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 $139.5, and the average expenditure in a sample survey of 37 female consumers was $65.76. Based on past surveys, the standard deviation for male consumers is assumed to be $35, 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? 11.46 b. At 99% confidence, what is the margin of error? 1.96 c. Develop a 99% confidence interval for the difference between the two population means. 66.41 to 67.65

Answer 1

1)

a)

difference in sample means = x̅1 - x̅2 =    139.5   -   65.76   =   73.74

b)

margin of error, E = Z*SE =    2.5758   *   5.696   =   14.67

c)

confidence interval is                   
Interval Lower Limit= (x̅1 - x̅2) - E =    73.740   -   14.671   =   59.07
Interval Upper Limit= (x̅1 - x̅2) + E =    73.740   +   14.671   =   88.41

2)

Degree of freedom, DF=   n1+n2-2 =    16              
t-critical value =    t α/2 =    1.7459   (excel formula =t.inv(α/2,df)          
                      
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    3.1038              
                      
std error , SE =    Sp*√(1/n1+1/n2) =    1.4723              
margin of error, E = t*SE =    1.7459   *   1.4723   =   2.5704  
                      
difference of means =    x̅1-x̅2 =    21.2000   -   22.800   = -1.6 or 1.6

confidence interval is                       
Interval Lower Limit=   (x̅1-x̅2) - E =    -1.6000   -   2.5704   =   -4.1704
Interval Upper Limit=   (x̅1-x̅2) + E =    -1.6000   +   2.5704   = 0.9704

-4170.4 and 970.4 (in 1000)

Please revert back in case of any doubt.

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