Question

Families USA, a monthly magazine that discusses issues related to health and health costs, surveyed 20 of its subscribers. It found that the annual health insurance premiums for a family with coverage through an employer averaged $10,950. The standard deviation of the sample was $920. (Use t Distribution Table.)

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.)

Answer: $10,594 to $11,306

How large a sample is needed to find the population mean within $230 at 98% confidence? (Round up your answer to the next whole number.) (I don't have the answer to this one and the one I through was right is wrong)

Answer 1

Solution:
Given in the question

the annual health insurance premiums for a family with coverage through an employer averaged = $10950
Sample standard deviation = 920
Sample size = 20
Degree of freedom = 20 -1 =19
Here we will use t test as population standard deviation is unknown and sample size is less than 30
So here alpha = 0.1, alpha/2 = 0.05
talpha/2 at df=19 is 1.73
90% Confidence interval can be calculated as
Mean +/- talpha/2*SD/sqrt(n)
10950 +/- 1.73*920/sqrt(20)
10950 +/- 355.89
So 90% confidence interval is 10594.11 to 11305.89
after round off 90% Confidence interval is 10594.11 to 11305.89
Solution(b)
Here Given Margin of error = 230
at 98% confidence interval so alpha = 1-0.98 = 0.02
alpha/2 = 0.02/2 = 0.01
Here we will use Z test as we need to find sample size so Zalpha/2 = 2.33
Sample size can be calculated as
Sample size n = (Zalpha/2 * S/Margin of error)^2 = ((2.33*920)/230)^2 = (9.32)^2 = 86.86 or 87
So 87 sample is needed to find the population mean within 230 at 98% confidence.