Question

A newspaper reports that the average expenditure on Easter is $100.89. Do male and female consumers differ in the amounts they spend? The average expenditure in a sample survey of 40 male consumers was $133.67, and the average expenditure in a sample survey of 30 female consumers was $65.64. Based on past surveys, the standard deviation for male consumers is assumed to be $45, and the standard deviation for female consumers is assumed to be $20.

(a) What is the point estimate (in dollars) of the difference between the population mean expenditure for males and the population mean expenditure for females? (Use male − female.)

(b) At 99% confidence, what is the margin of error (in dollars)? (Round your answer to the nearest cent.)

(c) Develop a 99% confidence interval (in dollars) for the difference between the two population means. (Use male − female. Round your answer to the nearest cent.)

Answer 1

Given,

Number of male consumers in the random survey : sample size for male consumers : n1 =40

Sample  average expenditure for male consumers : = 133.67

Population Standard deviation for male consumers : = 45

Number of female consumers in the random survey : sample size for female consumers : n2 =30

Sample  average expenditure for female consumers : = 65.64

Population Standard deviation for female consumers : = 20

(a) What is the point estimate (in dollars) of the difference between the population mean expenditure for males and the population mean expenditure for females

point estimate (in dollars) of the difference between the population mean expenditure for males and the population mean expenditure for females

= difference between the sample mean expenditure for males and the sample mean expenditure for females

= - = 133.67 - 65.64 = 68.03

point estimate (in dollars) of the difference between the population mean expenditure for males and the population mean expenditure for females = $68.03

(b) At 99% confidence, what is the margin of error (in dollars)

Margin of Error : E

for 99% confidence level = (100-99)/100 =0.01

/2 = 0.01/2 =0.005

Z0.005 = 2.5758

At 99% confidence, Margin of error = $20.60

(c) Develop a 99% confidence interval (in dollars) for the difference between the two population means.

99% confidence interval (in dollars) for the difference between the two population means:

= Point estimate Margin of Error = 68.03 20.60 = (47.43,88.63)

99% confidence interval (in dollars) for the difference between the two population means: = (47.43, 88.63)