A research study related to health and health costs, surveyed 30 families. It fo

Question

A research study related to health and health costs, surveyed 30 families. It found that the annual health insurance premiums for a family with coverage through an employer averaged $16,469. The standard deviation of the sample was $2,000. (a) What is the population mean? What is the best estimate of the population mean? (b) Based on this sample information, develop a 95% confidence interval for the population mean yearly premium. (c) How large of a sample is needed to assess the population mean with an allowable error of $600 at 99% confidence?

Explanation / Answer

a) The population mean is unknown. The best estimate of the population mean is the sample mean $16,469. b) The 95% critical z-value is plus or minus 1.96. Sample mean = 16469 Standard deviation = 2000 Standard error of mean = s / v n Standard error of mean = 2000 / v 30 SE = 2000/5.4772 Standard error of mean 365.1484 Confidence interval 16469-(365.1484)(1.96) and 16469+(365.1484)(1.96) 95% confidence interval (15753.31, 17184.69) c) The 99% z critical value is 2.575 Sample size n = (z sd /error)^2 = [(2.575) (2000) / 600]^2 = 73.67 =74

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