Question

Surgery U.S. insurers’ costs for knee replacement surgery range from $17,627 to $25,462. Estimate the population variance (standard deviation) in cost with 98% confidence based on a random sample of 10 persons who have had this surgery. The retail costs (for uninsured persons) for the same procedure range from $40,640 to $58,702. Estimate the population variance and standard deviation in cost with 98% confidence based on a sample of 10 persons, and compare your two intervals.

Answer 1

Answer:

Insured persons:

Population sd = $4,391.4

Population variance = 19,284,393.96 square dollars

Uninsured persons:

Population sd = $10,123.5

Population variance = 102,485,252.3 square dollars

The cost interval for insured persons is less than that of uninsured persons.

Explanation:

Insured persons:

98% confidence interval is ($17,627, $25,462)

Lower limit = $17,627

Upper limit = $25,462

Margin of error (E) = (upper limit - lower limit) ÷ 2 = (25,462 - 17,627) ÷ 2 = 7,835 ÷ 2 = $3917.5

n = 10

degree of freedom = n-1 = 10-1 = 9

confidence level (C) = 98% = 0.98

significance level = 1 - C = 1 - 0.98 = 0.02 = 2%

critical value (t) corresponding to 9 degrees of freedom and 2% significance level is 2.821

Population sd = E×√n/t = 3917.5×√10/2.821 = $4391.4

Population variance = (population sd)^2 = ($4391.4)^2 = 19,284,393.96 square dollars

Uninsured persons:

98% confidence interval is ($40,640, $58,702)

Lower limit = $40,640

Upper limit = $58,702

Margin of error (E) = (58,702 - 40,640) ÷ 2 = 18,062/2 = $9,031

n = 10

degree of freedom = n-1 = 10-1 = 9

significance level = 2%

critical value (t) = 2.821

Population sd = E×√n/t = 9,031×√10/2.821 = $10,123.5

Population variance = ($10,123.5)^2 = 102,485,252.3 square dollars.

The cost interval for insured persons is less than than that of uninsured persons.