Ready to put your stats savvy to the test? Our free Chi Square Test Questions Quiz is designed to challenge you with thought-provoking chi square test questions and questions on chi square that mirror real-world data scenarios. Whether you're seeking chi square test practice or craving a fun chi square quiz to sharpen analytical skills, you'll learn to interpret contingency tables, measure independence, and strengthen your grasp of hypothesis testing. Brush up fundamentals with our mathematics mcq questions , or explore more with an engaging statistics quiz . Designed for students, researchers, and data enthusiasts, this interactive challenge adapts to your level and delivers instant feedback. Ready to prove your prowess? Click Start Now and dive into your chi square adventure!
What is the main purpose of the chi-square goodness-of-fit test?
To estimate variance in a continuous dataset
To assess the strength of a linear relationship between two variables
To determine if observed frequencies match an expected distribution
To compare means between two groups
The chi-square goodness-of-fit test compares observed counts across categories to expected counts under a specified distribution. It evaluates whether the differences between observed and expected frequencies are due to random variation or if they indicate a significant departure. This test is specifically designed for categorical data and does not assess relationships between variables, test means, or measure variance. For more information see StatTrek.
In a chi-square test of independence, what does a contingency table display?
The frequency distribution of two categorical variables
The correlation coefficient between variables
The variance within a single category
The mean differences between groups
A contingency table organizes counts for each combination of categories from two variables, allowing the chi-square test of independence to evaluate whether the variables are related. It does not display means or correlation coefficients. By tabulating frequencies in rows and columns, researchers can observe patterns of association. For more details see Wikipedia.
Which of the following is an assumption of the chi-square test?
Homogeneity of regression slopes
Observations are independent
Variables are measured on an interval scale
Data are normally distributed
One key assumption of the chi-square test is that each observation is independent of the others, meaning no subject contributes to more than one cell count. It does not require normal distribution or interval measurements because it deals with categorical data. Additionally, it does not involve regression assumptions such as homogeneity of slopes. For more on assumptions see StatisticsByJim.
How are expected frequencies calculated in a goodness-of-fit test?
By dividing the observed frequencies by the number of categories
By taking the square root of observed frequencies
By adding observed and expected frequencies and halving
By multiplying the total sample size by the hypothesized category proportions
Expected frequencies in a goodness-of-fit test are computed by multiplying the total number of observations by the expected proportion for each category under the null hypothesis. This yields the counts you would anticipate if the observed distribution fits the theoretical model. Other methods, such as dividing observed counts or taking square roots, are incorrect for expected frequency calculations. For further explanation see GraphPad.
What is the formula for the chi-square test statistic?
?((O ? E)²)
?((O ? E) / E)
?((O ? E)² / E)
?((E ? O)² / E²)
The chi-square test statistic is calculated as the sum of squared differences between observed (O) and expected (E) frequencies divided by the expected frequencies: X² = ?((O ? E)² / E). This formula ensures that larger deviations from expectations contribute more to the statistic. Other expressions, such as summing raw differences or dividing by E², do not yield the correct test value. See Wikipedia for more details.
If you have 3 categories, what are the degrees of freedom for a goodness-of-fit test?
4
2
3
1
For a chi-square goodness-of-fit test, the degrees of freedom equal the number of categories minus one: df = k ? 1. With 3 categories, df = 3 ? 1 = 2. This value is used to determine the critical chi-square value or p-value. Further information is available at StatTrek.
When expected cell counts are less than 5, which correction is recommended?
Holm-Bonferroni method
Yates' continuity correction
Tukey's Honest Significant Difference
Bonferroni correction
When expected frequencies in a 2×2 contingency table fall below 5, Yates' continuity correction is often applied to the chi-square test to reduce bias. This adjustment subtracts 0.5 from the absolute difference between observed and expected counts before squaring. Other corrections like Bonferroni and Holm-Bonferroni address multiple comparisons, not small cell counts. For more details, see Wikipedia.
A researcher conducts a chi-square test of independence with a 2×2 table and obtains ?² = 4.5 with p < 0.05. What conclusion is appropriate?
Conclude the sample size is too small
Conclude there is no association between variables
Fail to reject the null hypothesis of independence
Reject the null hypothesis of independence
A p-value less than 0.05 indicates that the observed association is unlikely to have arisen by chance under the null hypothesis of independence. Therefore, the correct decision is to reject the null hypothesis and conclude that the variables are associated. Stating there is no association or commenting on sample size is not supported by the test result. See StatisticsByJim for more information.
What is the effect of combining categories on a chi-square test?
It does not affect degrees of freedom or expected frequencies
It invalidates the test entirely
It reduces the degrees of freedom and can increase expected frequencies
It increases the degrees of freedom and decreases expected frequencies
Combining two or more categories in a chi-square test reduces the number of categories (k), which in turn reduces the degrees of freedom (df = k ? 1). This action also tends to increase the expected frequency in the combined category, helping to meet the assumption of adequate expected counts. It does not invalidate the test if categories are logically merged, but it alters the test's df and power. For more details, see StatisticsByJim.
Which measure estimates effect size for a chi-square test?
Pearson's r
Cohen's d
Eta-squared
Cramer's V
Cramer's V is a standardized measure of association that quantifies effect size for chi-square tests, particularly in contingency tables of any dimension. It ranges from 0 (no association) to 1 (perfect association). Other metrics like Cohen's d and Pearson's r apply to different statistical tests, not chi-square. For further reading see Wikipedia.
In a chi-square test of independence, how is the expected count for a specific cell calculated?
(Observed count / Grand total) × 100
(Row total + Column total) / 2
(Row total × Column total) / Grand total
(Row total × Column total) / 100
The expected frequency for a cell in a contingency table is computed as (row total × column total) divided by the grand total. This formula ensures that the expected counts reflect marginal distributions under the null hypothesis of independence. Other formulas, such as averaging totals or scaling by 100, do not provide the correct expected count. More information can be found at Statisticshowto.
What does a chi-square statistic of zero indicate in a goodness-of-fit test?
There is a strong association between categories
All observed frequencies are zero
Observed frequencies perfectly match expected frequencies
The test is not applicable
A chi-square value of zero occurs when every observed frequency equals its corresponding expected frequency exactly, indicating a perfect fit. No deviation means the test statistic sums to zero across all categories. It does not imply zero counts or a measure of association, but rather a perfect distributional match. See StatTrek for more details.
Which extension of the chi-square test uses likelihood ratios instead of Pearson's statistic?
The G-test (likelihood ratio test)
Wilcoxon signed-rank test
Fisher's exact test
McNemar's test
The G-test, or likelihood ratio test, is an alternative to Pearson's chi-square that uses the log-likelihood ratio formula G = 2?O ln(O/E). It is particularly useful when comparing observed and expected counts and shares similar asymptotic properties. Fisher's exact test and McNemar's test serve different purposes, and the Wilcoxon signed-rank test applies to paired continuous data. For more on the G-test see Wikipedia.
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Study Outcomes
Apply Chi-Square Formulas -
Use chi square test formulas to calculate test statistics for both goodness-of-fit and independence scenarios, ensuring accurate chi square test practice.
Interpret Test Statistics -
Analyze the results of chi square test questions to determine statistical significance and draw valid conclusions from sample data.
Differentiate Test Types -
Distinguish between goodness-of-fit and test of independence procedures by identifying when to use each method in a chi square quiz framework.
Select Appropriate Degrees of Freedom -
Determine degrees of freedom for various chi square tests to correctly reference distribution tables in questions on chi square.
Evaluate Hypotheses -
Formulate and test null and alternative hypotheses for categorical data, refining your chi square test practice with instant feedback.
Enhance Problem-Solving Skills -
Build confidence in tackling complex chi square test questions by practicing with realistic scenarios and immediate result analysis.
Cheat Sheet
Chi-Square Test Formula -
Diving into chi square test questions starts with the core formula: χ² = Σ(O - E)² / E. Calculate each cell's (Observed - Expected)² divided by Expected, then sum these values to get your test statistic. For example, if O=20 and E=15, that term contributes (5²/15)=1.67 to χ².
Goodness-of-Fit vs. Test of Independence -
Goodness-of-fit tests evaluate if a single categorical variable matches a theoretical distribution, while tests of independence assess whether two variables are related. Remember, your chi square quiz may present a one-way table for goodness-of-fit or a two-way contingency table for independence analysis. This distinction guides your calculation of expected frequencies and degrees of freedom.
Degrees of Freedom -
Degrees of freedom (df) adjust for the number of categories or table dimensions in your chi square test practice. For goodness-of-fit, use df = k - 1 where k is the number of categories; for independence, use df = (rows - 1)×(columns - 1). A quick mnemonic is "subtract one per variable axis" to keep df straight in any chi square scenario.
Key Assumptions and Conditions -
Chi square tests require independent observations, mutually exclusive categories, and generally expect at least five observations per cell. If more than 20% of cells fall below this threshold, consider combining categories or using Fisher's exact test instead. Checking these conditions before you tackle questions on chi square ensures valid and reliable results.
Interpreting Results and Effect Size -
After computing χ², compare it to the critical value from the chi-square distribution table or calculate a p-value to test your hypothesis. A low p-value (usually < 0.05) signals a statistically significant result, helping you answer chi square test questions with confidence. To assess effect size, compute Cramér's V = √[χ² / (n×df_min)], which indicates the strength of association beyond mere significance.
