---
title: "AP Statistics Unit 3 Practice Test"
description: "A timed AP Statistics Unit 3 (Inference for Categorical Data: Proportions) practice test in the AP format, drawn from 5,150 multiple-choice questions and 25 FRQs written for the unit. The standard version takes about 88 minutes."
canonical: "https://fiveable.me/ap-stats/unit-test/3"
type: "unit-test"
subject: "AP Statistics"
unit: "Unit 3"
---

# AP Statistics Unit 3 Practice Test

## Overview

A timed AP Statistics Unit 3 (Inference for Categorical Data: Proportions) practice test in the AP format, drawn from 5,150 multiple-choice questions and 25 FRQs written for the unit. The standard version takes about 88 minutes.

## Test Format

- 20 Multiple Choice Questions (43 minutes)
- 1 FRQ – FRQ 2 – Multi-Focus on Practices 3 and 4 (23 minutes)
- 1 FRQ – FRQ 3 – Inference (23 minutes)

## Unit FRQs by Type

- [FRQ 2 – Multi-Focus on Practices 3 and 4](/ap-stats/frqs/frq-2?unit=3): 10 questions
- [FRQ 3 – Inference](/ap-stats/frq-practice?type=frq-3-inference&id=6a67c99a2de38de8e968cdb7): 10 questions
- [FRQ 4 – Multi-Focus on Practices 2, 3, and 4](/ap-stats/frqs/frq-4?unit=3): 5 questions

## Practice Questions

### Question 1

A school district surveys two high schools about a new schedule. At School A, 18 out of 150 students support the change. At School B, 16 out of 200 students support the change. Which expression calculates the 95% confidence interval for the difference in support proportions (School A minus School B)?

- (A) (0.12 - 0.08) ± 1.96 √((0.12)(0.88)/150 + (0.08)(0.92)/200)
- (B) (0.12 - 0.08) ± 1.96 √((0.10)(0.90)/150 + (0.10)(0.90)/200)
- (C) (0.12 - 0.08) ± 1.96 √((0.12)(0.88)/150 - (0.08)(0.92)/200)
- (D) (0.12 - 0.08) ± 1.96 (√((0.12)(0.88)/150) + √((0.08)(0.92)/200))

**Answer: (A)** Determining the interval parameters requires using the unpooled standard error formula with sample-specific proportions.

### Question 2

A quality control manager compares defect rates at two plants. In a sample of 50 items from Plant A, 5 are defective (proportion = 0.10). In a sample of 200 items from Plant B, 80 are defective (proportion = 0.40). What is the value of the standard error for the difference in sample proportions?

- (A) Approximately 0.055
- (B) Approximately 0.075
- (C) Approximately 0.077
- (D) Approximately 0.025

**Answer: (A)** Calculating the standard error requires summing the individual variances: √[(0.1)(0.9)/50 + (0.4)(0.6)/200] ≈ 0.055.

### Question 3

An analyst compares return rates for two product categories. For electronics, 30 of 200 items were returned (15%). For clothing, 10 of 200 items were returned (5%). Calculating a 90% confidence interval for the difference in proportions (Electronics minus Clothing) yields which result?

- (A) (0.051, 0.149)
- (B) (0.042, 0.158)
- (C) (0.067, 0.133)
- (D) (-0.049, 0.049)

**Answer: (A)** Constructing the interval uses z* = 1.645 and the unpooled standard error ≈ 0.030, resulting in 0.10 ± 0.049.

### Question 4

A researcher plans to estimate the difference in voter support between two districts with a 95% confidence level and a margin of error of 0.05. Assuming the proportions in both districts are 0.5 to maximize variability, what is the minimum sample size required for each district (assuming n₁ = n₂)?

- (A) 769 voters per district
- (B) 385 voters per district
- (C) 200 voters per district
- (D) 1537 voters per district

**Answer: (A)** Solving for n requires setting the margin of error equation 0.05 = 1.96√[(0.5)(0.5)/n + (0.5)(0.5)/n] and isolating n.

### Question 5

A quality control manager finds 15 defective widgets in a random sample of 150. He calculates a 95% confidence interval to evaluate the claim that the true defect rate is 0.05. Which conclusion is supported by the interval?

- (A) Not supported, because 0.05 is outside the interval (0.052, 0.148)
- (B) Supported, because 0.05 is inside the interval (0.037, 0.163)
- (C) Supported, because 0.05 is close to the interval (0.052, 0.148)
- (D) Not supported, because the sample proportion 0.10 exceeds 0.05

**Answer: (A)** Calculating the interval (0.10 ± 1.96 * 0.0245) yields (0.052, 0.148). The claim is not supported because 0.05 lies below the lower boundary of plausible values.

### Question 6

In a random sample of 200 seeds, 150 germinate. A botanist calculates a 95% confidence interval to determine if a germination rate of 0.80 is plausible. Which statement correctly interprets the interval?

- (A) Plausible, because 0.80 is within the interval (0.690, 0.810)
- (B) Not plausible, because 0.80 is outside the interval (0.719, 0.781)
- (C) Not plausible, because 0.80 is outside the interval (0.725, 0.775)
- (D) Not plausible, because the sample proportion 0.75 is below 0.80

**Answer: (A)** Calculating the interval (0.75 ± 1.96 * 0.0306) yields (0.690, 0.810). Since 0.80 is contained within this range, it remains a plausible value for the population proportion.

### Question 7

A market researcher surveys 400 people and finds that 216 prefer Brand X. She calculates a 99% confidence interval to test the claim that the majority (>0.50) prefer Brand X. Is the claim supported?

- (A) No, because the interval (0.476, 0.604) contains the value 0.50
- (B) Yes, because the interval (0.515, 0.565) is entirely above 0.50
- (C) Yes, because the sample proportion 0.54 is greater than 0.50
- (D) No, because the interval (0.476, 0.604) is centered at 0.54

**Answer: (A)** Calculating the interval (0.54 ± 2.576 * 0.0249) yields (0.476, 0.604). Because the interval includes 0.50, we cannot conclude with 99% confidence that the true proportion is strictly greater than 0.50.

### Question 8

In a random sample of 500 voters, 300 support a new policy. A political scientist calculates a 95% confidence interval to test the claim that the support rate differs from 0.55. What is the correct conclusion?

- (A) Yes, because 0.55 is outside the interval (0.557, 0.643)
- (B) No, because 0.55 is inside the interval (0.544, 0.656)
- (C) Yes, because 0.55 is outside the interval (0.578, 0.622)
- (D) No, because the sample proportion 0.60 is close to 0.55

**Answer: (A)** Calculating the interval (0.60 ± 1.96 * 0.0219) yields (0.557, 0.643). Since 0.55 is not contained within these boundaries, there is convincing evidence of a difference.

### Question 9

A clinical trial involves 600 patients divided equally into three groups: Medication A, Medication B, and Placebo. The results show that 150 patients showed no improvement, while 450 improved. What is the expected count of patients showing no improvement in the Medication A group?

- (A) 50
- (B) 75
- (C) 150
- (D) 200

**Answer: (A)** Calculates the expected count using the group size (600 / 3 = 200) and outcome total: (200 × 150) / 600 = 50.

### Question 10

In a city with 1,000 registered voters, 400 live in District North and 600 live in District South. A specific proposition is supported by 600 voters citywide. If district residence and support for the proposition are independent, how many voters in District North are expected to support the proposition?

- (A) 200
- (B) 240
- (C) 360
- (D) 400

**Answer: (B)** Calculates the expected count using the formula (Row Total × Column Total) / Table Total: (400 × 600) / 1,000 = 240.

## Review the Unit

- [3.1 Estimators](/ap-stats/unit-3/biased-unbiased-point-estimates/study-guide/eZ5sR9XOkLB1o9KKpMHF)
- [3.2 Sampling Distributions for Sample Proportions](/ap-stats/unit-3/sampling-distributions-for-sample-proportions/study-guide/Ezxev8MPpv3mFKjV4Gq3)
- [3.3 Constructing a Confidence Interval for a Population Proportion](/ap-stats/unit-3/constructing-confidence-interval-for-population-proportion/study-guide/rYCExLGlPYtMNFiJOWAR)
- [3.4 Justifying a Claim Based on a Confidence Interval for a Population Proportion](/ap-stats/unit-3/justifying-claim-based-on-confidence-interval-for-population-proportion/study-guide/YeTpyj6nyq03j0AJO3Bm)
- [3.5 Setting Up a Test for a Population Proportion](/ap-stats/unit-3/setting-up-test-for-population-proportion/study-guide/QLu7hUN0rwtnxLF7YdBT)
- [3.6 p-Values](/ap-stats/unit-3/interpreting-p-values/study-guide/b0FEXf5MDjyQtz4Skz70)
- [3.7 Carrying Out a Test for a Population Proportion](/ap-stats/unit-3/concluding-test-for-population-proportion/study-guide/THZeUpkm11DAwnNb6p4g)
- [3.8 Potential Errors When Performing Tests](/ap-stats/unit-3/potential-errors-when-performing-tests/study-guide/YxhrmoLje3YYOwcocJrS)
- [3.9 Sampling Distributions for the Difference Between Sample Proportions](/ap-stats/unit-3/sampling-distributions-for-differences-sample-proportions/study-guide/VOvA8du6YHMjhEwB7lEW)
- [3.10 Constructing a Confidence Interval for the Difference Between Two Population Proportions](/ap-stats/unit-3/confidence-intervals-for-difference-two-proportions/study-guide/YjPeyk5dGYBwzENoOU6S)
- [3.11 Justifying a Claim Based on a Confidence Interval for the Difference Between Two Population Proportions](/ap-stats/unit-3/justifying-claim-based-on-confidence-interval-for-difference-population-proportions/study-guide/m6d2wvAbSyMqmNwEyt43)
- [3.12 Setting Up a Test for the Difference Between Two Population Proportions](/ap-stats/unit-3/setting-up-test-for-difference-two-population-proportions/study-guide/sJt2F9NwsQ2gihYEHGjP)
- [3.13 Carrying Out a Test for the Difference Between Two Population Proportions](/ap-stats/unit-3/carrying-out-test-for-difference-two-population-proportions/study-guide/mkdLDKdZAuQ6pIsV3aOL)
- [3.14 Setting Up a Chi-Square Test for Homogeneity or Independence](/ap-stats/unit-3/setting-up-chi-square-test-for-homogeneity-or-independence/study-guide/tZimSvE0pNy9ak5fjCmh)
- [3.15 Carrying Out a Chi-Square Test for Homogeneity or Independence](/ap-stats/unit-3/carrying-out-chi-square-test-for-homogeneity-or-independence/study-guide/AmRuD3egBrZMCLGaBpMw)
- [Unit 3 Overview: Inference for Categorical Data: Proportions](/ap-stats/unit-3/review/study-guide/qzicHbvrrBrvuJWSB5xP)

## Other Unit Tests

- [Unit 1: Exploring One–Variable Data and Collecting Data](/ap-stats/unit-test/1)
- [Unit 2: Probability, Random Variables, and Probability Distributions](/ap-stats/unit-test/2)
- [Unit 4: Inference for Quantitative Data: Means](/ap-stats/unit-test/4)
- [Unit 5: Regression Analysis](/ap-stats/unit-test/5)

## About This Document

Canonical Fiveable pages are available as Markdown at the same path plus `.md`.

- [llms.txt](https://fiveable.me/llms.txt): index of Fiveable's sections and URL patterns
- [llms-full.txt](https://fiveable.me/llms-full.txt): complete subject and unit listing
- [MCP server](https://fiveable.me/mcp): call Fiveable as tools instead of fetching pages (`https://fiveable.me/api/mcp`)
- [MCP server for AP teachers](https://fiveable.me/mcp/teachers): a teacher's classes, assignments and AP-rubric grading (`https://fiveable.me/api/mcp/teacher`)
