1. **Problem 1: 90% Confidence Interval for Mean Annual Income** Given: - Sample size $n=20$ - Sample mean $\bar{x}=120000$ - Population standard deviation $\sigma=7500$ - Confidence level = 90%, so $z_{\alpha/2}=1.645$ Formula for confidence interval when $\sigma$ is known: $$\bar{x} \pm z_{\alpha/2} \frac{\sigma}{\sqrt{n}}$$ Calculate margin of error: $$1.645 \times \frac{7500}{\sqrt{20}} = 1.645 \times 1677.05 = 2759$$ Find limits: $$LL = 120000 - 2759 = 117241$$ $$UL = 120000 + 2759 = 122759$$ **Answer:** The 90% confidence interval estimate is $$[117241,\ 122759]$$ 2. **Problem 2: 95% Confidence Interval for Mean Assembly Time** Given: - Sample size $n=15$ - Sample mean $\bar{x}=12.2$ - Sample standard deviation $s=2.4$ - Confidence level = 95%, thus $t_{\alpha/2, n-1}=2.145$ Formula for confidence interval when $\sigma$ is unknown: $$\bar{x} \pm t_{\alpha/2, n-1} \frac{s}{\sqrt{n}}$$ Calculate margin of error: $$2.145 \times \frac{2.4}{\sqrt{15}} = 2.145 \times 0.619 = 1.33$$ Find limits: $$LL = 12.2 - 1.33 = 10.87$$ $$UL = 12.2 + 1.33 = 13.53$$ **Answer:** The 95% confidence interval estimate is $$[10.87, \ 13.53]$$ 3. **Problem 3: 95% Confidence Interval for Population Proportion** a.i. Parameter of interest is population proportion $p$ of people who consider television their major source. Point estimator is sample proportion $\hat{p} = \frac{120}{200} = 0.6$ a.ii. The formula for 95% confidence interval for a proportion is: $$\hat{p} \pm z_{\alpha/2} \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$$ Where $z_{\alpha/2} = 1.96$ for 95% confidence. **Answer:** Parameter: $p$, estimator: $\hat{p}=0.6$, and confidence interval formula given above.