Question

In​ 2003, an organization surveyed 1 comma 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 comma 508 adult Americans​ surveyed, 1 comma 085 stated the United States made the right decision. In​ 2008, the organization asked the same question of 1 comma 508 adult Americans and found that 575 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 minus p 2008.

Answer 1

Answer:

$(0.719-0.381) - 1.64 \sqrt{\frac{0.719(1-0.719)}{1508} +\frac{0.381(1-0.381)}{1508}}=0.310$  

$(0.719-0.381) + 1.64 \sqrt{\frac{0.719(1-0.719)}{1508} +\frac{0.381(1-0.381)}{1508}}=0.366$  

And the 90% confidence interval would be given (0.310;0.366).  

We are confident at 90% that the difference between the two proportions is between $0.310 \leq p_A -p_B \leq 0.366$

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

$p_A$ represent the real population proportion for 2003  

$\hat p_A =\frac{1085}{1508}=0.719$ represent the estimated proportion for 2003

$n_A=1508$ is the sample size required for 2003

$p_B$ represent the real population proportion for 2008

$\hat p_B =\frac{575}{1508}=0.381$ represent the estimated proportion for 2008

$n_B=1508$ is the sample size required for 2008

$z$ represent the critical value for the margin of error  

The population proportion have the following distribution  

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$  

The confidence interval for the difference of two proportions would be given by this formula  

$(\hat p_A -\hat p_B) \pm z_{\alpha/2} \sqrt{\frac{\hat p_A(1-\hat p_A)}{n_A} +\frac{\hat p_B (1-\hat p_B)}{n_B}}$  

For the 90% confidence interval the value of $\alpha=1-0.90=0.1$ and $\alpha/2=0.05$, with that value we can find the quantile required for the interval in the normal standard distribution.  

$z_{\alpha/2}=1.64$  

And replacing into the confidence interval formula we got:  

$(0.719-0.381) - 1.64 \sqrt{\frac{0.719(1-0.719)}{1508} +\frac{0.381(1-0.381)}{1508}}=0.310$  

$(0.719-0.381) + 1.64 \sqrt{\frac{0.719(1-0.719)}{1508} +\frac{0.381(1-0.381)}{1508}}=0.366$  

And the 90% confidence interval would be given (0.310;0.366).  

We are confident at 90% that the difference between the two proportions is between $0.310 \leq p_A -p_B \leq 0.366$