A recent poll of 85 randomly selected cable subscribers found that 39% would be willing to pay extra for a new nature channel. To the nearest percent, with a confidence level of 95% (z*-score 1.96), what is the confidence interval for the proportion of cable subscribers who would be willing to pay extra for the new nature channel?
Accepted Solution
A:
The mean is 85*0.39 = 33.15, while the standard error is sqrt(0.39*0.61/85) = 0.0529. Using the z-score of 1.96, the confidence interval is:33.15 +/- 1.96*0.0529 = (33.05, 33.25)By dividing by 85, this corresponds to a proportion of:(0.3888, 0.3912)