AP Statistics Chi-Square Tests Worksheet Solutions (PDF)

AP Statistics math worksheet solutions: Chi-Square Tests – Maths PDF

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This chi square answers for AP Statistics covers chi-square tests (Common Core AP STAT 8.1, AP STAT 8.2, AP STAT 8.3, AP STAT 8.4, AP STAT 8.5, AP STAT 8.6, AP STAT 8.7), with a free PDF to print.

Each solution lists hypotheses, expected counts, conditions, the chi-square statistic with df, the p-value, and a conclusion in context. P-values come from the chi-square distribution and are rounded to four decimals.

AP Statistics · Common Core State Standards: AP STAT 8.1, AP STAT 8.2, AP STAT 8.3, AP STAT 8.4, AP STAT 8.5, AP STAT 8.6, AP STAT 8.7

Chi square answers: exercises 1 to 10

Exercise 1 – Expected counts for GOF

\(0.30(200)=60\), \(0.30(200)=60\), \(0.20(200)=40\), \(0.20(200)=40\).

Red 60, blue 60, green 40, yellow 40.

Exercise 2 – Degrees of freedom

(a) \(6-1=5\). (b) \((3-1)(4-1)=6\). (c) \((2-1)(2-1)=1\).

5; 6; 1.

Exercise 3 – Compute chi-square

\(\chi^{2}=\frac{(68-60)^{2}}{60}+\frac{(56-60)^{2}}{60}+\frac{(42-40)^{2}}{40}+\frac{(34-40)^{2}}{40}\)

\(=1.067+0.267+0.100+0.900=2.333\).

chi-square = 2.333, df = 3.

Exercise 4 – p-value

Right-tail p-value.

\(p\text{-value}=P(\chi^{2}_{3} > 2.333)=0.5062\).

0.5062 > 0.05: fail to reject H0. There is not convincing evidence that the color mix differs from the company’s claim.

Exercise 5 – Largest contribution

Contributions: red 1.067, blue 0.267, green 0.100, yellow 0.900.

Red contributes the most (1.067): there were 8 more red candies than expected.

Exercise 6 – Hypotheses

A die has 6 faces.

H0: the probability of each face is 1/6. Ha: at least one face has a probability different from 1/6.

Exercise 7 – Check conditions

Expected counts: 14, 12, 8, 4, 2.

No: two expected counts (4 and 2) are less than 5.

Exercise 8 – Fair die

120 rolls of a die.

Expected 20 each (all at least 5). \(\chi^{2}=5.00,\ df=5\), p-value \(=0.4159\).

Fail to reject H0: not convincing evidence that the die is unfair.

Exercise 9 – Expected counts in a table

\(E=\frac{100\times 80}{200}=40\) (row total 100 for sophomores, column total 80 for the library).

40.

Exercise 10 – Chi-square for the table

Observed (expected) counts. Expected = row total times column total over grand total.

Contributions: 2.50, 0.62, 1.25, 2.50, 0.62, 1.25.

\(\chi^{2}=8.750,\ df=2\).

chi-square = 8.750; df = 2.

Exercises 11 and up

Exercise 11 – Conclusion for the table

0.0126 < 0.05.

Reject H0. There is convincing evidence of an association between class (sophomore or senior) and preferred study location.

Exercise 12 – Homogeneity

Homogeneity test: \(H_0:\) the distribution of opinion is the same in all regions. Expected counts: 43.3, 56.7, 43.3, 56.7, 43.3, 56.7 (all at least 5).

\(\chi^{2}=8.82,\ df=2\), p-value \(=0.0121\).

Reject H0: convincing evidence that opinion differs among regions.

Exercise 13 – Independence or homogeneity?

(a) One sample, two variables: independence. (b) Separate samples from three populations, one variable: homogeneity.

(a) chi-square test for independence; (b) chi-square test for homogeneity.

Exercise 14 – Missing expected count

Expected: 30, 30, 40. \(\chi^{2}=\frac{0^{2}}{30}+\frac{(25-30)^{2}}{30}+\frac{(45-40)^{2}}{40}=0+0.833+0.625=1.458\).

chi-square = 1.458 with df = 2; the p-value is 0.4823, which gives no convincing evidence against the null proportions.

Exercise 15 – Wrong test

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

Chi-square assumes independent observations; these are paired data, so the test is not appropriate.

Exercise 16 – Small expected counts

Two expected counts are below 5.

No; combine categories if reasonable or collect more data.

Exercise 17 – 2 by 2 table

Expected counts: 28.0, 42.0, 52.0, 78.0 (all at least 5). \(\chi^{2}=36.63,\ df=1\), p-value \(=0.000000\).

Reject H0: convincing evidence of an association between age group and streaming.

Exercise 18 – Which procedure?

(a) Two-sample t. (b) Chi-square goodness-of-fit (or one-proportion z). (c) One-proportion z interval. (d) Chi-square test for independence.

(a) two-sample t test; (b) goodness-of-fit; (c) one-proportion z interval; (d) chi-square independence.

Exercise 19 – Multi-step

Expected 22.5 each. Contributions: 2.500, 8.100, 0.100, 2.500. \(\chi^{2}=13.20,\ df=3\), p-value \(=0.0042\).

Reject H0; convincing evidence that seasons are not equally popular. Summer contributes the most.

Exercise 20 – Interpret the p-value

The p-value is computed assuming no association.

If class level and car type were independent in the population, there would be about a 2% chance of getting a chi-square statistic at least this large in a random sample.

Key facts behind this chi square answers

Key facts

chi-square = sum of (O – E)^2 / E; right-tailed p-value; distribution skewed right.
Goodness-of-fit: one categorical variable; df = number of categories – 1.
Two-way table: E = row total x column total / grand total; df = (r – 1)(c – 1).
Conditions: random sample or random assignment, and all expected counts at least 5.
Homogeneity: compare distributions across groups; independence: two variables in one sample.

How to use this chi square 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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