In a random sample of 110 customers of a high-speed internet provider, 63 said t

Question

In a random sample of 110 customers of a high-speed internet provider, 63 said that their service had been interrupted one or more times in the past month.

A. Find a 95% confidence interval for the proportion of customers whose service was interrupted one or more times in the past month. Round the answers to three decimal places.

B.Find a 99% confidence interval for the proportion of customers whose service was interrupted one or more times in the past month. Round the answers to three decimal places

C. Find the sample size needed for a 95% confidence interval to specify the proportion to within ±0.05. Round up the answer to the nearest integer

D.Find the sample size needed for a 99% confidence interval to specify the proportion to within ±0.05. Round up the answer to the nearest integer.

Explanation / Answer

Ans:

Sample proportion=63/110=0.573

n=110

a)95% confidence interval for true p

=0.573+/-1.96*sqrt(0.573*(1-0.573)/110)

=0.573+/-0.092

=(0.481, 0.665)

B)99% confidence interval for true p

=0.573+/-2.58*sqrt(0.573*(1-0.573)/110)

=0.573+/-0.122

=(0.451, 0.695)

C)

Sample size required,n=1.96^2*0.573*(1-0.573)/0.05^2

n=376

D)

Sample size required,n=2.58^2*0.573*(1-0.573)/0.05^2

n=651

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