A regional US commercial bank issues both Visa credit cards

A regional U.S. commercial bank issues both Visa credit cards and MasterCard credit cards. As a part of its annual review of the profitability of each type of credit card, the bank randomly samples 36 customers to measure the average annual charges per card. It has completed its analysis of the Visa card accounts and is now focused on its MasterCard customers. A random sample of 36 MasterCard accounts shows the following annual spending per account (rounded to the nearest dollar): Based on these randomly sampled accounts, what is the best point estimate of the true mean annual spending for MasterCard account holders? If the bank is interested in developing a 95% confidence interval estimate of mean annual spending, what distribution will be used to determine the critical value? Determine the standard error of the sampling distribution. Construct the 95% confidence interval estimate for the population mean of annual MasterCard spending for the bank\'s customers. If the bank desires to have a higher level of confidence in its interval estimate, what w ill happen to the margin of error?

Solution

a)

The point estimate is the sample mean here,

X = sample mean =    2776   [ANSWER]
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b)

We can use the z distribution, as n = 36 > 30 here.      
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c)

Here,

standard error = s/sqrt(n)

as

s = sample standard deviation =    556.801837          
n = sample size =    36          

Then

standard error = 92.80030617 [ANSWER]

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d)
              

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 =    2776          
z(alpha/2) = critical z for the confidence interval =    1.959963985          
s = sample standard deviation =    556.801837          
n = sample size =    36          
              
Thus,              
              
Lower bound =    2594.114742          
Upper bound =    2957.885258          
              
Thus, the confidence interval is              
              
(   2594.114742   ,   2957.885258   ) [ANSWER]

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e)

Larger confidence intervals will yield higher margins of error, as the critical z value will increase. [ANSWER]