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 $10,979. The standard deviation of the sample was $1,000.a. Based on this sample information, develop a 90 percent confidence interval for thepopulation mean yearly premium.b. How large a sample is needed to find the population mean within $250 at 99 percentconfidence?

Answer 1

(a) The 90 percent confidence interval for the population mean yearly premium is ($10,974.53, $10983.47).

(b) The sample size required is 107.

Explanation:

(a)

The (1 - α)% confidence interval for population mean is:

`CI=\bar x\pm t_(\alpha/2, (n-1))* (s)/(√(n))`

Given:

`\bar x=\$10,979\\s=\$1000\\n=20`

Compute the critical value of t for 90% confidence level as follows:

`t_(\alpha/2, (n-1))=t_(0.10/2, (20-1))=t_(0.05, 19)=1.729`

*Use a t-table.

Compute the 90% confidence interval for population mean as follows:

`CI=\bar x\pm t_(\alpha/2, (n-1))* (s)/(√(n))`

`=10979\pm 1.729* (1000)/(√(20))\\=10979\pm4.47\\ =(10974.53, 10983.47)`

Thus, the 90 percent confidence interval for the population mean yearly premium is ($10,974.53, $10983.47).

(b)

The margin of error is provided as:

MOE = $250

The confidence level is, 99%.

The critical value of z for 99% confidence level is:

`z_(\alpha/2)=z_(0.01/2)=z_(0.005)=2.58`

Compute the sample size as follows:

`MOE= z_(\alpha/2)* (s)/(√(n))`

`n=[(z_(\alpha/2)* s)/(MOE) ]^(2)`

`=[(2.58* 1000)/(250)]^(2)`

`=106.5024\\\approx107`

Thus, the sample size required is 107.