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Chapter 8: Problem 99

A bank manager wants to know the mean amount owed on credit card accounts that become delinquent. A random sample of 100 delinquent credit card accounts taken by the manager produced a mean amount owed on these accounts equal to \(\$ 2640\). The population standard deviation was \(\$ 578\). a. What is the point estimate of the mean amount owed on all delinquent credit card accounts at this bank? b. Construct a \(97 \%\) confidence interval for the mean amount owed on all delinquent credit card accounts for this bank.

Short Answer

Expert verified
a. The point estimate of the mean amount owed on all delinquent credit card accounts at this bank is $2640. b. The 97% confidence interval for the mean amount owed on all delinquent credit card accounts for this bank is between the range calculated in step 4.

Step by step solution

01

Determine point estimate

The point estimate of the population mean is simply the sample mean. In this case, the point estimate would be the mean of the 100 delinquent accounts, which is $2640.
02

Determine Z-score

The Z-score for a 97% confidence interval can be found using a standard normal z-score table. From the table, for a two-tailed test (because we want an estimate that the true mean is above or below our point estimate) with a 1.5% (100% - 97% / 2) significance level on each tail, the Z score is approximately 2.17.
03

Calculate Standard Error

The standard error (SE) can be calculated using the formula: SE = \(\frac{\sigma}{\sqrt{n}}\), where \(\sigma\) is the population standard deviation and n is the sample size. Substituting our values, SE = \(\frac{578}{\sqrt{100}}\).
04

Construct the confidence interval

The 97% confidence interval can be calculated using the formula: Estimate 卤 Z*SE. This gives a range from \($2640 - (2.17*\frac{578}{\sqrt{100}})\) to \($2640 + (2.17*\frac{578}{\sqrt{100}})\).

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Most popular questions from this chapter

A marketing researcher wants to find a \(95 \%\) confidence interval for the mean amount that visitors to a theme park spend per person per day. She knows that the standard deviation of the amounts spent per person per day by all visitors to this park is \(\$ 11\). How large a sample should the researcher select so that the estimate will be within \(\$ 2\) of the population mean?

a. Find the value of \(t\) for the \(t\) distribution with a sample size of 21 and area in the left tail equal to \(.10\). b. Find the value of \(t\) for the \(t\) distribution with a sample size of 14 and area in the right tail equal to \(.025\). c. Find the value of \(t\) for the \(t\) distribution with 45 degrees of freedom and \(.001\) area in the right tail. d. Find the value of \(t\) for the \(t\) distribution with 37 degrees of freedom and \(.005\) area in the left tail.

Suppose, for a sample selected from a population, \(\bar{x}=25.5\) and \(s=4.9\). a. Construct a \(95 \%\) confidence interval for \(\mu\) assuming \(n=47\). b. Construct a \(99 \%\) confidence interval for \(\mu\) assuming \(n=47\). Is the width of the \(99 \%\) confidence interval larger than the width of the \(95 \%\) confidence interval calculated in part a? If yes, explain why. c. Find a \(95 \%\) confidence interval for \(\mu\) assuming \(n=32\). Is the width of the \(95 \%\) confidence interval for \(\mu\) with \(n=32\) larger than the width of the \(95 \%\) confidence interval for \(\mu\) with \(n=47\) calculated in part a? If so, why? Explain.

A jumbo mortgage is a mortgage with a loan amount above the industry-standard definition of conyentional conforming loan limits. As of January 2009, approximately \(2.57 \%\) of people who took out a jumbo mortgage during the previous 12 months were at least 60 days late on their payments. Suppose that this percentage is based on a random sample of 1430 people who took out a jumbo mortgage during the previous 12 months. a. Construct a \(95 \%\) confidence interval for the proportion of all people who took out a jumbo mortgage during the previous 12 months and were at least 60 days late on their payments. b. Suppose the confidence interval obtained in part a is too wide. How can the width of this interval be reduced? Discuss all possible alternatives. Which alternative is the best?

a. How large a sample should be selected so that the margin of error of estimate for a \(98 \%\) confidence interval for \(p\) is \(.045\) when the value of the sample proportion obtained from a preliminary sample is \(.53\) ? b. Find the most conservative sample size that will produce the margin of error for a \(98 \%\) confidence interval for \(p\) equal to \(.045\).
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