/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 58 A well-known bank credit card fi... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A well-known bank credit card firm wishes to estimate the proportion of credit card holders who carry a nonzero balance at the end of the month and incur an interest charge. Assume that the desired margin of error is .03 at \(98 \%\) confidence. a. How large a sample should be selected if it is anticipated that roughly \(70 \%\) of the firm's card holders carry a nonzero balance at the end of the month? b. How large a sample should be selected if no planning value for the proportion could be specified?

Short Answer

Expert verified
a. 425; b. 602.

Step by step solution

01

Understanding the Problem

The problem asks us to determine the sample size needed for estimating a proportion with a specified margin of error and confidence level. Part (a) uses a given estimate of the proportion, while part (b) has no initial estimate.
02

Know the Formula for Sample Size

The formula for the sample size needed to estimate a proportion with a given margin of error (E) and confidence level is: \[ n = \left( \frac{Z^2 \cdot \hat{p} \cdot (1-\hat{p})}{E^2} \right)\] where \( \hat{p} \) is the estimated proportion and \( Z \) is the Z-value for the confidence level.
03

Calculate Z-value for 98% Confidence

For a 98% confidence level, the Z-value is determined from the standard normal distribution. The Z-value associated with 98% confidence is approximately 2.33.
04

Solve Part (a) with Estimated Proportion

Given \( \hat{p} = 0.7 \), \( E = 0.03 \), and \( Z = 2.33 \), substitute these values into the sample size formula: \[ n = \left( \frac{2.33^2 \cdot 0.7 \cdot (1-0.7)}{0.03^2} \right) \] Calculate: \[ n \approx \left( \frac{5.4289 \cdot 0.21}{0.0009} \right) \approx 424.28\] Rounding up, the sample size \( n = 425 \).
05

Solve Part (b) with No Estimated Proportion

If no estimate is known, use \( \hat{p} = 0.5 \) for maximum variability. Substitute \( \hat{p} = 0.5 \) into the formula with \( Z = 2.33 \) and \( E = 0.03 \): \[ n = \left( \frac{2.33^2 \cdot 0.5 \cdot (1-0.5)}{0.03^2} \right) \] Calculate: \[ n \approx \left( \frac{5.4289 \cdot 0.25}{0.0009} \right) \approx 601.34 \] Rounding up, the sample size \( n = 602 \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Confidence Interval
A confidence interval gives us a range that likely contains the true value of a population parameter. In this context, we're interested in the proportion of credit card holders who carry a nonzero balance at the end of the month. By calculating a confidence interval, the bank can make informed decisions knowing how much variation there might be from the sample to the entire customer base.
- The interval is dictated by a confidence level, which in this exercise is 98%. This means that if we were to take multiple samples and build intervals each time, 98% of them would contain the true population proportion. - The confidence interval is crucial as it helps the bank understand the reliability and precision of their estimate, thus aiding strategic financial planning.
A higher confidence level means a wider interval, but also greater assurance that it contains the actual population proportion. Understanding and applying confidence intervals is key in interpreting data and making predictions about larger populations from sample data.
Margin of Error
The margin of error represents the range within which the true population parameter is expected to fall. For this bank, having a margin of error of 0.03 means they can expect their estimate of the proportion to be within 3 percentage points of the true proportion.
- Smaller margins of error are desirable because they indicate more precise estimates. - However, a smaller margin requires a larger sample size, as seen in the calculations.
In practice, choosing an acceptable margin of error is a balance between desired precision and available resources. For example, a smaller margin may provide more reliable insights but requires more financial and time costs to gather larger samples. Balancing margin of error and sample size allows an organization to make data-driven decisions efficiently.
Proportion Estimation
Proportion estimation is a technique used to infer the percentage of a population that exhibits a particular characteristic based on a sample. In this exercise, the bank estimates the proportion of holders with a nonzero balance.
- A known proportion (\(\hat{p}\)) allows specific calculations, offering direct insight.- If no estimate exists, using 0.5 maximizes variability, ensuring the sample size is sufficient regardless of the true proportion. This approach ensures the results are meaningful even with uncertainty in initial estimates.
Proportion estimation is a fundamental statistical tool, crucial for business and research purposes alike. It helps provide meaningful insights from sample data, enabling strategic decisions based on statistical evidence alone.
Z-value
The Z-value indicates how many standard deviations an element is from the mean in a standard normal distribution. It's crucial in determining the sample size, as it directly affects the width of the confidence interval.
- For a 98% confidence level, the Z-value is approximately 2.33. This value reflects the desired level of certainty. - The Z-value is chosen based on the confidence level. Higher confidence levels result in larger Z-values, hence larger sample sizes are generally required.
Understanding Z-values and their relation to confidence levels is essential for conducting accurate statistical analysis. They help analysts assess the precision of their estimates and adjust their sample sizes accordingly to ensure reliable results.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Although airline schedules and cost are important factors for business travelers when choosing an airline carrier, a \(U S A\) Today survey found that business travelers list an airline's frequent flyer program as the most important factor. From a sample of \(n=1993\) business travelers who responded to the survey, 618 listed a frequent flyer program as the most important factor. a. What is the point estimate of the proportion of the population of business travelers who believe a frequent flyer program is the most important factor when choosing an airline carrier? b. Develop a \(95 \%\) confidence interval estimate of the population proportion. c. How large a sample would be required to report the margin of error of .01 at \(95 \%\) confidence? Would you recommend that \(U S A\) Today attempt to provide this degree of precision? Why or why not?

A simple random sample of 60 items resulted in a sample mean of \(80 .\) The population standard deviation is \(\sigma=15\) a. Compute the \(95 \%\) confidence interval for the population mean. b. Assume that the same sample mean was obtained from a sample of 120 items. Provide a \(95 \%\) confidence interval for the population mean. c. What is the effect of a larger sample size on the interval estimate?

In a survey, the planning value for the population proportion is \(p^{*}=.35 .\) How large a sample should be taken to provide a \(95 \%\) confidence interval with a margin of error of \(.05 ?\)

A survey conducted by the American Automobile Association showed that a family of four spends an average of \(\$ 215.60\) per day while on vacation. Suppose a sample of 64 families of four vacationing at Niagara Falls resulted in a sample mean of \(\$ 252.45\) per day and a sample standard deviation of \(\$ 74.50\) a. Develop a \(95 \%\) confidence interval estimate of the mean amount spent per day by a family of four visiting Niagara Falls. b. Based on the confidence interval from part (a), does it appear that the population mean amount spent per day by families visiting Niagara Falls differs from the mean reported by the American Automobile Association? Explain.

A simple random sample of 40 items resulted in a sample mean of \(25 .\) The population standard deviation is \(\sigma=5\) a. What is the standard error of the mean, \(\sigma_{z} ?\) b. \(\quad\) At \(95 \%\) confidence, what is the margin of error?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.