Question

.Alcohol Service Policy In a Gallup poll of 1004 adults, 93% indicated that restaurants and bars should refuse service to patrons who have had too much to drink. Construct a 95% confidence interval estimate of the percentage of all adults who believe that restaurants and bars should refuse service for those who have had too much to drink. Write a statement interpreting the confidence interval.

Determining Sample Size for a Survey See the survey described in Exercise 1. If you plan to conduct a new poll to confirm that the percentage continues to be correct, how many randomly selected adults must you survey if you want 95% confidence that the margin of error is four percentage points?

Answer 1

Given

p = percentage or proportion of adults believe that restaurants and bars should refuse service to patrons who have had too much to drink

p = 0.93 , n = sample size = 1004

95% confidence interval estimate of the percentage of all adults who believe that restaurants and bars should refuse service for those who have had too much to drink is

p^ - Z_{​​​​​​a/2}* sqrt( p^ *(1-p^)/n) < p < p^ + Z_{​​​​​​a/2}* sqrt(p^*(1-p^)/n)

For a = 0.05

Z_{​​​​​​a/2} = Z_{​​​​​0.025} = 1.96

0.93 - 1.96*sqrt( 0.93*0.07/1004) < p <

0.93 + 1.96*sqrt(0.93*0.07/1004)

0.93 - 0.0158 < p < 0.93 + 0.0158

0.91 < p < 0.95

Interpretation: We are 95% confident that the percentage of all adults who believe that restaurants and bars should refuse service for those who have had too much to drink is lies between 91% to 95%

We have to find Sample size n for c = confidence level = 0.95

E = margin of error = 0.04

n = p^ * (1 - p^ ) * ( Z_{​​​​​​c}​​​/ E)²

For c = 0.95 , Z_{​​​​​​0.95} = 1.96

n = 0.93 * 0.07 * ( 1.96/0.04)²

n = 157