#### Question

A 2011 Gallup poll found that 76% of Americans believe that high achieving high school students should be recruited to become teachers. This poll was based on a random sample of 1002 Americans. a) Find a 90% confidence interval for the proportion of Americans who would agree with this. b) Interpret your interval in this context. c) Explain what "90% confidence" means. d) Do these data refute a pundit 's claim that 2/ 3 of Americans believe this statement? Explain.

#### Solution

Verified#### Step 1

1 of 5Given:

$n=1002$

$\hat{p}=76\%=0.76$

$c=90\%=0.90$

(a) For confidence level $1-\alpha=0.90$, determine $z_{\alpha/2}=z_{0.05}$ using the normal probability table in the appendix (look up 0.05 in the table, the z-score is then the found z-score with opposite sign):

$z_{\alpha/2}=1.645$

The margin of error is then:

$E=z_{\alpha/2}\cdot \sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}=1.645\times \sqrt{\dfrac{0.76(1-0.76)}{1002}}\approx 0.0222$

The boundaries of the confidence interval are then:

$\hat{p}-E=0.76-0.0222= 0.7378=73.78\%$

$\hat{p}+E=0.76+0.0222= 0.7822=78.22\%$