{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 82.307906Related Questions
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