Madison Dulay Question 5, 11.1.28 Part 3 of 3 HW Score: $77.78 \%, 4.67$ of 6 points Points: 0.33 of 1 Save In 2003, an organization surveyed 1,500 adult Americans and asked about a certain war, "Do you believe the United States made the right or wrong decision to use military force?" Of the 1,500 adult Americans surveyed, 1,086 stated the United States made the right decision. In 2008, the organization asked the same question of 1,500 adult Americans and found that 580 believed the United States made the right decision. Construct and interpret a $90 \%$ confidence interval for the difference between the two population proportions, $p_{2003}-p_{2008}$. Interpret the $90 \%$ confidence interval for the difference between the two population proportions, $p_{2003}-p_{2008}$. Choose the correct answer below. A. There is $90 \%$ confidence that the difference in the portion of adult Americans from 2003 to 2008 who believe the United States made the right decision to use military force in the country In the country is greater than the lower bound. B. There is 10% confidence that the difference in the proportion of adult Americans from 2003 to 2008 who believe the Jnited States made the right decision to use military force in the country is between the lower and upper bounds of the interval. C. There is 90% confidence that the difference in the proportion of adult Americans from 2003 to 2008 who believe the United States made the right decision to use military force in the country is between the lower and upper bounds of the interval
Solution
Step 1 ：First, we calculate the proportions for each year. In 2003, the proportion of adult Americans who believed the United States made the right decision to use military force is \(\frac{1086}{1500} = 0.724\). In 2008, the proportion is \(\frac{580}{1500} = 0.38667\).
Step 2 ：Next, we calculate the difference between the two proportions: \(0.724 - 0.38667 = 0.33733\).
Step 3 ：We then calculate the standard error using the formula \(\sqrt{\frac{p_{2003}(1-p_{2003})}{n_{2003}} + \frac{p_{2008}(1-p_{2008})}{n_{2008}}} = \sqrt{\frac{0.724(1-0.724)}{1500} + \frac{0.38667(1-0.38667)}{1500}} = 0.01707\).
Step 4 ：For a 90% confidence level, the z-score is approximately 1.645. We use this to calculate the confidence interval using the formula \((p_{2003} - p_{2008}) \pm z \times se\). The lower bound of the confidence interval is \(0.33733 - 1.645 \times 0.01707 = 0.30926\) and the upper bound is \(0.33733 + 1.645 \times 0.01707 = 0.36541\).
Step 5 ：\(\boxed{\text{Final Answer: The 90% confidence interval for the difference between the two population proportions, } p_{2003}-p_{2008} \text{, is approximately } (0.309, 0.365)\). This means we are 90% confident that the true difference in the proportion of adult Americans from 2003 to 2008 who believe the United States made the right decision to use military force is between 0.309 and 0.365.
