The average expenditure on Valentines Day was expected to be

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 42 male consumers was $139, and the average expenditure in a sample survey of 33 female consumers was $67. 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 $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.

Solution

a)

X1 =    139
X2 =    67
Thus, the point estimate is

X1 - X2 = 72 [ANSWER]

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b)

Calculating the standard deviations of each group,              
              
s1 =    32          
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 =    42          
n2 = sample size of group 2 =    33          
Thus, df = n1 + n2 - 2 =    73          

Also, sD =    5.6690853          

Note that              

Margin of error = z*alpha/2 * sD
  
Thus,

Margin of error = 14.60259604 [ANSWER]

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c)
          
Lower Bound = X - z(alpha/2) * sD              
Upper Bound = X + z(alpha/2) * sD              
              
where              
              
alpha/2 = (1 - confidence level)/2 =    0.005          
              
Thus,              

X = sample mean =    72          

z(alpha/2) = critical z for the confidence interval =    2.575829304          

sD = standard error =    5.6690853          
              
              
Thus,              
              
Lower bound =    57.39740396          
Upper bound =    86.60259604          
              
Thus, the confidence interval is              
              
(   57.39740396   ,   86.60259604   ) [ANSWER]