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Chapter 9: Problem 22

Credit Card Debt A school administrator is concerned about the amount of credit card debt college students have. She wishes to conduct a poll to estimate the percentage of full-time college students who have credit card debt of \(\$ 2000\) or more. What size sample should be obtained if she wishes the estimate to be within 2.5 percentage points with \(94 \%\) confidence if (a) a pilot study indicates the percentage is \(34 \% ?\) (b) no prior estimates are used?

Short Answer

Expert verified
For (a): 1265. For (b): 1414.

Step by step solution

01

- Understand the Formula

To calculate the sample size for estimating a population proportion with a given margin of error and confidence level, use the formula: \[ n = \frac{Z^2 \times p \times (1 - p)}{E^2}\]where \(n\) is the sample size, \(Z\) is the Z-score for the confidence level, \(p\) is the estimated population proportion, and \(E\) is the margin of error.
02

- Determine the Z-score

For a confidence level of 94%, the corresponding Z-score is 1.88 (looked up from standard Z-tables).
03

- Calculate the Sample Size for (a)

Given that the pilot study indicates the percentage at 34%, we have: \( p = 0.34 \) and \( E = 0.025 \). Substitute these values into the sample size formula: \[ n = \frac{(1.88)^2 \times 0.34 \times (1 - 0.34)}{(0.025)^2} = \frac{3.5344 \times 0.34 \times 0.66}{0.000625} = \frac{0.790358}{0.000625} \] \[ n \that{= 1264.573} \]. Round up to the nearest whole number: \( n = 1265 \).
04

- Calculate the Sample Size for (b)

When no prior estimate is available, use \( p = 0.5 \) for maximizing variability: \( p = 0.5 \) and \( E = 0.025 \). Substitute these values into the sample size formula: \[ n = \frac{(1.88)^2 \times 0.5 \times (1 - 0.5)}{(0.025)^2} = \frac{3.5344 \times 0.25}{0.000625} = \frac{0.8836}{0.000625} \] \[ n \that{= 1413.76} \]. Round up to the nearest whole number: \( n = 1414 \).

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Key Concepts

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

credit card debt in students
College students often find themselves grappling with credit card debt. This can be due to several factors like tuition fees, textbooks, and living expenses. Credit card debt specifically refers to the outstanding balances that students may carry on their credit cards.
confidence level
When conducting surveys or polls, the confidence level indicates how sure we are about our results. A 94% confidence level, for instance, means that if we were to conduct the same survey 100 times, 94 of those times we would expect the results to be within a certain range around the true population value. The Z-score associated with a 94% confidence level is 1.88.
margin of error
The margin of error is crucial in surveys as it denotes the range within which we can expect the true population proportion to fall. A margin of error of 2.5% means our estimates can be 2.5 percentage points above or below the actual population percentage. Lower margins of error require larger sample sizes.
population proportion
The population proportion refers to the fraction of the population that exhibits a particular characteristic. In this context, it is the percentage of full-time students with $2000 or more in credit card debt. When no prior data is available, statisticians often use 0.5 as the estimated proportion to ensure maximum variability and accuracy in the sample size calculation.

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Most popular questions from this chapter

A simple random sample of size \(n\) is drawn from a population whose population standard deviation, \(\sigma,\) is known to be \(5.3 .\) The sample mean, \(\bar{x},\) is determined to be 34.2 (a) Compute the \(95 \%\) confidence interval about \(\mu\) if the sample size, \(n,\) is 35 (b) Compute the \(95 \%\) confidence interval about \(\mu\) if the sample size, \(n,\) is \(50 .\) How does increasing the sample size affect the margin of error, \(E ?\) (c) Compute the \(99 \%\) confidence interval about \(\mu\) if the sample size, \(n,\) is \(35 .\) Compare the results to those obtained in part (a). How does increasing the level of confidence affect the size of the margin of error, \(E ?\) (d) Can we compute a confidence interval about \(\mu\) based on the information given if the sample size is \(n=15 ?\) Why? If the sample size is \(n=15,\) what must be true regarding the population from which the sample was drawn?

Fifty rounds of a new type of ammunition were fired from a test weapon, and the muzzle velocity of the projectile was measured. The sample had a mean muzzle velocity of 863 meters per second, with a standard deviation of 2.7 meters per second. Construct and interpret a \(99 \%\) confidence interval for the mean muzzle velocity.

Many companies pay a portion of their profits to shareholders in the form of dividends. A stock's dividend yield is defined as the dividend paid by the company divided by its stock price. A stock analyst wants to estimate the mean dividend yield of financial stocks. He obtains a simple random sample of 11 financial stocks and obtains the following dividend yields. The data are in percent, so 3.17 represents \(3.17 \% .\) Construct and interpret a \(90 \%\) confidence interval for the dividend yield of financial stocks. An analysis of the data indicates that it is reasonable to conclude dividend yields follow a normal distribution. The data set has no outliers. $$\begin{array}{llllll} 3.19 & 1.86 & 0 & 0.54 & 1.76 & 2.08 \\ \hline 0.67 & 0.67 & 1.57 & 0 & 2.12 & \\ \hline \text {Source} \text { Morningstar } & & & & \\ \hline \end{array}$$

A simple random sample of size \(n\) is drawn from a population whose population standard deviation, \(\sigma,\) is known to be \(3.8 .\) The sample mean, \(\bar{x}\), is determined to be \(59.2 .\) (a) Compute the \(90 \%\) confidence interval about \(\mu\) if the sample size, \(n,\) is 45 (b) Compute the \(90 \%\) confidence interval about \(\mu\) if the sample size, \(n,\) is \(55 .\) How does increasing the sample size affect the margin of error, \(E ?\) (c) Compute the \(98 \%\) confidence interval about \(\mu\) if the sample size, \(n,\) is \(45 .\) Compare the results to those obtained in part (a). How does increasing the level of confidence affect the size of the margin of error, \(E ?\) (d) Can we compute a confidence interval about \(\mu\) based on the information given if the sample size is \(n=15 ?\) Why? If the sample size is \(n=15,\) what must be true regarding the population from which the sample was drawn?

Suppose a certain population, \(A,\) has standard deviation \(\sigma_{A}=5,\) and a second population, \(B,\) has standard deviation \(\sigma_{B}=10 .\) How many times larger than population \(A^{\prime} \mathrm{s}\) sample size does population \(B\) 's need to be to estimate \(\mu\) with the same margin of error? [Hint: Compute \(\left.n_{B} / n_{A} \cdot\right]\)
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