1. **State the problem:** We want to test if the mean daily sales difference between Store 1 and Store 2, denoted as $\mu_d$, is zero or not, based on the differences recorded over 12 days. 2. **Hypotheses:** - Null hypothesis $H_0$: $\mu_d = 0$ (no difference in mean daily sales) - Alternative hypothesis $H_1$: $\mu_d \neq 0$ (mean daily sales differ) 3. **Type of test statistic:** Since the population of differences is normally distributed and the sample size is small (12), we use a **t-test for the mean of paired differences**. 4. **Calculate the test statistic:** - Differences $d_i$ are given: 186, 223, 45, 198, -23, 194, -47, 161, 108, 21, 222, 11 - Sample size $n=12$ - Sample mean difference $\bar{d} = \frac{\sum d_i}{n} = \frac{186+223+45+198-23+194-47+161+108+21+222+11}{12} = \frac{1299}{12} = 108.25$ - Sample standard deviation $s_d$: Calculate each $(d_i - \bar{d})^2$: $(186-108.25)^2= 6,086.06$ $(223-108.25)^2= 13,196.56$ $(45-108.25)^2= 4,001.06$ $(198-108.25)^2= 7,882.56$ $(-23-108.25)^2= 17,302.56$ $(194-108.25)^2= 7,423.56$ $(-47-108.25)^2= 24,195.06$ $(161-108.25)^2= 2,778.06$ $(108-108.25)^2= 0.06$ $(21-108.25)^2= 7,544.06$ $(222-108.25)^2= 12,885.06$ $(11-108.25)^2= 9,414.06$ - Sum of squares $= 112,708.2$ - Sample variance $s_d^2 = \frac{112,708.2}{n-1} = \frac{112,708.2}{11} = 10,246.2$ - Sample standard deviation $s_d = \sqrt{10,246.2} = 101.22$ - Test statistic $t = \frac{\bar{d} - 0}{s_d / \sqrt{n}} = \frac{108.25}{101.22 / \sqrt{12}} = \frac{108.25}{29.22} = 3.704$ 5. **Critical values:** For a two-tailed test at $\alpha=0.05$ with $df = n-1 = 11$, the critical t-values are $\pm t_{0.025,11} = \pm 2.201$ 6. **Conclusion:** Since $|t| = 3.704 > 2.201$, we reject the null hypothesis. **Answer:** At the 0.05 significance level, there is sufficient evidence to conclude that the mean daily sales of the two stores differ.