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

95% Confidence interval: (0.73,0.79)

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{\frac{\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{(0.76(1-0.76))/(1021)})\\\\ = 0.76\pm 0.0261\\=(0.7339,0.7861)\\\approx (0.73,0.79)`