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 $10,385. The standard deviation of the sample was $1,087.

    

(a)	Based on this sample information, develop a 95% confidence interval for the population mean yearly premium. (Round up your answers to the next whole number.)

Confidence interval for the population mean yearly premium is between $   and $ .

(b)	How large a sample is needed to find the population mean within $235 at 99% confidence? (Round up your answer to the next whole number.)

Sample size

Answer 1

Solution :

Given that,

a) = 10385

s = 1087

n = 20

Degrees of freedom = df = n - 1 = 20 - 1 = 19

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t _/2,df = t_0.025,19 =2.09

Margin of error = E = t_/2,df * (s /n)

= 2.09 * (1087 / 20)

= 508.73

Margin of error = 508.73

The 99% confidence interval estimate of the population mean is,

- E < < + E

10385 - 508.73  < < 10385 + 508.73

9876.27 < < 10893.73

(9876.27, 10893.73 )

b)

Solution :

Given that,

standard deviation = = 1087

margin of error = E = 235

At 99% confidence level the z is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z_/2 = Z_0.005 = 2.576

Sample size = n = ((Z_/2 * ) / E)²

= ((2.576 * 1087) / 235)²

=142

Sample size = 142