AP Statistics Practice Quiz: Carrying Out a Chi-Square Test for Goodness of Fit

Written by AP Content Team, Verified for 2026 AP Exams, Last updated: May 2026

Test your understanding with short quizzes. This quiz has 16 questions to check your progress.

Question 1 of 16

A chi-square test for goodness of fit is conducted to see if the distribution of colors in a bag of candies matches the manufacturer's stated proportions for six different colors. How many degrees of freedom should be used for this test?

A)6B)5C)7D)Cannot be determined from the information given

All Questions (16)

A) 6

B) 5

C) 7

D) Cannot be determined from the information given

Correct Answer: B

The degrees of freedom (df) for a chi-square test for goodness of fit are calculated as the number of categories minus 1. In this case, there are 6 color categories, so df = 6 - 1 = 5.

In a chi-square goodness-of-fit test, one category has an observed count of 30 and an expected count of 20. What is the contribution of this single category to the overall chi-square test statistic?

A) 0.5

B) 1.0

C) 5.0

D) 10.0

Correct Answer: C

The contribution for each category to the chi-square statistic is calculated as (Observed - Expected)^2 / Expected. For this category, the calculation is (30 - 20)^2 / 20 = 10^2 / 20 = 100 / 20 = 5.0.

A researcher expects a 1:1:2 ratio for three different outcomes in a genetic cross. They observe counts of 25, 35, and 40, respectively, from a sample of 100. What is the chi-square test statistic?

A) 0.0

B) 4.0

C) 6.0

D) 8.0

Correct Answer: C

The total sample size is 100. The expected counts based on the 1:1:2 ratio are (1/4)*100=25, (1/4)*100=25, and (2/4)*100=50. The chi-square statistic is the sum of (O-E)^2/E for each category: (25-25)^2/25 + (35-25)^2/25 + (40-50)^2/50 = 0 + 100/25 + 100/50 = 0 + 4 + 2 = 6.0.

Which of the following best defines the p-value in the context of a chi-square test for goodness of fit?

A) The probability that the null hypothesis is true.

B) The probability of obtaining a test statistic as or more extreme than the observed value, assuming the null hypothesis is true.

C) The probability that the alternative hypothesis is true.

D) The probability of obtaining the observed counts exactly, assuming the null hypothesis is true.

Correct Answer: B

This is the standard definition of a p-value. It measures the strength of evidence against the null hypothesis (H0) by calculating the likelihood of the observed result (or a more extreme one) if H0 were actually true.

A chi-square test for goodness of fit was performed to determine if the days of the week for car accidents at a certain intersection occur with equal frequency. The test resulted in a p-value of 0.24. Assuming a significance level of α = 0.05, what is the correct interpretation of the p-value?

A) There is strong evidence that accidents occur with equal frequency.

B) There is strong evidence that accidents do not occur with equal frequency.

C) Assuming accidents occur with equal frequency, there is a 24% chance of observing a distribution as or more varied than the one in the sample.

D) There is a 24% chance that the null hypothesis of equal frequencies is true.

Correct Answer: C

The p-value is the probability of observing a test statistic as or more extreme than the one calculated, given the null hypothesis is true. This option correctly interprets this probability in the context of the problem. Options A and B state conclusions, and option D incorrectly defines the p-value as the probability of H0 being true.

A botanist conducts a chi-square goodness-of-fit test to see if the distribution of flower phenotypes in a cross follows the expected Mendelian ratio. The calculated p-value is 0.02. Using a significance level of α = 0.05, what is the appropriate decision?

A) Reject H0, because the p-value is less than α.

B) Fail to reject H0, because the p-value is less than α.

C) Reject H0, because the p-value is greater than α.

D) Fail to reject H0, because the p-value is greater than α.

Correct Answer: A

The decision rule for a hypothesis test is: If the p-value is less than or equal to the significance level (α), we reject the null hypothesis (H0). Here, 0.02 is less than 0.05, so the correct decision is to reject H0.

A chi-square test for goodness of fit comparing observed and expected frequencies of four categories resulted in a p-value of 0.03. Using a significance level of α = 0.05, which statement provides the best justification for a claim about the population?

A) We fail to reject H0. There is not convincing evidence that the true distribution differs from the expected distribution.

B) We reject H0. There is convincing evidence that the true distribution of outcomes differs from the expected distribution.

C) We reject H0. The data proves that the true distribution of outcomes is different from the expected distribution.

D) We fail to reject H0. There is convincing evidence that the true distribution of outcomes is the same as the expected distribution.

Correct Answer: B

Since the p-value (0.03) is less than α (0.05), we reject the null hypothesis. This provides convincing evidence for the alternative hypothesis, which states that the population distribution is different from the one specified in H0. Option C uses the word 'proves,' which is too strong for a statistical test. Option D incorrectly accepts H0.

The chi-square test statistic for a goodness-of-fit test is calculated using which of the following formulas?

A) The sum of (Observed - Expected) / Expected

B) The sum of (Observed - Expected)^2 / Observed

C) The sum of (Observed - Expected)^2 / Expected

D) The sum of (Observed + Expected)^2 / Expected

Correct Answer: C

The formula for the chi-square test statistic is the sum of the squared differences between observed and expected counts, divided by the expected counts for each category.

In a chi-square goodness-of-fit test, a very large value for the test statistic (χ²) indicates that:

A) The observed counts are very close to the expected counts.

B) The p-value will be large, leading to a failure to reject H0.

C) The differences between the observed and expected counts are large relative to the expected counts.

D) The null hypothesis is very likely to be true.

Correct Answer: C

The chi-square statistic formula, Σ[(Observed - Expected)^2 / Expected], shows that large differences between observed and expected counts result in a large test statistic. A large test statistic suggests the data does not fit the null hypothesis well.

A chi-square test for goodness of fit is performed, and the results provide statistical reasoning to reject the null hypothesis. This outcome suggests that:

A) The sample data provides convincing evidence that the population distribution matches the one specified in the null hypothesis.

B) The research question can be answered with certainty.

C) The sample data provides convincing evidence that the population distribution does not match the one specified in the null hypothesis.

D) The randomization distribution must have been used instead of the theoretical chi-square distribution.

Correct Answer: C

Rejecting the null hypothesis (H0) means there is sufficient statistical evidence to support the alternative hypothesis (Ha). In a goodness-of-fit test, Ha states that the population distribution is different from the one hypothesized in H0. This connects the test result to the research question.

A market researcher concludes that there is convincing evidence that customer preferences for four different packaging designs are not equally distributed. For this conclusion to be justified based on a chi-square goodness-of-fit test with α = 0.10, which of the following must be true?

A) The p-value was greater than 0.10.

B) The p-value was less than or equal to 0.10.

C) The degrees of freedom were equal to 4.

D) The observed counts were exactly equal to the expected counts.

Correct Answer: B

A conclusion of 'convincing evidence' implies that the null hypothesis was rejected. The decision rule is to reject H0 if the p-value is less than or equal to the significance level, alpha. Therefore, the p-value must have been ≤ 0.10.

According to the provided content, what are the two primary methods for determining the p-value for a chi-square test for goodness of fit?

A) Using a z-table or a t-table.

B) Using a randomization distribution or a normal distribution.

C) Using a chi-square distribution table or technology.

D) Calculating it directly from the (Observed - Expected) values.

Correct Answer: C

The content explicitly states, 'The p-value for a chi-square test is found using a table or technology.' For a chi-square test, the appropriate theoretical distribution is the chi-square distribution, which can be referenced via a table or calculated with technology.

In a chi-square goodness-of-fit test, if all other factors remain constant, what is the effect of increasing the magnitude of the differences between the observed and expected counts?

A) The chi-square statistic will decrease, and the p-value will increase.

B) The chi-square statistic will increase, and the p-value will decrease.

C) The chi-square statistic will increase, and the p-value will increase.

D) The chi-square statistic will decrease, and the p-value will decrease.

Correct Answer: B

Larger differences between observed and expected counts lead to a larger numerator in the chi-square formula, which increases the overall test statistic. A larger, more extreme test statistic corresponds to a smaller area in the tail of the chi-square distribution, resulting in a smaller p-value.

A city planner tests the claim that the usage of five city parks is not equal. A chi-square goodness-of-fit test yields a test statistic of 15.2 with 4 degrees of freedom, resulting in a p-value of 0.004. Which statement provides the most complete statistical justification for the planner's claim?

A) The p-value of 0.004 is very small, so we reject the null hypothesis.

B) Because the p-value of 0.004 is less than a standard alpha level like 0.05, we reject the null hypothesis that park usage is equal and conclude there is convincing evidence that park usage is not equal.

C) The test statistic of 15.2 is large, which proves that the distribution of park usage is not equal.

D) The null hypothesis that park usage is equal is rejected because the observed values were different from the expected values.

Correct Answer: B

This statement is the most complete because it follows the full process of statistical reasoning: it compares the p-value to a significance level (alpha), states the decision about the null hypothesis, and provides a conclusion in the context of the problem with appropriate statistical language ('convincing evidence').

The degrees of freedom for a chi-square test for goodness of fit are calculated as:

A) The sample size - 1.

B) The number of categories - 1.

C) The sample size - 2.

D) The number of categories - 2.

Correct Answer: B

This is the direct definition provided in the content for calculating degrees of freedom in a chi-square goodness-of-fit test.

A study was conducted to see if the distribution of M&M's colors in a large bag follows the proportions claimed by the company. A chi-square goodness-of-fit test produced a test statistic of χ² = 11.5. The resulting p-value is 0.042. What does this p-value of 0.042 represent?

A) There is a 4.2% chance that the company's claimed proportions are correct.

B) There is a 4.2% chance that the company's claimed proportions are incorrect.

C) Assuming the company's claimed proportions are correct, there is a 4.2% probability of getting a sample with color differences as large or larger than what was observed.

D) There is a 95.8% chance that the observed differences are due to random sampling variation alone.

Correct Answer: C

This is the precise interpretation of the p-value. It is a conditional probability, based on the assumption that the null hypothesis (the company's claimed proportions) is true. It represents the probability of observing a sample result as or more extreme than the one obtained.