1) Construct a 99% confidence interval to estimate the population proportion wit

Question

1) Construct a 99% confidence interval to estimate the population proportion with a sample proportion equal to 0.44 and a sample size equal to 100.

A 99% confidence interval estimates that the population proportion is between a lower limit of ___ and an upper limit of ____. (Round to three decimal places as needed.)

A) Determine the margin of error for a confidence interval to estimate the population proportion for the following confidence levels with a sample proportion equal to 0.28 and n=100.

a. The margin of error for a confidence interval to estimate the population proportion for the 90% confidence level is

b. The margin of error for a confidence interval to estimate the population proportion for the 95% confidence level is

c. The margin of error for a confidence interval to estimate the population proportion for the 98% confidence level is

B) Determine the sample size n needed to construct a 99% confidence interval to estimate the population mean when =82 and the margin of error equals 13.

n=

C) Determine the sample size n needed to construct a 95% confidence interval to estimate the population mean when =51 and the margin of error equals 5.

n=

Explanation / Answer

1)here p=0.44

and std error =(p(!-p)/n)1/2 =(0.0496

for 99% CI, z=2.5758

hence confidence interval =p -/+ z*Std error =0.312 ; 0.568

Aa) here std errror =(p(!-p)/n)1/2 = 0.0449

for 90% CI, z=1.6449

hence margin of error =z*Std error =0.074

b) for 95% , z=1.96

margin of error =0.088

c)for 99% CI, z=2.5758

margin of error =0.116

B)as sample size =(Z*std deviation/margin of error)2 =~264

C)as sample size =(Z*std deviation/margin of error)2 =400

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