Bank of America\'s Consumer Spending Survey collected data on annual credit card

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. 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 d = $881, and the sample standard deviation was sd = $1,102. a. 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. b. Use a.05 level of significance. What is the p-value? Can you conclude that the population means differ? c. Which category, groceries or dining out, has a higher population mean annual credit card charge? 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)?

Explanation / Answer

a)

Formulating the null and alternative hypotheses,              
              
Ho:   u   =   0  
Ha:    u   =/   0  

b)
              
As we can see, this is a    two   tailed test.      
Getting the test statistic, as              
              
X = sample mean =    881          
uo = hypothesized mean =    0          
n = sample size =    42          
s = standard deviation =    1102          
              
Thus, z = (X - uo) * sqrt(n) / s =    5.181064025          
              
Also, the p value is              
              
p =    2.20624*10^-7   [ANSWER, almost 0]

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c)

As z > 0, GROCERIES have a higher mean. [ANSWER]

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c-2.)

The point estimate is the sample mean,

d = 881 [ANSWER]

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c-3.

Note that              
Margin of Error E = z(alpha/2) * s / sqrt(n)              
Lower Bound = X - z(alpha/2) * s / sqrt(n)              
Upper Bound = X + z(alpha/2) * s / sqrt(n)              
              
where              
alpha/2 = (1 - confidence level)/2 =    0.025          
X = sample mean =    881          
z(alpha/2) = critical z for the confidence interval =    1.959963985          
s = sample standard deviation =    1102          
n = sample size =    42          
              
Thus,              
Margin of Error E =    333.2767675          
Lower bound =    547.7232325          
Upper bound =    1214.276767          
              
Thus, the confidence interval is              
              
(   547.7232325   ,   1214.276767   ) [ANSWER]

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