Question

Bank of America's Consumer Spending Survey collected data on annual credit card charges in seven different categories of expenditures: transportation, groceries, dining out, household expenses, home furnishings, apparel, and entertainment (U.S. Airways Attache, December 2003). Using data from a sample of 42 credit card accounts, assume that each account was used to identify the annual credit card charges for groceries (population 1) and the annual credit card charges for dining out (population 2). Using the difference data, the sample mean difference was     = $859, and the sample standard deviation was sd = $1,213.

What is the point estimate of the difference between the population means? $

What is the 95% confidence interval estimate of the difference between the population means (to the nearest whole number)?

Answer 1

point estimate of the difference between the population means = $859

-------------------------------------------------------------------

mean of difference ,    D̅ =   859
      
std dev of difference , Sd =        1213
Degree of freedom, DF=   n - 1 =    41
t-critical value =    t α/2,df =    2.0195 [excel function: =t.inv(α/2,df) ]
      
  
std error , SE =    Sd / √n =    187.17
margin of error, E =    t*SE =    377.997
      
mean of difference ,    D̅ =   859
95% confidence interval is       
Interval Lower Limit=   D̅ - E =   481.0026
Interval Upper Limit=   D̅ + E =   1236.9974
so,

confidence interval is (481 , 1237)