The average expenditure on Valentine\'s Day was expected to be $100.89 (USA Toda

Question

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 52 male consumers was $134, and the average expenditure in a sample survey of 34 female consumers was $69. 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 $20. The z value is 2.576 . Round your answers to 2 decimal places. a. What is the point estimate of the difference between the population mean expenditure for males and the population mean expenditure for females? b. At 99% confidence, what is the margin of error? c. Develop a 99% confidence interval for the difference between the two population means.

Explanation / Answer

a.
Point of estimate = 134-69 = 65
b.
Margin of Error = Z a/2 * Sqrt ( sd1 ^2 / n1 + sd2 ^2 /n2 )
Where,
sd1 = SD of Sample 1, sd2 = SD of sample2
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
Standard deviation( sd1 )=32
Sample Size(n1)=52
Standard deviation( sd2 )=20
Sample Size(n12=34
Margin of Error = [ Z a/2 * Sqrt( 1024/52+400/34)]
= [ (Z a/2 * Sqrt( 31.457) ]
= [ (2.576 * Sqrt( 31.457) ]
= 14.4479

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)=134
Standard deviation( sd1 )=32
Sample Size(n1)=52
Mean(x2)=69
Standard deviation( sd2 )=20
Sample Size(n12=34
CI = [ ( 134-69) ±Z a/2 * Sqrt( 1024/52+400/34)]
= [ (65) ± Z a/2 * Sqrt( 31.457) ]
= [ (65) ± 2.576 * Sqrt( 31.457) ]
= [50.5521 , 79.4479]

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