A restaurant 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 96​% confidence interval with a margin of error of no more than 4​%to estimate the proportion.The sample size needed is

Question
Asked Oct 20, 2019

A restaurant 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 96​% confidence interval with a margin of error of no more than 4​%to estimate the proportion.

The sample size needed is
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Expert Answer

Step 1

Given data

Error = 4% = 0.04

Confi...

Significance level = a = 1 - 0.96 = 0.04
Zo.04 = Zo.02 = -2.054
2
(From Excel = NORM. S.INV (0.02))
Since, population proportion is not given
Assuming Population proportion 0.5
Margin of error formula is given by
|P(1-р)
E Za X
п
Simplifying the above formula
2
Za
2
-2.054
n %3D p(1 — р) х
2
= 659.20 660
= 0.5 x 0.5 x
-
Е
0.04
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Significance level = a = 1 - 0.96 = 0.04 Zo.04 = Zo.02 = -2.054 2 (From Excel = NORM. S.INV (0.02)) Since, population proportion is not given Assuming Population proportion 0.5 Margin of error formula is given by |P(1-р) E Za X п Simplifying the above formula 2 Za 2 -2.054 n %3D p(1 — р) х 2 = 659.20 660 = 0.5 x 0.5 x - Е 0.04

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