Question

North American Weekly, a biweekly magazine that discusses topics related to Native American health, surveyed 19 of its subscribers. It found that the annual health insurance premiums for a family with coverage through an employer averaged $10,230. The standard deviation of the sample was $1,060.

(1.)	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 $___.

Answer 1

Solution :

Given that,

Point estimate = sample mean = = $10230

sample standard deviation = s = $1060

sample size = n = 19

Degrees of freedom = df = n - 1 = 19 - 1 = 18

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t _/2,df = t_0.025,18 = 2.101

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

= 2.101 * (1060 / 19)

= 511

The 95% confidence interval estimate of the population mean is,

- E < < + E

10230 - 511 < < 10230 + 511

9719 < < 10741

Confidence interval for the population mean yearly premium is between $9719 and $10741