Question

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A restuarant would like to estimate the proportion of tips that exceed 18% of its dinner bills. Without any knowledge of the population proportion, determine the sample size needed to construct a 97% confidence interval with a margin of error of no more than 6% to estimate the proportion.

A restuarant would like to estimate the proportion of tips that exceed 18% of its dinner bills. Without any knowledge of the population proportion, determine the sample size needed to construct a 97% confidence interval with a margin of error of no more than 6% to estimate the proportion.

Answer 1

Solution :

Given that,

$\hat p$ = 0.5

1 - $\hat p$ = 0.5

margin of error = E = 0.06

Z_$\alpha$/2 = 2.17

sample size = n = (Z_{$\alpha$ / 2} / E)² * $\hat p$ * (1 - $\hat p$ )

= (2.17 / 0.06)² * 0.5 * 0.5

= 327.007

sample size = n = 328