Question

An online survey reported that 60% of persons are not confident that they are saving enough for retirement. Suppose the Department of Economics would like to do a follow-up study to determine how many persons are saving each year toward retirement and want to use $100 as the desired margin of error for an interval estimate of the population mean. Use $1100 as a planning value for the standard deviation and recommend a sample size for each of the following situations: (a) A 90% confidence interval is desired for the mean amount saved. [10 marks] (b) A 95% confidence interval is desired for the mean amount saved. [10 marks] (c) When the desired margin of error is set, what happens to the sample size as the confidence level is increased? Would you recommend using a 99% confidence interval in this case? Discuss.

Answer 1

a)

for 90 % CI value of z=	1.645
standard deviation =	1100.00
margin of error E =	100
required sample size n=(z/E)2    =	328.0

b)

for 95 % CI value of z=	1.960
standard deviation =	1100.00
margin of error E =	100
required sample size n=(z/E)2    =	465.0

c)

as we increase  confidence level ; requirement of sample size also increases.

d)

for 99 % CI value of z=	2.576
standard deviation =	1100.00
margin of error E =	100
required sample size n=(z/E)2 =	803.0

as sample size requirement for 99% confidence is not very high ; we recommend using a 99% confidence interval in this case for better coverage of  population mean