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A restaurant would like to estimate the proportion of tips that exceed 18% of its dinner...


A restaurant would like to estimate the proportion of tips that exceed 18% of its dinner bills. Without any knowledge of the
my knowledge of the population proportion, determine the sample size needed to construct a 96% confidence interval with a mar
A restaurant would like to estimate the proportion of tips that exceed 18% of its dinner bills. Without any knowledge of the no more than 6% to estimate the proportion The sample size needed is (Round up to the nearest integer.)
my knowledge of the population proportion, determine the sample size needed to construct a 96% confidence interval with a margin of error of
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ReportAnswer #1

Given:

p = 0.18, q = 1 - 0.18 = 0.82

For 96% confidence, z = 2.05

E = 0.06

Hence,

Sample size required

n = (0.18)(0.82)(\frac{2.05}{0.06})^2

n = 173

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