Question

The distribution of blood cholesterol level in the population of young men aged 20 to 34 years is close to Normal with standard deviation σ = 45 milligrams per deciliter (mg/dL). You measure the blood cholesterol of 12 cross-country runners. The mean level is x = 172 mg/dL. Assume that σ is the same as in the general population.

How large a sample is needed to cut the margin-of-error in half for a 90% confidence interval of the mean level μ among the cross-country runners?

24    2448

Answer 1

Solution:

Given:

Population  standard deviation σ = 45 milligrams per deciliter (mg/dL)

Sample size = n= 12

We have to find sample size is needed to cut the margin-of-error in half for a 90% confidence interval of the mean level μ among the cross-country runners.

thus first find margin of error for 12 runners.

Z_c is z critical value for c = 90% confidence level.

Find Area = ( 1 + c ) / 2 = ( 1 + 0.90) / 2 = 1.90 / 2 = 0.9500

Look in z table for Area = 0.9500 or its closest area and find corresponding z value.

Area 0.9500 is in between 0.9495 and 0.9505 and both the area are at same distance from 0.9500

Thus we look for both area and find both z values

Thus Area 0.9495 corresponds to 1.64 and 0.9505 corresponds to 1.65

Thus average of both z values is : ( 1.64+1.65) / 2 = 1.645

Thus Z_c = 1.645

thus

thus half margin of error =

thus new sample size is:

thus answer is 48