1. **State the problem:** We want to find the 90% and 95% confidence intervals for the average weekly sales of a brand of soap based on a sample of 13 weeks: 123, 110, 95, 120, 87, 89, 100, 105, 98, 88, 75, 125, 101. 2. **Formula for confidence interval:** The confidence interval for the mean when the population standard deviation is unknown and sample size is small is given by: $$\bar{x} \pm t_{\alpha/2, n-1} \times \frac{s}{\sqrt{n}}$$ where: - $\bar{x}$ is the sample mean - $s$ is the sample standard deviation - $n$ is the sample size - $t_{\alpha/2, n-1}$ is the t-score from the t-distribution with $n-1$ degrees of freedom 3. **Calculate the sample mean $\bar{x}$:** $$\bar{x} = \frac{123 + 110 + 95 + 120 + 87 + 89 + 100 + 105 + 98 + 88 + 75 + 125 + 101}{13}$$ $$\bar{x} = \frac{1316}{13} = 101.23$$ 4. **Calculate the sample standard deviation $s$:** First, find each squared deviation from the mean, sum them, then divide by $n-1=12$, and take the square root. Sum of squared deviations: $$(123-101.23)^2 + (110-101.23)^2 + \cdots + (101-101.23)^2 = 2381.69$$ Sample variance: $$s^2 = \frac{2381.69}{12} = 198.47$$ Sample standard deviation: $$s = \sqrt{198.47} = 14.09$$ 5. **Find t-scores for 90% and 95% confidence intervals with 12 degrees of freedom:** - For 90% confidence, $t_{0.05,12} = 1.782$ - For 95% confidence, $t_{0.025,12} = 2.179$ 6. **Calculate margin of error for each confidence level:** $$ME_{90} = 1.782 \times \frac{14.09}{\sqrt{13}} = 1.782 \times 3.91 = 6.97$$ $$ME_{95} = 2.179 \times \frac{14.09}{\sqrt{13}} = 2.179 \times 3.91 = 8.52$$ 7. **Construct confidence intervals:** - 90% CI: $$101.23 \pm 6.97 = (94.26, 108.20)$$ - 95% CI: $$101.23 \pm 8.52 = (92.71, 109.75)$$ **Final answer:** - 90% confidence interval for average weekly sales is **(94.26, 108.20)** - 95% confidence interval for average weekly sales is **(92.71, 109.75)**