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Chapter 9: Problem 37

A researcher for the U.S. Department of the Treasury wishes to estimate the percentage of Americans who support abolishing the penny. What size sample should be obtained if he wishes the estimate to be within 2 percentage points with \(98 \%\) confidence if (a) he uses a 2006 estimate of \(15 \%\) obtained from a Coinstar National Currency Poll? (b) he does not use any prior estimate?

Short Answer

Expert verified
a) 1727, b) 3393.

Step by step solution

01

Identify the Given Information and Required Formula

For both parts (a) and (b), the margin of error (E) is 0.02 (2 percentage points), the confidence level is 98%, the critical value for 98% confidence is 2.33 (from the Z-table), and the formula for the sample size (n) is \[ n = \frac{{Z^2 \times p \times (1 - p)}}{{E^2}} \]
02

Step 2a: Calculate Sample Size Using the Prior Estimate

For part (a), use the prior estimate from 2006, where \( p = 0.15 \): \[ n = \frac{{2.33^2 \times 0.15 \times (1 - 0.15)}}{{0.02^2}} \] Calculate as follows: \[ n = \frac{{5.4289 \times 0.15 \times 0.85}}{{0.0004}} \] \[ n = \frac{{0.6905}}{{0.0004}} \] \[ n 鈮� 1726.25 \] Since the sample size needs to be a whole number, round up to 1727.
03

Step 2b: Calculate Sample Size Without a Prior Estimate

For part (b), use \( p = 0.5 \) (the most conservative estimate): \[ n = \frac{{2.33^2 \times 0.5 \times (1 - 0.5)}}{{0.02^2}} \] Calculate as follows: \[ n = \frac{{5.4289 \times 0.25}}{{0.0004}} \] \[ n = \frac{{1.3572}}{{0.0004}} \] \[ n 鈮� 3393 \]

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Key Concepts

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

margin of error
Imagine you are trying to estimate the percentage of people who support abolishing the penny. A margin of error helps gauge how close your survey results are likely to come to the true population value. For example, if the margin of error is 2%, and your survey says 15% support abolishing the penny, the true value is likely between 13% and 17%.
  • It represents the range of the estimation error.
  • It is denoted as E in formulas.
  • The formula for margin of error in a proportion estimate is \(E = Z \times \sqrt{\frac{p(1-p)}{n}}\)
The margin of error depends not just on sample size but also on the level of confidence and variability in the data. A smaller margin of error means more precision but often requires a larger sample size.
confidence level
The confidence level is another crucial concept in statistics. It indicates how certain you are that the true value lies within your margin of error.
  • A 98% confidence level means you are 98% sure the true percentage of people who support abolishing the penny is within your 2% margin of error.
  • Higher confidence levels lead to larger margins of error and require larger sample sizes.
  • Common confidence levels are 90%, 95%, and 99%.
The formula to find the required sample size combines the confidence level, margin of error, and the proportion of the population. For example, if your confidence level is 98%, then your critical value (Z-score) is 2.33, taken from statistical Z-tables.
critical value
The critical value is a key number used in estimating sample size. It is derived from the confidence level and shows how many standard deviations away from the mean you need to look.
  • For a 98% confidence level, the critical value is 2.33.
  • This value comes from a Z-table in statistics.
  • The critical value affects the margin of error: higher confidence levels mean larger critical values and larger required sample sizes.
The critical value is found using Z-tables and is part of the formula for calculating sample size: \(n = \frac{Z^2 \times p \times (1 - p)}{E^2}\). By understanding and using the correct critical value, margin of error, and confidence level, one can accurately estimate the required sample size for surveys.

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Most popular questions from this chapter

How much time do Americans spend eating or drinking? Suppose for a random sample of 1001 Americans age 15 or older, the mean amount of time spent eating or drinking per day is 1.22 hours with a standard deviation of 0.65 hour. Source: American Time Use Survey conducted by the Bureau of Labor Statistics (a) A histogram of time spent eating and drinking each day is skewed right. Use this result to explain why a large sample size is needed to construct a confidence interval for the mean time spent eating and drinking each day. (b) There are over 200 million Americans age 15 or older. Explain why this, along with the fact that the data were obtained using a random sample, satisfies the requirements for constructing a confidence interval. (c) Determine and interpret a \(95 \%\) confidence interval for the mean amount of time Americans age 15 or older spend eating and drinking each day. (d) Could the interval be used to estimate the mean amount of time a 9-year- old American spends eating and drinking each day? Explain.

Construct the appropriate confidence interval. A simple random sample of size \(n=210\) is drawn from a population. The sample mean is found to be \(\bar{x}=20.1,\) and the sample standard deviation is found to be \(s=3.2\). Construct a \(90 \%\) confidence interval for the population mean.

A jar of peanuts is supposed to have 16 ounces of peanuts. The filling machine inevitably experiences fluctuations in filling, so a quality-control manager randomly samples 12 jars of peanuts from the storage facility and measures their contents. She obtains the following data: $$ \begin{array}{llllll} \hline 15.94 & 15.74 & 16.21 & 15.36 & 15.84 & 15.84 \\ \hline 15.52 & 16.16 & 15.78 & 15.51 & 16.28 & 16.53 \\ \hline \end{array} $$ (a) Verify that the data are normally distributed by constructing a normal probability plot. (b) Determine the sample standard deviation. (c) Construct a \(90 \%\) confidence interval for the population standard deviation of the number of ounces of peanuts. (d) The quality control manager wants the machine to have a population standard deviation below 0.20 ounce. Does the confidence interval validate this desire?

Construct the appropriate confidence interval. A simple random sample of size \(n=785\) adults was asked if they follow college football. Of the 785 surveyed, 275 responded that they did follow college football. Construct a \(95 \%\) confidence interval for the population proportion of adults who follow college football.

The following data represent the repair cost for a low-impact collision in a simple random sample of mini- and micro-vehicles (such as the Chevrolet Aveo or Mini Cooper). $$ \begin{array}{lrlrr} \hline \$ 3148 & \$ 1758 & \$ 1071 & \$ 3345 & \$ 743 \\ \hline \$ 2057 & \$ 663 & \$ 2637 & \$ 773 & \$ 1370 \\ \hline \end{array} $$ (a) Draw a normal probability plot to determine if it is reasonable to conclude the data come from a population that is normally distributed. (b) Draw boxplot to check for outliers. (c) Construct and interpret a \(95 \%\) confidence interval for the population mean cost of repair. (d) Suppose you obtain a simple random sample of size \(n=10\) of a Mini Cooper that was in a low-impact collision and determine the cost of repair. Do you think a \(95 \%\) confidence interval would be wider or narrower? Explain.
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