The average expenditure on Valentine\'s Day was expected to be $100.89 ( USA Tod
ID: 3158389 • Letter: T
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 40 male consumers was $134, and the average expenditure in a sample survey of 35 female consumers was $66. Based on past surveys, the standard deviation for male consumers is assumed to be $34, and the standard deviation for female consumers is assumed to be $16. 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.
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Explanation / Answer
a)
The best estimate is the difference in sample means,
X1 - X2 = 134 - 66 = $68 [ANSWER]
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b)
Calculating the means of each group,
X1 = 134
X2 = 66
Calculating the standard deviations of each group,
s1 = 34
s2 = 16
Thus, the standard error of their difference is, by using sD = sqrt(s1^2/n1 + s2^2/n2):
n1 = sample size of group 1 = 40
n2 = sample size of group 2 = 35
Also, sD = 6.017830649
For the 0.99 confidence level, then
alpha/2 = (1 - confidence level)/2 = 0.005
Z(alpha/2) = 2.575829304
hence,
margin of error = Z(alpha/2) * sD = 15.50090453 [ANSWER]
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c)
Alsp,
lower bound = [X1 - X2] - Z(alpha/2) * sD = 52.49909547
upper bound = [X1 - X2] + Z(alpha/2) * sD = 83.50090453
Thus, the confidence interval is
( 52.49909547 , 83.50090453 ) [ANSWER]
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