Question

Suppose a 2011 Gallup poll revealed that 76% of Americans believe that high achieving students should be recruited to become teachers. If this poll was taken from a voting district with sample size of 1021 at random, where there are 154,000 residents. Calculate the 95 percent confidence interval.

Answer 1

Answer:

95% Confidence interval:  (0.73,0.79)

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 1021

Percentage of Americans believe that high achieving students should be recruited to become teachers

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

95% Confidence interval:

$\hat{p}\pm z_{stat}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$

$z_{critical}\text{ at}~\alpha_{0.05} = \pm 1.96$

Putting the values, we get:

$0.76\pm 1.96(\sqrt{\dfrac{0.76(1-0.76)}{1021}})\\\\ = 0.76\pm 0.0261\\=(0.7339,0.7861)\\\approx (0.73,0.79)$