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A commercial bank issues several types of credit cards. As a part of its annual

ID: 3205030 • Letter: A

Question

A commercial bank issues several types of credit cards. As a part of its annual review of the profitability of each type of credit card, the bank randomly samples 30 customers to measure the average annual charges. For EmeraldCard customers, a random sample of 30 card accounts shows the following annual spending per account (rounded to the nearest dollar):

$28,426

$21,299

$22,493

$23,395

$24,729

$31,741

$22,432

$24,254

$20,763

$26,404

$22,251

$31,592

$14,654

$16,452

$19,431

$18,041

$29,121

$27,791

$27,035

$20,931

$19,594

$24,924

$21,227

$25,903

$29,635

$17,382

$20,155

$16,065

$25,633

$26,021

(a)   Based on these randomly sample accounts, what is the best estimate of the true mean annual spending for the EmeraldCard account holders?                                                                                                               

(b) If the bank is interested in developing a 90% confidence interval estimate of mean annual spending, what distribution will be used to determine the confidence limits? Explain.

(c)   Find the standard error (of the mean) of the sampling distribution in (b).                                                                                                         

(d) Construct the 90% confidence interval estimate for the population mean of annual spending for the bank’s EmeraldCard customers.                                                                                              

(e)   To improve estimation accuracy, the bank plans to increase the level of confidence in its interval estimate in (d) from 90% to 95%. What will be the change in the bound of the estimation error (or margin of error)?

$28,426

$21,299

$22,493

$23,395

$24,729

$31,741

$22,432

$24,254

$20,763

$26,404

$22,251

$31,592

$14,654

$16,452

$19,431

$18,041

$29,121

$27,791

$27,035

$20,931

$19,594

$24,924

$21,227

$25,903

$29,635

$17,382

$20,155

$16,065

$25,633

$26,021

Explanation / Answer

a)Mean=23325.8, SD=4576.4688

b)We use a t distribution as the sample size is small.

c)SE(mean)=4576.4688/sqrt(30)=835.545

d)t(90%,29dof)=1.6991

CI=[23325.8-1.6991*835.545,23325.8+1.6991*835.545]=[21906.1,24745.5]

e)t(95%,29dof)=2.045

CI=[23325.8-2.045*835.545,23325.8+2.045*835.545]=[21616.92,25034.68]

Hence margin of error=25034.68-24745.5=289.18

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