Suppose thirty-nine communities gave an average of xbar = 147.8 reported cases o

Question

Suppose thirty-nine communities gave an average of xbar = 147.8 reported cases of larceny per year. Assume that sigma is known to be 39.9 cases per year. Find a 90%, 95%, and 98% confidence interval for the population mean annual number of reported larceny cases in such communities. Compare the margins of error. As the confidence level increase, do the margins of error increase?

a) The 90% confidence level has a margin of error of 65.6; the 95% confidence level has a margin of error of 78.2; and the 98% confidence level has a margin of error of 93.0. As the confidence level increases, the margins of error increases.

b) The 90% confidence level has a margin of error of 10.5; the 95% confidence level has a margin of error of 12.5; and the 98% confidence level has a margin of error of 14.9. As the confidence level increases, the margins of error increases.

c) The 90% confidence level has a margin of error of 1.7; the 95% confidence level has a margin of error of 2.0; and the 98% confidence level has a margin of error of 2.4. As the confidence level increases, the margins of error increases.

d) The 90% confidence level has a margin of error of 14.9; the 95% confidence level has a margin of error of 12.5; and the 98% confidence level has a margin of error of 10.5. As the confidence level increases, the margins of error decreases.

e) The 90% confidence level has a margin of error of 93.0; the 95% confidence level has a margin of error of 78.2; and the 98% confidence level has a margin of error of 65.6. As the confidence level increases, the margins of error decreases.

Explanation / Answer

We use the following z values for the different confidence intervals obtained from normal tables

Standard Margin of error is given by z*sigma/ sqrt(n)

Hence, answer is b

Calculation for z= 1.645 is 1.645* 39.9/Sqrt(39) =10.51

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