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 56 male consumers was $139, and the average expenditure in a sample survey of 38 female consumers was $61. Based on past surveys, the standard deviation for male consumers is assumed to be $30, and the standard deviation for female consumers is assumed to be $10. 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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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 56 male consumers was $139, and the average expenditure in a sample survey of 38 female consumers was $61. Based on past surveys, the standard deviation for male consumers is assumed to be $30, and the standard deviation for female consumers is assumed to be $10. 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)

The point estimate is

X1 - X2 = 139 - 61 = $78 [ANSWER]

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b)

Calculating the means of each group,              
              
X1 =    139          
X2 =    61          
              
Calculating the standard deviations of each group,              
              
s1 =    30          
s2 =    10          
              
Thus, the standard error of their difference is, by using sD = sqrt(s1^2/n1 + s2^2/n2):              
              
n1 = sample size of group 1 =    56          
n2 = sample size of group 2 =    38          

Also, sD =    4.32469739          
              
For the   0.99   confidence level, then      
              
alpha/2 = (1 - confidence level)/2 =    0.005          
z(alpha/2) =    2.576          
              

Margin of error = z(alpha/2) * sD = 2.756*4.32469739 = 11.919 [ANSWER]

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c)

lower bound = [X1 - X2] - z(alpha/2) * sD =    66.85957952          
upper bound = [X1 - X2] + z(alpha/2) * sD =    89.14042048          
              
Thus, the confidence interval is              
              
(   66.85957952   ,   89.14042048   ) [ANSWER]

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