Question

In​ 2003, an organization surveyed 1,508 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 1,508 adult Americans​ surveyed, 1,085 stated the United States made the right decision. In​ 2008, the organization asked the same question of 1,508 adult Americans and found that 570 believed the United States made the right decision.
1. Construct and interpret a​ 90% confidence interval for the difference between the two population​ proportions, p 2003 - p 2008.

Answer 1

Answer:

The confidence interval for the difference of proportions of the population p2003-p2008 is $0.31\leq\Delta \pi \leq 0.37$.

Step-by-step explanation:

In this question we have to construct a 90% confidence interval (CI) for a difference of proportions.

The proportion for 2003 is $p_1=1085/1508=0.72$.

The proportion for 2008 is $p_2=570/1508=0.38$.

The difference in proportions is $\Delta p=p_1-p_2=0.72-0.38=0.34$.

The standard deviation of the difference of proportions can be estimated as:

$s=\sqrt{\frac{p_1(1-p_1)}{n_1}+\frac{p_2(1-p_2)}{n_2} }=\sqrt{\frac{0.72(1-0.72)}{1508}+\frac{0.38(1-0.38)}{1508} }=\sqrt{0.00029} =0.017$

For a 90% CI, the z-value is 1.64.

Then we can construct the CI as

$\Delta p-z\sigma\leq\Delta \pi \leq\Delta p+z\sigma\\\\0.34-1.64*0.017\leq\Delta \pi \leq0.34+1.64*0.017\\\\ 0.34-0.03\leq\Delta \pi \leq0.34+0.03\\\\0.31\leq\Delta \pi \leq 0.37$