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 44 male consumers was $135.67, and the average expenditure in a sample survey of 34 female consumers was $68.64. Based on past surveys, the standard deviation for male consumers is assumed to be $39, and the standard deviation for female consumers is assumed to be $24.

a) 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)?

b)At 99% confidence, what is the margin of error (to 2 decimals)?

c)Develop a 99% confidence interval for the difference between the two population means (to 2 decimals).

Explanation / Answer

a)

The point estimate is the diffrence in sample means,

X1 - X2 = 135.67 - 68.64 = 67.03 [ANSWER]

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b)

Calculating the means of each group,              
              
X1 =    135.67          
X2 =    68.64          
              
Calculating the standard deviations of each group,              
              
s1 =    39          
s2 =    24          
              
Thus, the standard error of their difference is, by using sD = sqrt(s1^2/n1 + s2^2/n2):              
              
n1 = sample size of group 1 =    44          
n2 = sample size of group 2 =    34          
Thus, df = n1 + n2 - 2 =    76          
Also, sD =    7.17700204          
              
              
For the   0.99   confidence level, then      
              
alpha/2 = (1 - confidence level)/2 =    0.005          
t(alpha/2) =    2.642078313          

Hence,

Margin of error = t(alpha/2) * sD = 18.96220144 [ANSWER]

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c)

Also,
              
lower bound = [X1 - X2] - t(alpha/2) * sD =    48.06779856          
upper bound = [X1 - X2] + t(alpha/2) * sD =    85.99220144          
              
Thus, the confidence interval is              
              
(   48.06779856   ,   85.99220144   ) [ANSWER]

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Hi! If you use another method/formula in calculating the degrees of freedom in this t-test, please resubmit this question together with the formula/method you use in determining the degrees of freedom. That way we can continue helping you! Thanks!

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