Question

(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:____ to____ %

Answer 1

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,

$\mu = M\pm Z(SE)$

where,

M = sample mean

Z  = Z statistic determined by confidence level

SE = standard error of mean

Calculating the values,

$M=\frac{x}{n} =\frac{242}{600} =0.4033$

$Z=1.96$ from the tables

$SE=\sqrt{\frac{M\times (1-M)}{n}}$

$\Rightarrow SE=\sqrt{\frac{0.4033\times (1-0.4033)}{600}}=0.02$

Putting all the values in the formula,

$\mu = M\pm Z(SE)$

$\mu = 0.4033\pm 1.96(0.02)$

$=0.4033\pm 0.0392$

$=0.4425, 0.3641$

$= 44.25\%,36.41\%$