Question

Families USA, a monthly magazine that discusses issues related to
health and health costs, surveyed 22 of its subscribers. It found that
the annual health insurance premiums for a family with coverage
through an employer averaged $10,800. The standard deviation of the
sample was $1,060. (Use t Distribution Table.)

A.Based on this sample information, develop a 90% confidence interval
for the population mean yearly premium. (Round your answers to the
nearest whole dollar amount.)

B.How large a sample is needed to find the population mean within $240
at 90% confidence? (Round up your answer to the next whole number.)

Answer 1

A.

90% confidence interval = [mean - 1.64*SD, mean + 1.64*SD]

= [10800 - 1.64*1060, 10800 + 1.64*1060]

= [$ 9061.6 , $ 12538.4]

= [$ 9062 , $ 12538] {rounded to whole dollars}

B.

SD population = SD*(n^0.5)

1060 is SD for 22 people

SDpopulation = 1060*(22^0.5)

SD of sample = 1060/(n^0.5)

range of confidence interval 90% = 2*1.64*SD

2*1.64*SD = $240

2*1.64*($1060/(n^0.5)) = $240

n = [2*1.64*1060/240]^2 = 209.86 = 210 {rounded to next whole number}

n=210

(please UPVOTE)