Chi-Squared Tests - IB Mathematics Applications and Interpretation

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?
Answer: 48 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: $50 + 20 + 30 = 100$ ha. The proportion of the total area that is Wetland is $\frac{20}{100} = 0.2$ . 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: $E = 240 \times 0.2 = 48$ .
What is the best interpretation of this large component value?
Answer: The number of Model Z customers with High Satisfaction is very different from what independence would predict. The chi-squared test statistic is the sum of the components $\frac{(O-E)^2}{E}$ 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.
What are the null hypothesis (H₀) and the number of degrees of freedom (df) for this test?
Answer: H₀: The resident distribution is uniform; df = 4 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 $df = k - 1$ , where $k$ is the number of categories. Here, there are 5 districts, so $k=5$ . Therefore, $df = 5 - 1 = 4$ .
Assuming the theory is correct, what is the expected frequency of the phenotype corresponding to the '3' in the ratio?
Answer: 60 First, sum the parts of the ratio: $9 + 3 + 3 + 1 = 16$ . The proportion for the phenotype corresponding to '3' is $\frac{3}{16}$ . The total number of offspring is 320. The expected frequency is this proportion multiplied by the total number: $E = \frac{3}{16} \times 320 = 60$ . There are two phenotypes with this expected frequency.
What is the null hypothesis for this test?
Answer: H₀: The number of defects is uniformly distributed across the four machines. 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.
For a chi-squared test of independence on this data, what is the correct number of degrees of freedom?
Answer: 4 The number of degrees of freedom for a chi-squared test of independence is calculated as $df = (r - 1)(c - 1)$ , where $r$ is the number of rows and $c$ 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, $df = (3 - 1)(3 - 1) = 2 \times 2 = 4$ .
If the significance level was changed from $\alpha = 0.05$ to $\alpha = 0.10$ , how would the conclusion of the test change?
Answer: The conclusion would change from not rejecting H₀ to rejecting H₀. The p-value of 0.057 is constant regardless of the chosen significance level. At $\alpha = 0.05$ , the condition $p < \alpha$ is not met (since $0.057 > 0.05$ ), so we would fail to reject H₀. At $\alpha = 0.10$ , the condition $p < \alpha$ is met (since $0.057 < 0.10$ ), so we would reject H₀. Therefore, changing the significance level from 5% to 10% changes the conclusion from 'fail to reject' to 'reject'.
Which of the following describes the null hypothesis (H₀) for the appropriate chi-squared test?
Answer: The choice of news source is not associated with the student's faculty. 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."
What is the expected value of the new chi-squared test statistic?
Answer: 33.6 The formula for the chi-squared statistic is $\chi^2 = \sum \frac{(O - E)^2}{E}$ . Both observed (O) and expected (E) frequencies are proportional to the total sample size (N). Let $O = p_o N$ and $E = p_e N$ . Then a single term is $\frac{(p_o N - p_e N)^2}{p_e N} = \frac{N^2(p_o - p_e)^2}{p_e N} = N \frac{(p_o - p_e)^2}{p_e}$ . This shows that the $\chi^2$ statistic is directly proportional to the sample size N, provided the observed proportions remain the same. The sample size increased by a factor of $\frac{800}{200} = 4$ . Therefore, the new $\chi^2$ statistic will be $4 \times 8.4 = 33.6$ .
What has occurred and what is the probability of this type of error on any given test where H₀ is true?
Answer: A Type I error has occurred; the probability is 0.05. The null hypothesis (H₀) is that there is no difference (independence). The reality is that H₀ is true. The test concluded there is a difference, which means H₀ was rejected. Rejecting a true H₀ is the definition of a Type I error. The significance level, $\alpha$ , is the probability of making a Type I error. Since $\alpha = 0.05$ , the probability of this error is 0.05.
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?
Answer: 60 The formula for expected frequency in a test for independence is $E = \frac{(\text{row total}) \times (\text{column total})}{\text{grand total}}$ . 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: $E = \frac{150 \times 200}{500} = \frac{30000}{500} = 60$ .
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?
Answer: Combine the row or column containing that cell with an adjacent one, then re-calculate expected frequencies. 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.
What is the real-world consequence of this Type I error?
Answer: The company concludes the drug is associated with the side effect when, in fact, it is not. 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.
What is the expected number of car drivers who report high stress levels, assuming transport mode and stress are independent?
Answer: 40 To calculate the expected frequency, we use the formula: $E = \frac{(\text{row total}) \times (\text{column total})}{\text{grand total}}$ . The row for 'Car' has a total of 80. The column for 'High Stress' has a total of $30 + 70 = 100$ . The grand total is 200. Therefore, the expected frequency is $E = \frac{80 \times 100}{200} = \frac{8000}{200} = 40$ .
What is the correct interpretation of this result?
Answer: Since $0.35 > 0.10$ , we fail to reject H₀, meaning there is insufficient evidence to claim an association exists. The p-value is 0.35 and the significance level $\alpha$ is 0.10. Since $p \ge \alpha$ , we fail to reject the null hypothesis (H₀). H₀ states that the variables are independent. Failing to reject H₀ does not prove independence; it simply means that the data does not provide enough evidence to conclude that an association exists at the 10% significance level. We cannot be certain if an error was made.
Based on the result of the test, which of the following is the most appropriate conclusion?
Answer: 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. The decision rule for a hypothesis test is to compare the p-value to the significance level ( $\alpha$ ). Here, $p = 0.023$ and $\alpha = 0.05$ . Since $p < \alpha$ , 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.
Given that the critical value for this test at the 1% significance level is 15.086, what is the correct conclusion?
Answer: Do not reject H₀ because $12.8 < 15.086$ . There is insufficient evidence to conclude the die is not fair. 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 $\chi^2 = 12.8$ and the critical value is 15.086. Since $12.8 < 15.086$ , 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.
Which statement accurately describes the relationship between $p_A$ and $p_B$ ?
Answer: $p_A > p_B$ 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 $\chi^2_A = 10.5$ and $\chi^2_B = 18.2$ , 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 $p_B < p_A$ , making $p_A > p_B$ . Option A suggests $p_A < p_B$ , which reverses the correct relationship between test statistics and p-values. Option B incorrectly claims you need the significance level $\alpha$ 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: $p_A > p_B$ . 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.
What is the correct number of degrees of freedom for this chi-squared test?
Answer: 5 The formula for degrees of freedom in a goodness-of-fit test where parameters are estimated from the data is $df = k - 1 - m$ , where $k$ is the number of categories (bins) and $m$ is the number of estimated parameters. Here, there are $k=8$ bins. Two parameters, the mean ( $\mu$ ) and the standard deviation ( $\sigma$ ), were estimated from the data. Thus, $m=2$ . The degrees of freedom are $df = 8 - 1 - 2 = 5$ .

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