0.42)(0.58)}{1,500}}. Were conditions for inference met?

A

Yes, because there was a random sample, 1,500 is less than 10\% of all college students, and 1,500(0.45), 1,500(0.55), 1,500(0.42), and 1,500(0.58) are all 10\geq 10.

B

No, because there was no random assignment between female and male college students.

C

No, because the independence assumption is violated.

D

No, because the random sample may not be truly representative of the population.

3.10 Constructing a Confidence Interval for the Difference Between Two Population Proportions question 3

[Maximum number: 1]

Two brands of golf cart batteries, R and S, were tested to investigate whether there is a difference between the proportions of fully charged batteries that could power a golf cart for at least thirty miles. A random sample of 50 brand R batteries was selected and a random sample of 50 brand S batteries was selected. Batteries were randomly assigned to one of 100 different golf carts. Each cart was driven until the battery failed. Of the carts with brand R batteries, 40 traveled over thirty miles, and of the carts with brand S batteries, 35 traveled over thirty miles. Which of the following is a 95 percent confidence interval for the difference in the proportions of fully charged batteries that can power a golf cart for at least thirty miles for the two brands?

Statistics 2021Intl MCQ

A

(0.80.7)±1.645(0.8)(0.2)50+(0.7)(0.3)50(0.8-0.7) \pm 1.645 \sqrt{\frac{(0.8)(0.2)}{50}+\frac{(0.7)(0.3)}{50}}

B

(0.80.7)±1.645(0.75)(0.25)(150+150)(0.8-0.7) \pm 1.645 \sqrt{(0.75)(0.25)\left(\frac{1}{50}+\frac{1}{50}\right)}

C

(0.80.7)±1.645(0.8)(0.2)50(0.7)(0.3)50(0.8-0.7) \pm 1.645 \sqrt{\frac{(0.8)(0.2)}{50}-\frac{(0.7)(0.3)}{50}}

D

(0.80.7)±1.960(0.8)(0.2)50+(0.7)(0.3)50(0.8-0.7) \pm 1.960 \sqrt{\frac{(0.8)(0.2)}{50}+\frac{(0.7)(0.3)}{50}}

E

(0.80.7)±1.960(0.75)(0.25)(150+150)(0.8-0.7) \pm 1.960 \sqrt{(0.75)(0.25)\left(\frac{1}{50}+\frac{1}{50}\right)}

All question bank results loaded