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Chapter 9: Problem 40

A sociologist wishes to conduct a poll to estimate the percentage of Americans who favor affirmative action programs for women and minorities for admission to colleges and universities. What sample size should be obtained if she wishes the estimate to be within 4 percentage points with \(90 \%\) confidence if (a) she uses a 2003 estimate of \(55 \%\) obtained from a Gallup Youth Survey? (b) she does not use any prior estimates? (c) Why are the results from parts (a) and (b) so close?

Short Answer

Expert verified
(a) 419, (b) 423, (c) Results are close because p 鈮� 0.5 maximizes p(1-p).

Step by step solution

01

- Set up Margin of Error and Confidence Level

The margin of error (E) is 4 percentage points, so E = 0.04. The confidence level is 90%, so the critical value (z-value) for a 90% confidence level is approximately 1.645.
02

- Calculate Sample Size with Prior Estimate

Use the formula for sample size with a prior estimate: \[ n = \left( \frac{z^2 \cdot p \cdot (1-p)}{E^2} \right) \] Substituting the given values: \[ n = \left( \frac{(1.645)^2 \cdot 0.55 \cdot (1-0.55)}{(0.04)^2} \right) \] Calculate the value: \[ n 鈮� \left( \frac{2.706025 \cdot 0.55 \cdot 0.45}{0.0016} \right) \] \[ n 鈮� \frac{0.670425425}{0.0016} \] \[ n 鈮� 419.015 \] Thus, rounding to the nearest whole number, the sample size needed is 419.
03

- Calculate Sample Size without Prior Estimate

Use the formula for sample size without a prior estimate: \[ n = \left( \frac{z^2 \cdot 0.25}{E^2} \right) \] Substituting the given values: \[ n = \left( \frac{(1.645)^2 \cdot 0.25}{(0.04)^2} \right) \] Calculate the value: \[ n 鈮� \left( \frac{2.706025 \cdot 0.25}{0.0016} \right) \] \[ n 鈮� \frac{0.67650625}{0.0016} \] \[ n 鈮� 422.81640625 \] Thus, rounding to the nearest whole number, the sample size needed is 423.
04

- Explain Why Results are Close

The results from parts (a) and (b) are close because the estimate from the 2003 Gallup Youth Survey, 55%, is near 50%. When p is near 0.5, the product of p and (1-p) is maximized (\[ 0.55 \times 0.45 = 0.2475 \]), and this closely approximate the worst-case scenario of 0.25 used in part (b). Hence, the sample size calculations are similar.

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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 (E) is a measure of the range within which we expect the true population parameter to lie. In our exercise, the sociologist wants the estimate to be within 4 percentage points of the true value. This means E = 0.04. The smaller the margin of error, the more precise the estimate, but it also requires a larger sample size. The margin of error is crucial because it directly affects the size of the sample needed to achieve a certain level of confidence in your results.
Confidence Level
The confidence level represents the degree of certainty that the true population parameter lies within the margin of error. Common confidence levels are 90%, 95%, and 99%. In this case, the confidence level is 90%. To translate this into the sample size calculation, we use a critical value (z-value). For a 90% confidence level, the z-value is approximately 1.645. This value helps quantify the 'wiggle room' we have around our sample estimate to ensure we capture the true population parameter within our specified confidence.
Prior Estimate
A prior estimate (denoted as p) is a previously obtained proportion from a similar study. This can help refine the sample size calculation. In our example, the sociologist uses a 2003 Gallup Youth Survey estimate of 55% (p = 0.55). Using a prior estimate generally reduces the required sample size as it offers a more accurate starting point compared to assuming no prior knowledge. Without a prior estimate, a conservative approach using p = 0.5 is often used, which maximises the product p(1-p) and hence the required sample size.
Sample Size Formula
The formula for determining sample size varies depending on whether a prior estimate is used. With a prior estimate, the formula is: \[ n = \frac{z^2 \times p \times (1-p)}{E^2} \] In our case, substituting the values of z (1.645), p (0.55) and E (0.04), the formula becomes: \[ n 鈮� \frac{(1.645)^2 \times 0.55 \times 0.45}{(0.04)^2} \] This calculation results in a sample size of approximately 419. Without a prior estimate, we use p = 0.5: \[ n = \frac{z^2 \times 0.25}{E^2} \] Substituting our z and E values, we get a sample size of approximately 423. The close values in our calculations show that using a prior estimate only slightly adjusts the needed sample size.
Critical Value
The critical value is a factor that determines the range of the confidence interval. For a 90% confidence level, this value is 1.645. It is derived from the standard normal distribution and corresponds to the point beyond which 5% of the area under the curve lies in either tail of the distribution. The higher the confidence level, the larger the critical value and consequently, a larger sample size is required to maintain a defined margin of error. Understanding the critical value and its role is essential in correctly calculating sample sizes and ensuring validity of statistical estimates.

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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?

A USA Today/Gallup poll asked 1006 adult Americans how much it would bother them to stay in a room on the 13 th floor of a hotel. Interestingly, \(13 \%\) said it would bother them. The margin of error was 3 percentage points with \(95 \%\) confidence. Which of the following represents a reasonable interpretation of the survey results? For those not reasonable, explain the flaw. (a) We are \(95 \%\) confident that the proportion of adult Americans who would be bothered to stay in a room on the 13th floor is between 0.10 and 0.16 . (b) We are between \(92 \%\) and \(98 \%\) confident that \(13 \%\) of adult Americans would be bothered to stay in a room on the 13th floor. (c) In \(95 \%\) of samples of adult Americans, the proportion who would be bothered to stay in a room on the 13 th floor is between 0.10 and 0.16 . (d) We are \(95 \%\) confident that \(13 \%\) of adult Americans would be bothered to stay in a room on the 13 th floor.

You Explain It! Hours Worked In a survey conducted by the Gallup Organization, 1100 adult Americans were asked how many hours they worked in the previous week. Based on the results, a \(95 \%\) confidence interval for mean number of hours worked was lower bound: 42.7 and upper bound: \(44.5 .\) Which of the following represents a reasonable interpretation of the result? For those that are not reasonable, explain the flaw. (a) There is a \(95 \%\) probability the mean number of hours worked by adult Americans in the previous week was between 42.7 hours and 44.5 hours. (b) We are \(95 \%\) confident that the mean number of hours worked by adult Americans in the previous week was between 42.7 hours and 44.5 hours. (c) \(95 \%\) of adult Americans worked between 42.7 hours and 44.5 hours last week. (d) We are \(95 \%\) confident that the mean number of hours worked by adults in Idaho in the previous week was between 42.7 hours and 44.5 hours.

A simple random sample of size \(n\) is drawn. The sample mean, \(\bar{x},\) is found to be \(35.1,\) and the sample standard deviation, \(s,\) is found to be \(8.7 .\) (a) Construct a \(90 \%\) confidence interval for \(\mu\) if the sample size, \(n,\) is \(40 .\) (b) Construct a \(90 \%\) confidence interval for \(\mu\) if the sample size, \(n\), is \(100 .\) How does increasing the sample size affect the margin of error, \(E ?\) (c) Construct a \(98 \%\) confidence interval for \(\mu\) if the sample size, \(n\), is \(40 .\) Compare the results to those obtained in part (a). How does increasing the level of confidence affect the margin of error, \(E ?\) (d) If the sample size is \(n=18\), what conditions must be satisfied to compute the confidence interval?

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?
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