IB Mathematics: Applications and Interpretation Quiz: Chi Squared Tests
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Chi Squared TestsQuestion 1 of 20

A city council claims that the distribution of residents across its five districts (A, B, C, D, E) is uniform. To test this claim, a sociologist surveys a random sample of 600 residents. They plan to conduct a chi-squared goodness-of-fit test.

What are the null hypothesis (H₀) and the number of degrees of freedom (df) for this test?

H₀: The resident distribution is not uniform; df = 5
H₀: The resident distribution is uniform; df = 4
H₀: The resident distribution is uniform; df = 599
H₀: The resident distribution is not uniform; df = 4
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IB Mathematics: Applications and Interpretation Quiz

IB Mathematics: Applications and Interpretation Quiz: Chi Squared Tests

Practice Chi Squared Tests in IB Mathematics: Applications and Interpretation with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Chi Squared Tests, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Mathematics: Applications and Interpretation.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

A city council claims that the distribution of residents across its five districts (A, B, C, D, E) is uniform. To test this claim, a sociologist surveys a random sample of 600 residents. They plan to conduct a chi-squared goodness-of-fit test.

What are the null hypothesis (H₀) and the number of degrees of freedom (df) for this test?

  1. H₀: The resident distribution is not uniform; df = 5
  2. H₀: The resident distribution is uniform; df = 4 (correct answer)
  3. H₀: The resident distribution is uniform; df = 599
  4. H₀: The resident distribution is not uniform; df = 4
Explanation: The null hypothesis (H₀) for a goodness-of-fit test is that the observed distribution fits the expected distribution. In this case, the expected distribution is uniform. So, H₀ is that the resident distribution is uniform. The number of degrees of freedom for a goodness-of-fit test is calculated as df=k−1df = k - 1, where kk is the number of categories. Here, there are 5 districts, so k=5k=5. Therefore, df=5−1=4df = 5 - 1 = 4.

Question 2

A survey was conducted to investigate a possible association between a person's preferred holiday type (Beach, City Break, Adventure) and their primary mode of transport (Plane, Train, Car). The results were collected from 300 people.

For a chi-squared test of independence on this data, what is the correct number of degrees of freedom?

  1. 2
  2. 4 (correct answer)
  3. 8
  4. 9
Explanation: The number of degrees of freedom for a chi-squared test of independence is calculated as df=(r−1)(c−1)df = (r - 1)(c - 1), where rr is the number of rows and cc is the number of columns in the contingency table. In this scenario, there are 3 holiday types (rows) and 3 modes of transport (columns). Therefore, df=(3−1)(3−1)=2×2=4df = (3 - 1)(3 - 1) = 2 \times 2 = 4.

Question 3

An investment firm surveys 500 clients, classifying them by risk tolerance (Low, Medium, High) and preferred investment type (Stocks, Bonds, Real Estate). There are 150 clients with High risk tolerance and 200 clients who prefer Stocks.

To perform a chi-squared test for independence, the expected frequency for High risk tolerance clients who prefer Stocks must be calculated. What is this value?

  1. 30
  2. 50
  3. 60 (correct answer)
  4. 70
Explanation: The formula for expected frequency in a test for independence is E=(row total)×(column total)grand totalE = \frac{(\text{row total}) \times (\text{column total})}{\text{grand total}}. Here, the row total for 'High risk' is 150, the column total for 'Stocks' is 200, and the grand total is 500. Plugging these values into the formula gives: E=150×200500=30000500=60E = \frac{150 \times 200}{500} = \frac{30000}{500} = 60.

Question 4

A survey asks 400 university students to name their faculty (Arts, Science, Engineering, Business) and their primary source of news (Social Media, Online Newspapers, Television). The goal is to see if these two variables are independent.

Which of the following describes the null hypothesis (H₀) for the appropriate chi-squared test?

  1. The proportion of students in each faculty is equal.
  2. The distribution of news sources is the same as the national average.
  3. Science students are more likely to read online newspapers than Arts students.
  4. The choice of news source is not associated with the student's faculty. (correct answer)
Explanation: When you encounter a chi-squared test for independence, you're examining whether two categorical variables are related to each other. The null hypothesis in these tests always states that the variables are independent—meaning one variable doesn't influence the distribution of the other. In this survey, you have two categorical variables: faculty (Arts, Science, Engineering, Business) and news source (Social Media, Online Newspapers, Television). The chi-squared test for independence will determine whether a student's faculty choice affects their preferred news source, or whether these choices are unrelated. The null hypothesis must state that faculty and news source are independent variables. This means the distribution of news source preferences should be roughly the same across all faculties—if 40% of all students prefer social media, then about 40% of Arts students, 40% of Science students, etc. should prefer social media. Option D correctly captures this: "The choice of news source is not associated with the student's faculty." Option A is wrong because it describes a goodness-of-fit test comparing faculty proportions, not a test of independence between two variables. Option B suggests comparing to external data (national averages), which isn't what this survey examines. Option C describes a specific alternative hypothesis about Science versus Arts students, but null hypotheses in independence tests never make claims about specific relationships—they always state no association exists. Remember: In chi-squared independence tests, the null hypothesis always claims the two variables are unrelated. Look for phrases like "not associated," "independent," or "no relationship."

Question 5

In a chi-squared test of independence, the calculated test statistic is χ2=7.5\chi^2 = 7.5. The test has 3 degrees of freedom. The p-value is found to be 0.057.

If the significance level was changed from α=0.05\alpha = 0.05 to α=0.10\alpha = 0.10, how would the conclusion of the test change?

  1. The conclusion would not change; H₀ would not be rejected in either case.
  2. The conclusion would not change; H₀ would be rejected in both cases.
  3. The conclusion would change from not rejecting H₀ to rejecting H₀. (correct answer)
  4. The conclusion would change from rejecting H₀ to not rejecting H₀.
Explanation: The p-value of 0.057 is constant regardless of the chosen significance level. At α=0.05\alpha = 0.05, the condition p<αp < \alpha is not met (since 0.057>0.050.057 > 0.05), so we would fail to reject H₀. At α=0.10\alpha = 0.10, the condition p<αp < \alpha is met (since 0.057<0.100.057 < 0.10), so we would reject H₀. Therefore, changing the significance level from 5% to 10% changes the conclusion from 'fail to reject' to 'reject'.

Question 6

Two chi-squared tests for independence are performed, each with 6 degrees of freedom. Test A results in a test statistic of χA2=10.5\chi^2_A = 10.5, and Test B results in a test statistic of χB2=18.2\chi^2_B = 18.2. Let pAp_A and pBp_B be the corresponding p-values.

Which statement accurately describes the relationship between pAp_A and pBp_B?

  1. pA<pBp_A < p_B
  2. The relationship cannot be determined without the significance level α\alpha.
  3. pA=pBp_A = p_B because the degrees of freedom are the same.
  4. pA>pBp_A > p_B (correct answer)
Explanation: When you encounter chi-squared test questions comparing p-values, remember that p-values have an inverse relationship with test statistics: larger test statistics produce smaller p-values, indicating stronger evidence against the null hypothesis. The p-value represents the probability of observing a test statistic as extreme or more extreme than what you calculated, assuming the null hypothesis is true. Since both tests have identical degrees of freedom (6), they use the same chi-squared distribution for comparison. With χA2=10.5\chi^2_A = 10.5 and χB2=18.2\chi^2_B = 18.2, Test B has a larger test statistic, meaning it falls further into the tail of the distribution. This translates to a smaller area under the curve beyond that point, hence pB<pAp_B < p_A, making pA>pBp_A > p_B. Option A suggests pA<pBp_A < p_B, which reverses the correct relationship between test statistics and p-values. Option B incorrectly claims you need the significance level α\alpha to compare p-values, but p-values are calculated independently of any chosen significance threshold. Option C makes the fundamental error of thinking equal degrees of freedom means equal p-values, ignoring that the test statistics themselves are different. Therefore, D is correct: pA>pBp_A > p_B. Study tip: For chi-squared tests (and all hypothesis tests), memorize this key relationship: larger test statistic → smaller p-value → stronger evidence against null hypothesis. This inverse relationship is crucial for interpreting results and comparing different tests.

Question 7

A market analyst wants to determine if customer satisfaction (categorized as Low, Medium, High) is independent of the product model purchased (Model X, Model Y, Model Z). They perform a chi-squared test and find that the component (O−E)2E\frac{(O-E)^2}{E} for the cell 'High Satisfaction' and 'Model Z' is unusually large compared to the other cells.

What is the best interpretation of this large component value?

  1. Model Z is the most popular product among all customers.
  2. Most customers who bought Model Z reported High Satisfaction.
  3. The number of Model Z customers with High Satisfaction is very different from what independence would predict. (correct answer)
  4. There must have been a calculation error for this specific cell's expected value.
Explanation: The chi-squared test statistic is the sum of the components (O−E)2E\frac{(O-E)^2}{E} for all cells. Each component measures the squared difference between the observed (O) and expected (E) counts, relative to the expected count. A large component value indicates a large discrepancy between what was observed and what would be expected if the two variables were independent. It pinpoints the specific category combination that contributes most to rejecting the null hypothesis.

Question 8

A goodness-of-fit test is used to determine if data on monthly sales follows a normal distribution. The data is grouped into 8 bins. The mean and standard deviation of the distribution are estimated from the sample data before calculating the expected frequencies for each bin.

What is the correct number of degrees of freedom for this chi-squared test?

  1. 5 (correct answer)
  2. 6
  3. 7
  4. 8
Explanation: The formula for degrees of freedom in a goodness-of-fit test where parameters are estimated from the data is df=k−1−mdf = k - 1 - m, where kk is the number of categories (bins) and mm is the number of estimated parameters. Here, there are k=8k=8 bins. Two parameters, the mean (μ\mu) and the standard deviation (σ\sigma), were estimated from the data. Thus, m=2m=2. The degrees of freedom are df=8−1−2=5df = 8 - 1 - 2 = 5.

Question 9

A chi-squared test for independence is planned. One of the conditions for the validity of this test is that all expected frequencies must be sufficiently large. The generally accepted minimum value is 5.

If, after calculating the expected frequencies from a contingency table, one cell has an expected frequency of 4.2, what is the most appropriate course of action?

  1. Proceed with the test, as the value is only slightly below the minimum threshold.
  2. Remove the data from that cell's row and column and recalculate everything.
  3. Combine the row or column containing that cell with an adjacent one, then re-calculate expected frequencies. (correct answer)
  4. Immediately reject the null hypothesis, as the condition for the test has not been met.
Explanation: When an expected frequency is less than 5, the chi-squared distribution may not be a good approximation, making the test results unreliable. The standard procedure is to combine categories. This involves merging the row or column with the low expected frequency with an adjacent one. This reduces the degrees of freedom but makes the test valid. Simply proceeding, removing data, or making a conclusion without a valid test are all incorrect actions.

Question 10

A pharmaceutical company tests a new drug to see if it causes a particular side effect. The null hypothesis is that the incidence of the side effect is independent of taking the drug. The test results lead to a Type I error.

What is the real-world consequence of this Type I error?

  1. The company concludes the drug has no association with the side effect when, in fact, it does.
  2. The company concludes the drug is associated with the side effect when, in fact, it is not. (correct answer)
  3. The company correctly concludes that the drug is associated with the side effect.
  4. The company fails to detect a real association between the drug and the side effect, potentially harming patients.
Explanation: A Type I error occurs when the null hypothesis (H₀) is rejected when it is actually true. In this context, H₀ is that the drug and the side effect are independent. Rejecting a true H₀ means concluding there is an association (dependence) when there isn't one. This could lead to the company issuing unnecessary warnings or even withdrawing a safe drug from the market.

Question 11

A genetic theory predicts that the offspring of a certain cross will exhibit four phenotypes in the ratio 9:3:3:1. An experiment yields 320 offspring.

Assuming the theory is correct, what is the expected frequency of the phenotype corresponding to the '3' in the ratio?

  1. 20
  2. 60 (correct answer)
  3. 80
  4. 180
Explanation: First, sum the parts of the ratio: 9+3+3+1=169 + 3 + 3 + 1 = 16. The proportion for the phenotype corresponding to '3' is 316\frac{3}{16}. The total number of offspring is 320. The expected frequency is this proportion multiplied by the total number: E=316×320=60E = \frac{3}{16} \times 320 = 60. There are two phenotypes with this expected frequency.

Question 12

In a survey of 200 commuters, 80 drove cars and 120 used public transport. Of the car drivers, 30 reported high stress levels, while of the public transport users, 70 reported high stress levels. A chi-squared test is used to test for independence between mode of transport and stress level.

What is the expected number of car drivers who report high stress levels, assuming transport mode and stress are independent?

  1. 30
  2. 40 (correct answer)
  3. 50
  4. 80
Explanation: To calculate the expected frequency, we use the formula: E=(row total)×(column total)grand totalE = \frac{(\text{row total}) \times (\text{column total})}{\text{grand total}}. The row for 'Car' has a total of 80. The column for 'High Stress' has a total of 30+70=10030 + 70 = 100. The grand total is 200. Therefore, the expected frequency is E=80×100200=8000200=40E = \frac{80 \times 100}{200} = \frac{8000}{200} = 40.

Question 13

A market analyst wants to determine if customer satisfaction (categorized as Low, Medium, High) is independent of the product model purchased (Model X, Model Y, Model Z). They perform a chi-squared test and find that the component (O−E)2E\frac{(O-E)^2}{E} for the cell 'High Satisfaction' and 'Model Z' is unusually large compared to the other cells.

What is the best interpretation of this large component value?

  1. Model Z is the most popular product among all customers.
  2. Most customers who bought Model Z reported High Satisfaction.
  3. The number of Model Z customers with High Satisfaction is very different from what independence would predict. (correct answer)
  4. There must have been a calculation error for this specific cell's expected value.
Explanation: The chi-squared test statistic is the sum of the components (O−E)2E\frac{(O-E)^2}{E} for all cells. Each component measures the squared difference between the observed (O) and expected (E) counts, relative to the expected count. A large component value indicates a large discrepancy between what was observed and what would be expected if the two variables were independent. It pinpoints the specific category combination that contributes most to rejecting the null hypothesis.

Question 14

An ecologist is studying the habitat preference of a bird species in a park with three distinct areas: Woodland, Wetland, and Grassland. The total areas are 50 ha, 20 ha, and 30 ha, respectively. The ecologist observes 240 birds.

If the birds show no preference and are distributed proportionally to the area of the habitats, what is the expected number of birds in the Wetland?

  1. 48 (correct answer)
  2. 80
  3. 96
  4. 120
Explanation: This is a goodness-of-fit problem where the expected distribution is based on the area of the habitats, not a uniform distribution. First, find the total area: 50+20+30=10050 + 20 + 30 = 100 ha. The proportion of the total area that is Wetland is 20100=0.2\frac{20}{100} = 0.2. The total number of birds observed is 240. The expected number of birds in the Wetland is the total number of birds multiplied by this proportion: E=240×0.2=48E = 240 \times 0.2 = 48.

Question 15

A manager wishes to test if the number of defects produced by four different machines (A, B, C, D) is the same. Over one week, the machines produce 22, 17, 28, and 25 defects, respectively. A goodness-of-fit test is performed.

What is the null hypothesis for this test?

  1. H₀: The average number of defects is 23.
  2. H₀: There is a significant difference in the number of defects between machines.
  3. H₀: The number of defects is independent of the machine.
  4. H₀: The number of defects is uniformly distributed across the four machines. (correct answer)
Explanation: The manager wants to test if the number of defects is the same for each machine. This is a goodness-of-fit test comparing the observed defect counts to a theoretical distribution. The hypothesis that the counts are the same is equivalent to stating that the defects are uniformly distributed across the machines. H₀ must state the specific distribution being tested against. 'Independent' relates to a test of independence, not goodness-of-fit.

Question 16

A researcher is investigating whether the choice of smartphone operating system (OS) is independent of a person's age group. They survey a large sample and categorize respondents into four age groups and three OS categories. A chi-squared test for independence is performed at a 5% significance level, yielding a p-value of 0.023.

Based on the result of the test, which of the following is the most appropriate conclusion?

  1. Since the p-value is less than 0.05, the null hypothesis is accepted, proving that OS choice and age group are independent.
  2. Since the p-value is less than 0.05, there is insufficient evidence to suggest an association between OS choice and age group.
  3. Since the p-value is less than 0.05, the null hypothesis is rejected, suggesting there is a significant association between OS choice and age group. (correct answer)
  4. Since the p-value is very small, it is proven that a person's age group causes them to choose a particular smartphone OS.
Explanation: The decision rule for a hypothesis test is to compare the p-value to the significance level (α\alpha). Here, p=0.023p = 0.023 and α=0.05\alpha = 0.05. Since p<αp < \alpha, we reject the null hypothesis (H₀). For a test of independence, H₀ states that the two variables are independent (have no association). Rejecting H₀ means we have found significant evidence of an association between the variables. Therefore, there is a significant association between OS choice and age group.

Question 17

A city council claims that the distribution of residents across its five districts (A, B, C, D, E) is uniform. To test this claim, a sociologist surveys a random sample of 600 residents. They plan to conduct a chi-squared goodness-of-fit test.

What are the null hypothesis (H₀) and the number of degrees of freedom (df) for this test?

  1. H₀: The resident distribution is not uniform; df = 5
  2. H₀: The resident distribution is uniform; df = 4 (correct answer)
  3. H₀: The resident distribution is uniform; df = 599
  4. H₀: The resident distribution is not uniform; df = 4
Explanation: The null hypothesis (H₀) for a goodness-of-fit test is that the observed distribution fits the expected distribution. In this case, the expected distribution is uniform. So, H₀ is that the resident distribution is uniform. The number of degrees of freedom for a goodness-of-fit test is calculated as df=k−1df = k - 1, where kk is the number of categories. Here, there are 5 districts, so k=5k=5. Therefore, df=5−1=4df = 5 - 1 = 4.

Question 18

A pharmaceutical company tests a new drug to see if it causes a particular side effect. The null hypothesis is that the incidence of the side effect is independent of taking the drug. The test results lead to a Type I error.

What is the real-world consequence of this Type I error?

  1. The company concludes the drug has no association with the side effect when, in fact, it does.
  2. The company concludes the drug is associated with the side effect when, in fact, it is not. (correct answer)
  3. The company correctly concludes that the drug is associated with the side effect.
  4. The company fails to detect a real association between the drug and the side effect, potentially harming patients.
Explanation: A Type I error occurs when the null hypothesis (H₀) is rejected when it is actually true. In this context, H₀ is that the drug and the side effect are independent. Rejecting a true H₀ means concluding there is an association (dependence) when there isn't one. This could lead to the company issuing unnecessary warnings or even withdrawing a safe drug from the market.

Question 19

A genetic theory predicts that the offspring of a certain cross will exhibit four phenotypes in the ratio 9:3:3:1. An experiment yields 320 offspring.

Assuming the theory is correct, what is the expected frequency of the phenotype corresponding to the '3' in the ratio?

  1. 20
  2. 60 (correct answer)
  3. 80
  4. 180
Explanation: First, sum the parts of the ratio: 9+3+3+1=169 + 3 + 3 + 1 = 16. The proportion for the phenotype corresponding to '3' is 316\frac{3}{16}. The total number of offspring is 320. The expected frequency is this proportion multiplied by the total number: E=316×320=60E = \frac{3}{16} \times 320 = 60. There are two phenotypes with this expected frequency.

Question 20

A goodness-of-fit test is conducted to see if a six-sided die is fair. The die is rolled 180 times. The calculated chi-squared test statistic is χ2=12.8\chi^2 = 12.8. The test is conducted at a 1% significance level.

Given that the critical value for this test at the 1% significance level is 15.086, what is the correct conclusion?

  1. Reject H₀ because 12.8>0.0112.8 > 0.01. There is evidence the die is not fair.
  2. Do not reject H₀ because 12.8<15.08612.8 < 15.086. There is insufficient evidence to conclude the die is not fair. (correct answer)
  3. Reject H₀ because 12.8<15.08612.8 < 15.086. There is sufficient evidence to conclude the die is fair.
  4. Do not reject H₀ because the p-value is less than 0.01. There is insufficient evidence to conclude the die is not fair.
Explanation: The decision rule using the critical value method is: if the test statistic is greater than the critical value, reject H₀. Here, the test statistic χ2=12.8\chi^2 = 12.8 and the critical value is 15.086. Since 12.8<15.08612.8 < 15.086, the test statistic does not fall in the rejection region. Therefore, we do not reject the null hypothesis (H₀). This means there is not enough evidence at the 1% significance level to conclude that the die is unfair.