Questions: 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, 1,080 stated the United States made the right decision. In 2008, the organization asked the same question of 1,500 adult Americans and found that 573 believed the United States made the right decision. Construct and interpret a 90% confidence interval for the difference between the two population proportions.

Solution Steps

For the year 2003, the sample proportion of adults who believed the United States made the right decision is calculated as follows:

\[ \hat{p}_1 = \frac{x_1}{n_1} = \frac{1080}{1500} = 0.72 \]

For the year 2008, the sample proportion is:

\[ \hat{p}_2 = \frac{x_2}{n_2} = \frac{573}{1500} = 0.382 \]

The difference in sample proportions between the two years is:

\[ \hat{p}_1 - \hat{p}_2 = 0.72 - 0.382 = 0.338 \]

To construct the 90% confidence interval for the difference between the two population proportions, we use the formula:

\[ (\hat{p}_1 - \hat{p}_2) \pm z \sqrt{\frac{\hat{p}_1(1 - \hat{p}_1)}{n_1} + \frac{\hat{p}_2(1 - \hat{p}_2)}{n_2}} \]

Where \( z \) is the z-score corresponding to a 90% confidence level, which is approximately \( 1.645 \).

Substituting the values:

\[ 0.72 - 0.382 \pm 1.645 \cdot \sqrt{\frac{0.72(1 - 0.72)}{1500} + \frac{0.382(1 - 0.382)}{1500}} \]

Calculating the standard error:

\[ \sqrt{\frac{0.72(0.28)}{1500} + \frac{0.382(0.618)}{1500}} \approx \sqrt{0.0001344 + 0.0001576} \approx \sqrt{0.000292} \approx 0.0171 \]

Now, substituting back into the confidence interval formula:

\[ 0.338 \pm 1.645 \cdot 0.0171 \]

Calculating the margin of error:

\[ 1.645 \cdot 0.0171 \approx 0.0281 \]

Thus, the confidence interval is:

\[ (0.338 - 0.0281, 0.338 + 0.0281) = (0.3099, 0.3661) \]

Final Answer