In Cleveland, a random sample of 73 mail carriers showed that 10 had been bitten by an animal during one week. In Philadelphia, in a random sample of 80 mail carriers, 16 had received animal bites. Find the 95% confidence interval for the difference of the two proportions. **Find the value of each statistic.****Express as a decimal rounded to four decimal places**- **Proportion one (Cleveland):** - [Select an answer] = ______ - (Calculate using a TI-84)- **Proportion two (Philadelphia):** - [Select an answer] = ______ - (Calculate using a TI-84)\[ \hat{q}_1 = 1 - \hat{p}_1 = 1 - \_\_\_\_ = \_\_\_\_ \] (Cleveland)\[ \hat{q}_2 = 1 - \hat{p}_2 = 1 - \_\_\_\_ = \_\_\_\_ \] (Philadelphia)- **Sample Size One (Cleveland):** - [Select an answer] = ______- **Sample Size Two (Philadelphia):** - [Select an answer] = ______**Round answer(s) to two decimal places**\[ \text{(α / 2)} = [Select an answer] \left( \_\_\_\_, \_\_\_\_, \_\_\_\_ \right) \]\[ \text{(α / 2)} = \_\_\_\_ \]**Round answer(s) to two decimal places**

Given, n1 = 73 x1 = 10 n2 = 80 x2 = 16 confidence level = 1 - α = 0.95 α = 0.05α2 = 0.025 To find, the 95% confidence interval for the difference of the two proportions. Answer: p1 = x1n1 = 1073 = 0.136983 p2 = x2n2 = 1680 = 0.2 The difference is: p1 - p2 = 0.136983 - 0.2 = -0.06301 The formula for confidence interval for proportion difference is: p1 - p2 ± Zα/2 × p1 (1 - p1)n1 + p2 (1-p2)n2 The 95% confidence is: For confidence level 0.95 the value of Zα/2 = 1.96 -0.06301 ± 1.96 × 0.136983 (1- 0.136983)73 + 0.2 (1 - 0.2)80-0.06301 ± 1.96 × 0.060162-0.06301 ± 0.117918 Lower limit = -0.06301 - 0.117918 = -0.18093 Upper limit = -0.06301 + 0.117918 = 0.054904 The 95% confidence interval is (-0.18093, 0.054904)