Question

A newspaper 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 40 male consumers was $133.67, and the average expenditure in a sample survey of 30 female consumers was $69.64. 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.) $ to $

Answer 1

	Male Consumers	Female Consumers
Sample Size	n1 = 40	n2 = 30
Sample Mean	X̅1 = $133.67	X̅2 =$69.64
Population Standard Deviation	σ1 = $45	σ2 = $20

a) Point estimate in dollars of the difference between the population mean expenditure for males       
and the population mean expenditure for females is given by      
      
= 133.67 - 69.64      
= 64.03      
Point Estimate = $64.03      
       
b) For 99% confidence interval, margin of error is given by      
         
For 99%, α = 0.01, α/2 = 0.005      
From the z-tables, or Excel function NORM.S.INV(α/2)      
z' = NORM.S.INV(0.005) = 2.576   (We take the positive value for calculations)   
         
ME = 20.6013      
Margin of Error = $20.60      
       
c) 99% Confidence interval is given by       
       
   = 64.03 ± 20.60      
= ($ 43.43, $ 84.63)      
99% confidence interval (in dollars) for the difference between the two population means is   ($ 43.43, $ 84.63)