(1 point) a random sample of 600 movie goers in flagstaff found 252 movie goers who had bought popcorn on their last visit. find a 95% confidence interval for the true percent of movie goers in flagstaff who have bought popcorn on their last visit. express your results to the nearest hundredth of a percent.%

Answer-

A 95% confidence interval for the true percent of movie goers is 36.41% to 44.25%

Solution-

Given,

n = 600 (sample size)

x = 252 (number of people who bought)

Confidence interval = 95%, so z = 1.96

We know that,

where,

M = sample mean

Z  = Z statistic determined by confidence level

SE = standard error of mean

Calculating the values,

from the tables

Putting all the values in the formula,

how am i supposed to answer

step-by-step explanation: