Chi-Square Goodness-of-Fit - College Statistics

1. A researcher tests if a coin is fair ($H_0: p_{heads}=0.5$) by flipping it 80 times and getting 50 heads. This can be analyzed with a one-proportion z-test or a chi-square goodness-of-fit test. If the one-proportion z-test yields a statistic of $z = 2.236$, what will be the value of the chi-square statistic?
   Answer: $\chi^2 = 5.000$ For a categorical variable with only two categories, the chi-square goodness-of-fit test is mathematically equivalent to the two-sided one-proportion z-test. The relationship between the test statistics is $\chi^2 = z^2$. Given $z = 2.236$, the chi-square statistic would be $\chi^2 = (2.236)^2 \approx 5.000$. The degrees of freedom for this chi-square test would be $k-1 = 2-1 = 1$.
2. An ecologist is testing whether the distribution of five species of trees in a forest tract matches a theoretical distribution. The initial analysis of a random sample of 200 trees reveals that the expected counts for two of the species are 4.2 and 3.8, respectively. The other expected counts are all above 30. What is the most appropriate course of action?
   Answer: Combine the two species with small expected counts into a single 'rare species' category before performing the test. The 'Large Counts' condition for a chi-square test requires all expected counts to be at least 5. When this condition is violated, the standard procedure is to combine the categories with small expected counts into a new, meaningful category. This ensures the condition is met while retaining the data. Proceeding with the test (A) is incorrect as the results would be unreliable. Collecting more data (B) might work, but it addresses observed, not expected, counts and may not be practical. Removing data (D) is inappropriate as it alters the research question and discards valid information.
3. A city is composed of four ethnic groups. A sociologist wants to determine if the distribution of ethnic groups among residents who use public parks is the same as the distribution in the city as a whole. The sociologist collects data on a random sample of 500 park users. Which is the most appropriate statistical test for this research question?
   Answer: A chi-square goodness-of-fit test The research question involves one categorical variable (ethnicity) from a single sample (park users) and compares its distribution to a known, hypothesized distribution (the city's overall ethnic distribution). This is the exact scenario for which a chi-square goodness-of-fit test is designed. A test for homogeneity would compare the distribution of a single categorical variable across multiple populations (e.g., comparing ethnicity of park users in City A vs. City B). A test for independence would examine the association between two categorical variables within a single population (e.g., testing if ethnicity and frequency of park use are independent).
4. A chi-square goodness-of-fit test for a variable with 8 categories produced a test statistic of $\chi^2=13.5$. The p-value for this test would be found by calculating the area to the right of 13.5 in a chi-square distribution with how many degrees of freedom?
   Answer: 7 For a chi-square goodness-of-fit test, the degrees of freedom (df) are calculated as $k-1$, where $k$ is the number of categories. In this problem, there are 8 categories, so the degrees of freedom are $8-1=7$. The p-value is the probability of obtaining a test statistic as extreme or more extreme than the observed one, which for a chi-square test is the area under the curve to the right of the test statistic. Therefore, the calculation would be based on a $\chi^2$ distribution with 7 degrees of freedom. The sample size is needed to calculate the test statistic, but not to determine the degrees of freedom.
5. A researcher conducts a chi-square goodness-of-fit test to see if the student body at a large university is 25% freshmen, 25% sophomores, 25% juniors, and 25% seniors. In a random sample of 200 students, she finds 60 freshmen, 45 sophomores, 55 juniors, and 40 seniors. Which of the following is a necessary condition for the validity of the p-value obtained from this test?
   Answer: The number of students expected in each class must be at least 5. When you encounter chi-square goodness-of-fit test questions, focus on the key assumptions that make the test statistically valid. This test compares observed frequencies to expected frequencies to determine if sample data fits a hypothesized distribution. The correct answer is D because chi-square tests require that expected frequencies (not observed frequencies) be at least 5 in each category. Let's verify this condition: with 200 total students and an expected distribution of 25% in each class, the expected frequency for each class is $200 \times 0.25 = 50$ students. Since 50 > 5 for all categories, this condition is satisfied. When expected frequencies are too small, the chi-square distribution doesn't provide a good approximation for the test statistic. Option A is incorrect because it focuses on observed frequencies rather than expected frequencies. The actual counts you observe (60, 45, 55, 40) don't need to meet the "at least 5" rule—it's the theoretical expected counts that matter. Option B is wrong because chi-square goodness-of-fit tests don't require the population to be normally distributed. This test works with categorical data and makes no assumptions about the shape of the underlying population distribution. Option C is incorrect because there's no general "greater than 30" sample size requirement for chi-square tests. The sample size requirement depends on having adequate expected frequencies in each category, not an arbitrary total. Study tip: Remember the "5 rule" for chi-square tests always applies to expected frequencies, never observed frequencies. Calculate expected frequencies first when checking test conditions.
6. A company produces widgets at three factories. Historically, Factory A produces 50% of the total output, Factory B produces 30%, and Factory C produces 20%. To test if this production distribution still holds, a random sample of 400 widgets is selected. The test yields a $\chi^2$ statistic of 6.50. What conclusion can be drawn at the 5% significance level?
   Answer: Reject $H_0$; there is evidence the production distribution has changed. To make a conclusion, we need to compare the test statistic to the critical value from the $\chi^2$ distribution. The degrees of freedom are $df = k-1 = 3-1 = 2$. The critical value for a $\chi^2$ distribution with 2 degrees of freedom at $\alpha = 0.05$ is 5.99. Since the calculated test statistic (6.50) is greater than the critical value (5.99), we reject the null hypothesis. This means there is sufficient statistical evidence to conclude that the production distribution has changed from the historical proportions.
7. A researcher is testing whether a six-sided die is fair. The null hypothesis is that the probability of rolling each number is $1/6$. After 120 rolls, a chi-square goodness-of-fit test produces a test statistic of $\chi^2 = 12.0$. Using the provided table of critical values, what is the correct conclusion at the $\alpha=0.05$ significance level?
   Answer: Reject $H_0$, as 12.0 is greater than 11.07. First, determine the degrees of freedom: for a goodness-of-fit test, $df = k-1$, where $k$ is the number of categories. For a six-sided die, $k=6$, so $df = 6-1=5$. From the provided table, the critical value for $df=5$ and $\alpha=0.05$ is 11.07. The decision rule is to reject the null hypothesis if the test statistic is greater than the critical value. Since the calculated test statistic (12.0) is greater than the critical value (11.07), we reject $H_0$. This provides sufficient evidence to conclude the die is not fair.
8. A chi-square goodness-of-fit test is being performed. Which of the following changes would be most likely to result in a smaller p-value, assuming the sample proportions in each category remain the same?
   Answer: Increasing the sample size A smaller p-value indicates stronger evidence against the null hypothesis. If the sample proportions remain the same, increasing the sample size $n$ will increase the magnitude of the differences between observed counts (O) and expected counts (E). This increases the $\chi^2$ statistic (it increases linearly with n), which in turn leads to a smaller p-value. Decreasing the sample size would have the opposite effect. Increasing the number of categories changes the degrees of freedom and the distribution, but does not have a predictable directional effect on the p-value without more information. Decreasing $\alpha$ changes the threshold for significance but does not change the p-value itself.
9. A biologist studies a plant species expected to produce offspring with four different petal colors in a 9:3:3:1 ratio. After performing a cross, the biologist counts the number of offspring in each color category. To test the genetic theory, a chi-square goodness-of-fit test is planned. However, the biologist estimates the proportion of the most common color from the sample data itself before proceeding with the test on the remaining three categories. How many degrees of freedom should be used for the chi-square test?
   Answer: 2 The standard degrees of freedom for a goodness-of-fit test is $df = k - 1$, where $k$ is the number of categories. In this case, there are 4 categories, so one might expect $df = 3$. However, the degrees of freedom must be reduced by one for each population parameter estimated from the sample data. The problem states that the proportion of one category was estimated from the data. Therefore, the degrees of freedom are $df = k - 1 - (\text{number of estimated parameters}) = 4 - 1 - 1 = 2$.
10. A goodness-of-fit test is used to determine if a set of observed frequencies fits a theoretical distribution. If the calculated chi-square statistic is very close to zero, what does this imply about the 'goodness of fit'?
    Answer: An excellent fit, because the observed frequencies are very close to the expected frequencies. The chi-square statistic, $\chi^2 = \sum \frac{(O - E)^2}{E}$, measures the total discrepancy between observed (O) and expected (E) frequencies. A $\chi^2$ value of zero would mean that the observed frequencies perfectly match the expected frequencies. Therefore, a value very close to zero indicates a very small discrepancy and an excellent fit between the data and the hypothesized model.
11. A chi-square goodness-of-fit test yields a p-value of 0.008. Which of the following is the most accurate interpretation of this result?
    Answer: There is strong evidence that the observed distribution of the variable is different from the hypothesized distribution. A small p-value (typically < 0.05 or 0.01) indicates that the observed data are unlikely to have occurred if the null hypothesis were true. This provides strong evidence against the null hypothesis. Therefore, we conclude that the actual distribution of the variable is different from the one specified in $H_0$. Distractor A is a common misinterpretation of the p-value. Distractor C is incorrect because the chi-square test is an omnibus test; it tells us that at least one proportion is different, but not necessarily all of them. Distractor D is incorrect because the p-value is the probability of observing a result as extreme or more extreme than the sample result, not the probability of the exact sample result.
12. A study tested whether the days of the week for car accidents at a particular intersection were uniformly distributed. A chi-square goodness-of-fit test was performed on a sample of accidents, resulting in a test statistic of $\chi^2 = 18.5$ with 6 degrees of freedom. The p-value was found to be 0.005. What is the most appropriate conclusion?
    Answer: There is strong evidence that the distribution of accidents across the days of the week is not uniform. A small p-value (0.005) leads to the rejection of the null hypothesis. The null hypothesis was that the distribution is uniform. Therefore, we have strong evidence that the distribution is not uniform. Choice A is too strong a conclusion; the omnibus nature of the chi-square test means we only know that at least one day's proportion is different, not necessarily all of them. Choice C is vague; 'significantly different' should refer to the overall pattern, not necessarily each individual day's count. Choice D uses the word 'proven', which is too absolute for statistical inference.
13. A researcher performs a chi-square goodness-of-fit test on a sample of data with 6 categories. They find that one of the observed counts is 3. An initial calculation shows that one of the expected counts is 4.5. The researcher is concerned about the validity of the test. Which of the following statements correctly addresses the conditions for this test?
    Answer: The test is invalid because at least one expected count is less than 5. The 'Large Counts' condition for a chi-square goodness-of-fit test requires that all expected counts be at least 5. The observed counts can be less than 5. Since the problem states that one expected count is 4.5, this condition is violated, making the results of the test unreliable. Therefore, the test is invalid as described. Choice D is incorrect because the condition applies to all expected counts; if even one is less than 5, the condition is not met.
14. A researcher conducts a chi-square goodness-of-fit test using a sample of size $n$. The resulting test statistic is $\chi^2_1$. A second researcher replicates the study with a new sample of size $2n$ but observes the exact same proportions in each category as the first study. What will be the value of the test statistic, $\chi^2_2$, for the second study?
    Answer: $\chi^2_2 = 2 \cdot \chi^2_1$ Let $O_{1i}$ and $E_{1i}$ be the observed and expected counts for category $i$ in the first study. Then $\chi^2_1 = \sum \frac{(O_{1i} - E_{1i})^2}{E_{1i}}$. In the second study, the sample size is $2n$ and the proportions are the same. This means the new observed counts are $O_{2i} = 2 O_{1i}$ and the new expected counts are $E_{2i} = 2 E_{1i}$. The new test statistic is $\chi^2_2 = \sum \frac{(2 O_{1i} - 2 E_{1i})^2}{2 E_{1i}} = \sum \frac{4(O_{1i} - E_{1i})^2}{2 E_{1i}} = 2 \sum \frac{(O_{1i} - E_{1i})^2}{E_{1i}} = 2 \chi^2_1$. Thus, doubling the sample size while keeping the proportions constant doubles the chi-square statistic.
15. In a chi-square goodness-of-fit test with 5 categories, the null hypothesis states that all categories are equally likely. In a sample of 100 individuals, the calculated test statistic is χ² = 9.55. The individual chi-square components for the first four categories are 1.25, 0.45, 2.00, and 4.05. The observed count for the fifth category was greater than its expected count. What was the observed count for the fifth category?
    Answer: 26 First, find the expected count for each category under the null hypothesis: E = n/k = 100/5 = 20. Next, find the chi-square component for the fifth category: χ²₅ = Total χ² - (sum of first four components) = 9.55 - (1.25 + 0.45 + 2.00 + 4.05) = 9.55 - 7.75 = 1.8. Now solve for the observed count using the component formula: (O₅ - E₅)²/E₅ = 1.8, so (O₅ - 20)²/20 = 1.8. This gives (O₅ - 20)² = 36, so O₅ - 20 = ±6. The possible values are O₅ = 26 or O₅ = 14. Since the problem states the observed count was greater than the expected count (20), we have O₅ = 26.
16. Which statement best explains the purpose of squaring the difference between observed and expected counts in the chi-square goodness-of-fit test formula, $\chi^2 = \sum \frac{(O - E)^2}{E}$?
    Answer: To ensure the test statistic cannot be negative and to give more weight to larger differences. There are two primary reasons for squaring the difference $(O-E)$. First, the sum of the differences $\sum (O-E)$ is always zero. Squaring makes all terms non-negative, preventing positive and negative differences from canceling each other out. Second, squaring gives greater weight to categories where the observed count is further from the expected count, meaning larger deviations from the null hypothesis have a disproportionately larger effect on the test statistic.
17. A marketing firm claims that a new advertisement will result in the following market shares for four competing brands: Brand A (50%), Brand B (25%), Brand C (15%), and Brand D (10%). A study is conducted on 300 consumers to test this claim. Which of the following is the correct null hypothesis for a chi-square goodness-of-fit test?
    Answer: $H_0: p_A=0.50, p_B=0.25, p_C=0.15, p_D=0.10$ Chi-square goodness-of-fit tests examine whether observed data matches an expected distribution. When you encounter these problems, focus on what the null hypothesis should claim about the population parameters - specifically, that the true population proportions equal the claimed values. The correct null hypothesis states that the true population proportions (denoted by $p$) equal the marketing firm's claimed values. Answer D correctly uses $p_A=0.50, p_B=0.25, p_C=0.15, p_D=0.10$, representing the population parameters we're testing against. Answer A uses $\hat{p}$ (p-hat), which represents sample proportions - statistics calculated from your data. The null hypothesis should make claims about population parameters, not sample statistics. This is a common notation trap. Answer B suggests all brands have equal market share (25% each). This would be appropriate if you were testing whether market shares were equal across brands, but that's not what the marketing firm claimed. You're testing their specific predicted percentages, not equality. Answer C states the hypothesis in terms of expected counts ($E_A=150$, etc.) rather than proportions. While these expected counts will be used in your chi-square calculations, the null hypothesis should be expressed in terms of population proportions, not expected frequencies. Remember: In goodness-of-fit tests, your null hypothesis should always state that the true population proportions ($p$) equal the claimed or theoretical values. Watch for the distinction between population parameters ($p$) and sample statistics ($\hat{p}$) - hypotheses test parameters, not statistics.
18. A chi-square goodness-of-fit test is performed on data with k=5 categories at a significance level of α=0.05. The critical value from the chi-square distribution is 9.49. Which of the following sets of individual chi-square components would lead to a decision to reject the null hypothesis?
    Answer: {3.8, 2.9, 1.5, 0.8, 0.6} To make a decision, we calculate the chi-square test statistic by summing the individual components and compare it to the critical value of 9.49. We reject the null hypothesis if the calculated statistic is greater than the critical value. Choice A: 3.8+2.9+1.5+0.8+0.6 = 9.6. Since 9.6 > 9.49, we reject H₀. Choice B: 3.1+2.4+2.0+1.2+0.7 = 9.4 < 9.49, fail to reject. Choice C: 2.5+2.1+1.9+1.5+1.4 = 9.4 < 9.49, fail to reject. Choice D: 4.0+3.0+1.0+0.5+0.9 = 9.4 < 9.49, fail to reject.
19. A city's population is known to be 50% Group A, 30% Group B, and 20% Group C. A random sample of 200 city employees is taken to see if the distribution within city employment mirrors the city's population. The sample contains 90 from Group A, 68 from Group B, and 42 from Group C. What is the chi-square test statistic for a goodness-of-fit test?
    Answer: 2.17 First, calculate the expected counts based on the city's population proportions and the sample size of 200. $E_A = 200 \times 0.50 = 100$, $E_B = 200 \times 0.30 = 60$, $E_C = 200 \times 0.20 = 40$. The observed counts are $O_A = 90, O_B = 68, O_C = 42$. Now, calculate the chi-square statistic: $\chi^2 = \sum \frac{(O - E)^2}{E} = \frac{(90-100)^2}{100} + \frac{(68-60)^2}{60} + \frac{(42-40)^2}{40} = \frac{100}{100} + \frac{64}{60} + \frac{4}{40} = 1 + 1.067 + 0.1 = 2.167 \approx 2.17$.

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