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stellarik [79]
3 years ago
7

(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____ %
Mathematics
1 answer:
garik1379 [7]3 years ago
5 0

<u>Answer-</u>

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

<u>Solution-</u>

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\%

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