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

Answer:

(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.

Step-by-step explanation:

(a)

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

$CI=\bar x\pm t_{\alpha/2, (n-1)}\times \frac{s}{\sqrt{n}}$

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:

$CI=\bar x\pm t_{\alpha/2, (n-1)}\times \frac{s}{\sqrt{n}}$

     $=10979\pm 1.729\times \frac{1000}{\sqrt{20}}\\=10979\pm4.47\\ =(10974.53, 10983.47)$

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:

$MOE= z_{\alpha/2}\times \frac{s}{\sqrt{n}}$

      $n=[\frac{z_{\alpha/2}\times s}{MOE} ]^{2}$

         $=[\frac{2.58\times 1000}{250}]^{2}$

         $=106.5024\\\approx107$

Thus, the sample size required is 107.