Exercise 1023 Algorithmic Bank of Americas Consumer Spending

{Exercise 10.23 (Algorithmic)}

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 = $838, and the sample standard deviation was sd = $1,131.

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   _________________   
Ha: d   _________________   

Use a .05 level of significance. What is the p-value?
The p-value is _________________   

Can you conclude that the population means differ?
_________________   

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)?
( ______ , ______ )

Solution

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 = 0
Ha: d =/= 0

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Use a .05 level of significance. What is the p-value?

Here,

z = d * sqrt(n) / sigma

z = 838*sqrt(42)/1131 = 4.801822021 [answer]

Using technology/table to get the right tailed area, then multiplying by 2 (As it is two tailed),

P = 0.00000157228 [ANSWER]

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Can you conclude that the population means differ?

As the P value is so low, then yes, we can conclude that they differ.

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Which category, groceries or dining out, has a higher population mean annual credit card charge?

It is groceries, as z > 0 here.

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

$838, as given [ANSWER]

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

Note that              
              
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 =    838          
z(alpha/2) = critical z for the confidence interval =    1.959963985          
s = sample standard deviation =    1131          
n = sample size =    42          
              
Thus,              
              
Lower bound =    495.9527913          
Upper bound =    1180.047209          
              
Thus, the confidence interval is              
              
(   495.9527913   ,   1180.047209   ) [ANSWER]