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

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 48 male consumers was $131, and the average expenditure in a sample survey of 32 female consumers was $63. 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 $14. 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.
to

Explanation / Answer

a.
Given that,
mean(x)=131
standard deviation , 1 =31
number(n1)=48
y(mean)=63
standard deviation, 2 =14
number(n2)=32
point estimate = 131-63 = 68
b.
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 = [ ( 131-63) ±Z a/2 * Sqrt( 961/48+196/32)]
= [ (68) ± Z a/2 * Sqrt( 26.1458) ]
= [ (68) ± 2.576 * Sqrt( 26.1458) ]
= [54.8281 , 81.1719]
c.
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 )=31
Sample Size(n1)=48
Standard deviation( sd2 )=14
Sample Size(n12=32
Margin of Error = [ Z a/2 * Sqrt( 961/48+196/32)]
= [ (Z a/2 * Sqrt( 26.1458) ]
= [ (2.576 * Sqrt( 26.1458) ]
= 13.1719

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