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{Exercise 10.05 (Algorithmic)} The USA Today reports that the average expenditur

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Question

{Exercise 10.05 (Algorithmic)}

The USA Today reports that the average expenditure on Valentine's Day is $100.89. Do male and female consumers differ in the amounts they spend? The average expenditure in a sample survey of 49 male consumers was $135.67, and the average expenditure in a sample survey of 37 female consumers was $68.64. Based on past surveys, the standard deviation for male consumers is assumed to be $31, and the standard deviation for female consumers is assumed to be $24.

What is the point estimate of the difference between the population mean expenditure for males and the population mean expenditure for females (to 2 decimals)?

At 99% confidence, what is the margin of error (to 2 decimals)?

Develop a 99% confidence interval for the difference between the two population means (to 2 decimals). Use z-table.
(  ,   )

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{Exercise 10.05 (Algorithmic)}

The USA Today reports that the average expenditure on Valentine's Day is $100.89. Do male and female consumers differ in the amounts they spend? The average expenditure in a sample survey of 49 male consumers was $135.67, and the average expenditure in a sample survey of 37 female consumers was $68.64. Based on past surveys, the standard deviation for male consumers is assumed to be $31, and the standard deviation for female consumers is assumed to be $24.

What is the point estimate of the difference between the population mean expenditure for males and the population mean expenditure for females (to 2 decimals)?

At 99% confidence, what is the margin of error (to 2 decimals)?

Develop a 99% confidence interval for the difference between the two population means (to 2 decimals). Use z-table.
(  ,   )

Explanation / Answer

The statistical software output for this problem is:

Two sample Z summary confidence interval:
1 : Mean of population 1 (Std. dev. = 31)
2 : Mean of population 2 (Std. dev. = 24)
1 - 2 : Difference between two means

99% confidence interval results:

Hence,

Point estimate = 67.03

Margin of error = 67.03 - 51.75 = 15.28

99% confidence interval: (51.75, 82.31)

Difference n1 n2 Sample mean Std. err. L. limit U. limit 1 - 2 49 37 67.03 5.9312572 51.752094 82.307906