Answer: Chi-square test; reject the null hypothesis

## Explanation This is a hypothesis test about **population variance/standard deviation**, so we should use the **Chi-square test** (not the t-test). ### Given: - Sample size: n = 24 months (2 years × 12 months) - Sample standard deviation: s = 5% - Population standard deviation under null hypothesis: σ₀ = 4% - Significance level: α = 5% (two-tailed test) ### Test Statistic: The chi-square test statistic for variance is: \[\chi^2 = \frac{(n-1)s^2}{\sigma_0^2} = \frac{(24-1) \times (0.05)^2}{(0.04)^2} = \frac{23 \times 0.0025}{0.0016} = \frac{0.0575}{0.0016} = 35.9375\] ### Critical Values: For a two-tailed test at 5% significance level: - Lower critical value: χ²(23, 0.975) = 11.6886 (not provided in table) - Upper critical value: χ²(23, 0.025) = 38.0757 (from table) ### Decision: Since our test statistic (35.9375) is **less than** the upper critical value (38.0757), we **do not reject** the null hypothesis. Therefore, the correct answer is: **Chi-square test; do not reject the null hypothesis** (Option D). **Note:** The provided answer in the text (Option A) appears to be incorrect based on the statistical analysis.

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