The average expenditure on Valentines Day was expected to be

The average expenditure on Valentine\'s Day was expected to be $100.89 (USA Today, February 13, 2006). 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 40 female consumers was $68.64. Based on past surveys, the standard deviation for male consumers is assumed to be $32, 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

Solution

a)
Point of estimate = 135.67-68.64 = 67.03
b)
M.E = Z a/2 * Sqrt( 1024/49+576/40) = 2.576 * Sqrt( 35.298) = 15.31
c)
CI = x1 - x2 ± Z a/2 * Sqrt ( sd1 ^2 / n1 + sd2 ^2 /n2 )
Where,
x1 = Mean of Sample 1, x2 = Mean of sample2
sd1 = SD of Sample 1, sd2 = SD of sample2
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
Mean(x1)=135.67
Standard deviation( sd1 )=32
Sample Size(n1)=49
Mean(x2)=68.64
Standard deviation( sd2 )=24
Sample Size(n12=40
CI = [ ( 135.67-68.64) ±Z a/2 * Sqrt( 1024/49+576/40)]
= [ (67.03) ± Z a/2 * Sqrt( 35.298) ]
= [ (67.03) ± 2.576 * Sqrt( 35.298) ]
= [51.7254 , 82.3346]