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 $10,979. The standard deviation of the sample was $1,000.

a. Based on this sample information, develop a 90 percent confidence interval for the population mean yearly premium.

b. How large a sample is needed to find the population mean within $250 at 99 percent confidence?

Given Information:

Standard deviation = σ = $1000

Mean = μ = $10,979

Confidence level = 90%

Margin of error = $250

Sample size = n = 20

Required Information:

a) 90% confidence interval = ?

b) Sample size at 99% confidence = n = ?

Answer:

a) 90% confidence interval = ($10,611.17, $11,346.83)

b) Sample size at 99% confidence = n = 107

Step-by-step explanation:

a) The confidence interval is given by

confidence interval = μ ± z*σ/√n

Where μ is the mean annual health insurance premium, σ is the standard deviation, n is number of subscribers that were surveyed, and z is the corresponding z-score for 90% confidence level.

The z-score for 90% confidence level is 1.645

Confidence interval = 10,979 ± (1.645*1000) / √20

Confidence interval = 10,979 ± 367.83

upper limit = 10,979 + 367.83 = 11,346.83

lower limit = 10,979 - 367.83 = 10,611.17

Confidence interval = ($10,611.17, $11,346.83)

Which means that we are 90% confident that the annual health insurance premiums for a family with coverage through an employer is between ($10,611.17, $11,346.83)

b) To find the required sample size

Me = z * (σ/√n)

√n = z*σ/Me

n = (z*σ/Me) ²

Where z is the corresponding z-score, n is the required sample size σ is the standard deviation and Me is the margin of error that is $250

The z-score for 99% confidence level is 2.576

n = (2.576*1000/250) ²

n = 106.17

n = 107

Therefore, a sample size of 107 is needed to find the population mean within $250 at 99% confidence.