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    = $852, and the sample standard deviation was s_d= $1,143.

1. Formulate the null and alternative hypotheses to test for no difference between the population mean credit card charges for groceries and the population mean credit card charges for dining out.
   H₀:  _d Selectgreater than or equal to 0greater than 0less than or equal to 0less than 0equal to 0not equal to 0Item 1
   H_a:  _d Selectgreater than or equal to 0greater than 0less than or equal to 0less than 0equal to 0not equal to 0Item 2
   
2. Use a .05 level of significance. What is the p-value?
   The p-value is Selectless than .01between .01 and .02between .02 and .05between .05 and .10between .10 and .20between .20 and .40greater than .40Item 3
   
   Can you conclude that the population means differ?
   SelectThere is a difference between the annual mean expendituresCannot conclude there is a difference between the annual mean expendituresItem 4
   
3. Which category, groceries or dining out, has a higher population mean annual credit card charge?
   SelectGroceriesDining outItem 5
   
   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

x̅d = 852, sd = 1143, n = 42

a)

Null and Alternative hypothesis:

Ho : µd = 0

H1 : µd ≠ 0

b)

Test statistic:

t = (x̅d)/(sd/√n) = (852)/(1143/√42) = 4.8308

df = n-1 = 41

p-value = T.DIST.2T(ABS(4.8308), 41) = 0.0000

The p-value is less than .01

There is a difference between the annual mean expenditures.

c)

Groceries has a higher population mean.

Point estimate = 852

95% Confidence interval :

At α = 0.05 and df = n-1 = 41, two tailed critical value, t-crit = T.INV.2T(0.05, 41) = 2.020

Lower Bound = x̅d - t-crit*sd/√n = 852 - 2.02 * 1143/√42 = 496

Upper Bound = x̅d + t-crit*sd/√n = 852 + 2.02 * 1143/√42 = 1208