Grade 9 Chi-Square Test Worksheet
Calculate expected counts and Pearson’s chi-square statistic for a two-way table. Includes a worked example, printable practice, and an answer key with explanations.
The answer key follows the practice below. In the editor, remove the final answer-key section before sharing a student copy.
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Standards
Grade 9 Chi-Square Test Worksheet
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Learning goal and worked example
Calculate expected counts and Pearson’s chi-square statistic for a two-way table.
Expected count = row total × column total ÷ grand total. A cell with observed 12 and expected 10 contributes (12−10)²/10 = 0.4 to χ².
Statistics extension, not assumed core Grade 9 content. Teach independence, expected counts, hypotheses, and degrees of freedom first. A significant test is not proof of causation.
Read each question carefully and provide your answers in the spaces provided. Show all your work for calculations.
1. A chi-square test of independence looks for evidence of an between two categorical variables.
2. The null hypothesis (H₀) in a Chi-square test typically states that there is association between the variables.
3. The degrees of freedom for a contingency table are calculated as (rows - 1) × ( - 1).
1. Which of the following is NOT an assumption of the Chi-square test?
Random sampling
Normally distributed data
Independent observations
Sufficiently large expected cell counts (use at least 5 in each cell for this worksheet)
2. What happens to the critical Chi-square value as the degrees of freedom increase (for a constant significance level)?
It decreases
It increases
It stays the same
It depends on the data
1. A researcher wants to investigate if there is a relationship between gender and preference for a certain type of music (Pop, Rock, Classical). What would be the null hypothesis for this study?
2. Explain the concept of 'expected frequency' in the context of a Chi-square test.
1. A larger chi-square statistic means a larger discrepancy between observed counts and counts expected under independence; it does not by itself measure the strength of association.
True
False
2. If the p-value from a Chi-square test is less than the significance level (e.g., 0.05), we fail to reject the null hypothesis.
True
False
In an illustrative random sample, each person contributes one independent observation. Recorded gender categories and drink preference have the counts below. Use Pearson’s chi-square statistic without a continuity correction.
Coffee Tea Total
Males 40 20 60
Females 30 50 80
Total 70 70 140
1. Calculate the expected frequency for 'Males' who prefer 'Coffee'.
2. Calculate the Chi-square statistic (χ²) for this data. (Show your steps and formula).
3. Determine the degrees of freedom for this test.
Answer key and worked solutions
Fill in the blanks: 1 association; 2 no; 3 columns.
Multiple choice: 1 b (normally distributed data is not required); 2 b (the critical value increases for a fixed significance level).
Short answer 1: The recorded gender category and music preference are independent in the population. 2: An expected frequency is the count predicted for a cell under independence: row total × column total / grand total.
True/false: 1 True. The statistic measures discrepancy, not effect size by itself. 2 False: p < α leads to rejecting the null hypothesis.
Calculation 1: Expected male/coffee count = 60×70/140 = 30. The four expected counts are 30,30,40,40 in row order.
Calculation 2: χ² = (40−30)²/30 + (20−30)²/30 + (30−40)²/40 + (50−40)²/40 = 100/30 + 100/30 + 100/40 + 100/40 = 35/3 ≈ 11.67.
Calculation 3: df = (2−1)(2−1) = 1. All expected counts exceed 5. These calculations use Pearson’s statistic without a continuity correction.
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