Question

Determine the sample size n needed to construct a 90​% confidence interval to estimate the population proportion when p overbar=0.66 and the margin of error equals 8​%.

n=_______​(Round up to the nearest​ integer.)

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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 98​% confidence interval with a margin of error of no more than 6​% to estimate the proportion.

The sample size needed is ________. ​(Round up to the nearest​ integer.)

Answer 1

Solution :

Given that,

= 0.66

1 - = 1 - 0.66 = 0.34

margin of error = E =8 % = 0.08

At 90% confidence level z

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z_{0.05 = 1.645}

Sample size = n = (Z_/2 / E)² * * (1 - )

= (1.645 / 0.08)² * 0.66 * 0.34

=95

Sample size = 95

(B)

Solution :

Given that,

= 0.18

1 - = 1 - 0.18 = 0.82

margin of error = E =6 % = 0.06

At 90% confidence level z

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z_{0.05 = 1.645}

Sample size = n = (Z_/2 / E)² * * (1 - )

= (1.645 / 0.06)² * 0.18 * 0.82

=111

Sample size = 111