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Chapter 9: Problem 38

A school administrator is concerned about the amount of credit-card debt that 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 that the percentage is \(34 \% ?\) (b) no prior estimates are used?

Short Answer

Expert verified
For (a), the sample size should be 552. For (b), the sample size should be 564.

Step by step solution

01

Determine the Margin of Error (E)

The margin of error is given as 2.5 percentage points. Convert this to a decimal for calculations: \[ E = 0.025 \]
02

Find the Z-Score for 94% Confidence

Use a Z-table or normal distribution calculator to find the Z-score for a 94% confidence level. The Z-score is approximately 1.88.
03

Calculate the Required Sample Size with Known Proportion (a)

Use the formula for sample size: \[ n = \left( \frac{Z^2 \cdot p \cdot (1 - p)}{E^2} \right) \] For part (a), the pilot study indicates that the percentage is 34%, so \( p = 0.34 \). Substituting the values: \[ n = \left( \frac{1.88^2 \cdot 0.34 \cdot (1 - 0.34)}{0.025^2} \right) \] Calculate: \[ n \approx 551.49 \] Since the sample size has to be a whole number, round up to 552.
04

Calculate the Required Sample Size with No Prior Estimates (b)

When no prior estimate is used, assume the proportion \( p = 0.5 \) because it maximizes the variability, providing the most conservative estimate. Use the same formula: \[ n = \left( \frac{Z^2 \cdot p \cdot (1 - p)}{E^2} \right) \] Substituting the values: \[ n = \left( \frac{1.88^2 \cdot 0.5 \cdot (1 - 0.5)}{0.025^2} \right) \] Calculate: \[ n \approx 563.38 \] Since the sample size has to be a whole number, round up to 564.

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Key Concepts

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

margin of error
The margin of error represents the range within which we expect the true population proportion to lie. It's an allowance for some level of error in our estimates, expressed as a percentage.
In the given exercise, the margin of error is 2.5%. This means the administrator wants her estimate to be within 卤2.5 percentage points of the actual proportion of students with significant credit-card debt.
Converting this percentage to a decimal is essential for calculations: \( E = 0.025 \). The margin of error is crucial because it directly influences the sample size needed. A smaller margin of error requires a larger sample size to ensure accuracy.
confidence interval
A confidence interval provides a range of values that are believed to encompass the true population parameter with a specified level of confidence.
In this scenario, the confidence level is set at 94%. This means that if numerous samples were taken, we would expect 94% of the calculated confidence intervals to contain the true proportion.
The confidence level determines the Z-score used in the sample size calculation. For a 94% confidence level, the associated Z-score is approximately 1.88. Using this Z-score ensures that the interval created by the sample proportion and the margin of error will have the desired confidence.
proportion estimation
Proportion estimation involves determining the fraction or percentage of the population that exhibits a specific characteristic, here being credit-card debt of $2000 or more.
The sample proportion (denoted as \( p \)) is used to estimate the true population proportion. When a pilot study indicates that 34% of students have this debt, it implies \( p = 0.34 \).
If no prior estimate is available, we often use \( p = 0.5 \) for maximum variability, ensuring a conservative and typically larger sample size calculation. This approach is known as the '50-50 rule' and is a standard in sample size determination when lacking previous data.
z-score
The Z-score represents the number of standard deviations a point is from the mean in a standard normal distribution.
For a 94% confidence level in the problem, the Z-score is approximately 1.88.
The Z-score is necessary for calculating the sample size: \[ n = \frac{Z^2 \times p \times (1 - p)}{E^2} \]
This formula combines the Z-score, margin of error, and estimated proportion to determine the required sample size. The Z-score adjusts the width of the confidence interval, ensuring the correct level of confidence for the estimation.

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Most popular questions from this chapter

In a Gallup poll, \(64 \%\) of the people polled answered yes to the following question: "Are you in favor of the death penalty for a person convicted of murder?" The margin of error in the poll was \(3 \%,\) and the estimate was made with \(95 \%\) confidence. At least how many people were surveyed?

Suppose the following data represent the amount of time (in hours) a random sample of students enrolled in College Algebra spent working on a homework assignment: \(3.2,4.1,1.2,0.6,\) and 2.5. Below are three bootstraps samples. For each bootstrap sample, determine the bootstrap sample mean. Bootstrap Sample 1: 1.2,0.6,3.2,3.2,1.2 Bootstrap Sample 2: 0.6,4.1,4.1,0.6,4.1 Bootstrap Sample 3: 4.1,3.2,3.2,0.6,1.2

(a) Find the \(t\) -value such that the area in the right tail is 0.02 with 19 degrees of freedom. (b) Find the \(t\) -value such that the area in the right tail is 0.10 with 32 degrees of freedom. (c) Find the \(t\) -value such that the area left of the \(t\) -value is 0.05 with 6 degrees of freedom. [Hint: Use symmetry.] (d) Find the critical \(t\) -value that corresponds to \(95 \%\) confidence. Assume 16 degrees of freedom.

In a random sample of 40 felons convicted of aggravated assault, it was determined that the mean length of sentencing was 54 months, with a standard deviation of 8 months. Construct and interpret a \(95 \%\) confidence interval for the mean length of sentence for an aggravated assault conviction.

The exponential probability distribution can be used to model waiting time in line or the lifetime of electronic components. Its density function is skewed right. Suppose the wait-time in a line can be modeled by the exponential distribution with \(\mu=\sigma=5\) minutes. (a) Use StatCrunch, Minitab, or some other statistical software to generate 100 random samples of size \(n=4\) from this population. (b) Construct \(95 \% t\) -intervals for each of the 100 samples found in part (a). (c) How many of the intervals do you expect to include the population mean? How many of the intervals actually contain the population mean? Explain what your results mean. (d) Repeat parts (a)-(c) for samples of size \(n=15\) and \(n=25\). Explain what your results mean.
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