1. **Problem Statement:** Calculate the point estimate for the difference in population means, the margin of error for a 90% confidence interval, and then find the 90% confidence interval for the difference in means between Location A and Location B. 2. **Step (a): Calculate the point estimate (difference in sample means).** - Find the mean of Location A data: $87, 138, 149, 91, 114, 139, 108, 119, 150, 81$ $$\bar{x}_1 = \frac{87 + 138 + 149 + 91 + 114 + 139 + 108 + 119 + 150 + 81}{10} = \frac{1176}{10} = 117.6$$ - Find the mean of Location B data: $107, 105, 109, 117, 101, 103, 119, 118, 106, 100$ $$\bar{x}_2 = \frac{107 + 105 + 109 + 117 + 101 + 103 + 119 + 118 + 106 + 100}{10} = \frac{1085}{10} = 108.5$$ - Point estimate for difference in means: $$\bar{x}_1 - \bar{x}_2 = 117.6 - 108.5 = 9.1$$ 3. **Step (b): Calculate the margin of error (M) for 90% confidence interval.** - Formula: $$M = t^* \times \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}$$ - Calculate sample variances: - For Location A: $$s_1^2 = \frac{\sum (x_i - \bar{x}_1)^2}{n_1 - 1}$$ Calculations: Differences squared: $(87-117.6)^2=930.76$, $(138-117.6)^2=416.16$, $(149-117.6)^2=992.16$, $(91-117.6)^2=702.76$, $(114-117.6)^2=12.96$, $(139-117.6)^2=462.76$, $(108-117.6)^2=92.16$, $(119-117.6)^2=1.96$, $(150-117.6)^2=1057.96$, $(81-117.6)^2=1345.96$ Sum: $5915.6$ $$s_1^2 = \frac{5915.6}{9} = 657.29$$ - For Location B: Differences squared: $(107-108.5)^2=2.25$, $(105-108.5)^2=12.25$, $(109-108.5)^2=0.25$, $(117-108.5)^2=72.25$, $(101-108.5)^2=56.25$, $(103-108.5)^2=30.25$, $(119-108.5)^2=110.25$, $(118-108.5)^2=90.25$, $(106-108.5)^2=6.25$, $(100-108.5)^2=72.25$ Sum: $452.5$ $$s_2^2 = \frac{452.5}{9} = 50.28$$ - Sample sizes: $n_1 = n_2 = 10$ - Degrees of freedom (approximate using smaller of $n_1-1$ and $n_2-1$): $df = 9$ - From t-distribution table for 90% confidence and $df=9$, two-tailed critical value: $$t^* = 1.833$$ - Calculate margin of error: $$M = 1.833 \times \sqrt{\frac{657.29}{10} + \frac{50.28}{10}} = 1.833 \times \sqrt{65.729 + 5.028} = 1.833 \times \sqrt{70.757} = 1.833 \times 8.41 = 15.41$$ 4. **Step (c): Calculate the 90% confidence interval for the difference in means.** - Confidence interval formula: $$\left( (\bar{x}_1 - \bar{x}_2) - M, (\bar{x}_1 - \bar{x}_2) + M \right)$$ - Substitute values: $$\left(9.1 - 15.41, 9.1 + 15.41\right) = (-6.31, 24.51)$$ **Final answers:** - Point estimate difference: $9.1$ - Margin of error: $15.410$ - 90% confidence interval: $(-6.310, 24.510)$