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    = $875, and the sample standard deviation was s_d= $1,070.

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

a)

null hypothesis: HO: μ		is equal to	0
Alternate Hypothesis: Ha: μ		is not equal to	0

b)

population mean μ=	0
sample mean 'x̄=	875.000
sample size   n=	42.00
sample std deviation s=	1070.000
std error 'sx=s/√n=	165.105
test stat t ='(x-μ)*√n/sx=	5.300
p value      =	0.0000

he p-value is  less than .01

we can conclude that There is a difference between the annual mean expenditures

c)Groceries has higher mean

point estimate of the difference =875

		for 95% CI; and 41 df, value of t=		2.0200
		margin of error E=t*std error    =		333.511
		lower bound=sample mean-E =		541
		Upper bound=sample mean+E =		1209

from above 95% confidence interval for population mean =(541 to 1209 )