--- title: "Chi-square: opinion change before election" subject: "Probability and Statistics" chapter: 7 key: "semester_5/probability_and_statistics:polls-opinion-change-chi-square" source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/polls-opinion-change-chi-square site: BCT Engineering --- # Chi-square: opinion change before election Probability and Statistics · Chapter 7: Inference concerning proportion - Asked in: 2070 Chaitra, 5 marks - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/probability_and_statistics/polls-opinion-change-chi-square - Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-7 ## Question The results of polls conducted two weeks and four weeks before an election are shown in the following table: | | Two weeks before election | Four weeks before election | |---|---|---| | For republican candidate | 79 | 91 | | For democratic candidate | 84 | 66 | | Undecided | 37 | 43 | Use the 0.05 level of significance to test whether there has been change in opinion during the 2 weeks between the polls. ## Short answer Chi-square test of homogeneity: χ² = 3.457, df = 2. Critical value at 0.05 is 5.991. Since 3.457 < 5.991, fail to reject H₀. No significant change in opinion distribution between the two polls; differences are due to sampling variability. ## Answer A chi-square test of homogeneity (equivalently independence) is used: the opinion distribution is compared across the two polls (3 × 2 table, N = 400). 1. H0: the distribution of opinion is the same in both polls (no change in opinion). H1: the distribution of opinion differs between the polls (there has been a change). 2. Level of significance: α = 0.05 3. Test statistic: χ² = Σ (O − E)²/E, where E = (row total × column total)/grand total. Observed frequencies with totals: ```text 2 wk 4 wk Total Republican 79 91 170 Democratic 84 66 150 Undecided 37 43 80 Total 200 200 400 ``` Expected frequencies: ```text 2 wk 4 wk Republican 85.00 85.00 Democratic 75.00 75.00 Undecided 40.00 40.00 ``` Terms (O − E)²/E: ```text 2 wk 4 wk Republican 0.424 0.424 Democratic 1.080 1.080 Undecided 0.225 0.225 ``` χ² = Σ (O − E)²/E = **3.457** 4. Degrees of freedom = (r − 1)(c − 1) = (3 − 1)(2 − 1) = **2**. Critical value: χ²(0.05; 2) = **5.991**. Reject H0 if χ² > 5.991. 5. Decision: calculated χ² < tabulated χ² (3.457 < 5.991, p-value = 0.1775), so **fail to reject H0**. **Conclusion:** there is no significant change in opinion between the two polls; the difference is due to sampling variation. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is H0 for the two polls? A. Opinion distribution is the same in both polls (no change) B. Republican support is higher in the later poll C. Undecided voters are zero D. Opinion changed between the polls Answer: A. The test of homogeneity assumes the opinion distribution is identical in both polls. (Key point: Null hypothesis) 2. What are the degrees of freedom for the 3 x 2 table? A. 1 B. 2 C. 5 D. 6 Answer: B. (3 - 1)(2 - 1) = 2. (Key point: Degrees of freedom) 3. What is the expected count for Republican in the 2-week poll? A. 79 B. 100 C. 85 D. 170 Answer: C. E = 170 x 200 / 400 = 85. (Key point: Expected frequency) 4. Calculated chi-square is 3.457 versus critical 5.991. What is concluded? A. Reject H0; opinion changed B. Fail to reject H0; no significant change in opinion C. Accept H1 at 5% D. Increase the sample size to conclude Answer: B. 3.457 < 5.991, so the differences are attributed to sampling variation. 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