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Chapter 7: Problem 34

A sociologist plans to conduct a survey to estimate the percentage of adults who believe in astrology. How many people must be surveyed if we want a confidence level of \(99 \%\) and a margin of error of four percentage points? a. Assume that nothing is known about the percentage to be estimated. b. Use the information from a previous Harris survey in which \(26 \%\) of respondents said that they believed in astrology.

Short Answer

Expert verified
a. 1036 people. b. 798 people.

Step by step solution

01

Understanding the formula

Use the formula for sample size in surveys: \[ n = \frac{Z^2 \times p \times (1 - p)}{E^2} \]Where: - \( Z \) is the Z-score corresponding to the confidence level- \( p \) is the estimated proportion of the population that has the characteristic- \( E \) is the margin of error.
02

Determine Z-score for 99% confidence level

For a 99% confidence level, find the Z-score. From Z-tables, \( Z = 2.576 \).
03

Part a: Assume nothing is known about the percentage

Use \( p = 0.5 \) for maximum variability (worst-case scenario). Apply the values to the formula:\[ n = \frac{(2.576)^2 \times 0.5 \times (1 - 0.5)}{0.04^2} \]Simplify the calculation step-by-step:1. \( Z^2 = (2.576)^2 = 6.635776 \)2. \( p \times (1 - p) = 0.5 \times 0.5 = 0.25 \)3. \( E^2 = 0.04^2 = 0.0016 \)4. Putting it all together:\[ n = \frac{6.635776 \times 0.25}{0.0016} \ n \approx 1036 \]Therefore, 1036 people must be surveyed.
04

Part b: Use the percentage from the previous survey

Here, \( p = 0.26 \). Apply the values to the formula:\[ n = \frac{(2.576)^2 \times 0.26 \times (1 - 0.26)}{0.04^2} \]Simplify the calculation step-by-step:1. \( Z^2 = (2.576)^2 = 6.635776 \)2. \( p \times (1 - p) = 0.26 \times 0.74 = 0.1924 \)3. \( E^2 = 0.04^2 = 0.0016 \)4. Putting it all together:\[ n = \frac{6.635776 \times 0.1924}{0.0016} \ n \approx 798 \]Therefore, 798 people must be surveyed.

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Key Concepts

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

confidence level
When conducting a survey, the confidence level represents the probability that the sample accurately reflects the true population parameter. For example, a 99% confidence level means that if you were to repeat the survey 100 times, the true percentage would fall within your margin of error 99 times out of 100.Confidence levels are crucial in determining the reliability of survey results. Common confidence levels include 90%, 95%, and 99%. As the confidence level increases, the required sample size also increases. This is because higher confidence levels require a more precise estimate to ensure reliability.
margin of error
The margin of error quantifies the range of uncertainty in your survey results. It shows how much the survey result can differ from the true population value. For instance, a margin of error of four percentage points means the survey result could be four points higher or lower than the true value.To calculate the required sample size in your survey, you need to decide on an acceptable margin of error. A smaller margin of error requires a larger sample size to increase the survey's precision. Conversely, a larger margin of error can be achieved with a smaller sample size. In the given exercise, a margin of error of four percentage points is used.
Z-score
In statistics, the Z-score is a measure that describes the number of standard deviations a data point is from the mean. Z-scores are crucial for translating confidence levels into sample size requirements. From Z-tables, we can find the Z-score corresponding to a specific confidence level. For a 99% confidence level, the Z-score is 2.576. This means that our sample estimate should be within 2.576 standard deviations of the true population parameter to achieve a 99% confidence level.The Z-score increases as the confidence level increases, which in turn increases the required sample size. This factor ensures that the estimates are more reliable and closer to the true population value.
population proportion
The population proportion, denoted as p, is the fraction of the population that possesses a particular attribute. For example, if 26% of respondents believe in astrology, then the population proportion p is 0.26. Without prior knowledge, we often assume maximum variability by setting p to 0.5. This worst-case scenario ensures the largest possible sample size for cautious and reliable estimates.Knowing the population proportion helps in calculating the sample size. In the original exercise, different values of p are used: 0.5 for the worst-case scenario and 0.26 from a previous survey. By substituting these values into the sample size formula, we see how the sample size requirement changes.

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Most popular questions from this chapter

In a survey of 1002 people, \(70 \%\) said that they voted in a recent presidential election (based on data from ICR Research Group). Voting records show that \(61 \%\) of eligible voters actually did vote. a. Find a \(98 \%\) confidence interval estimate of the proportion of people who say that they voted. b. Are the survey results consistent with the actual voter turnout of \(61 \%\) ? Why or why not?

If a simple random sample of size \(n\) is selected without replacement from a finite population of size \(N\), and the sample size is more than \(5 \%\) of the population size \((n>0.05 N)\), better results can be obtained by using the finite population correction factor, which involves multiplying the margin of error \(E\) by \(\sqrt{(N-n) /(N-1)}\). For the sample of 100 weights of M\&M candies in Data Set 27 "M\&M Weights" in Appendix B, we get \(\bar{x}=0.8565 \mathrm{~g}\) and \(s=0.0518 \mathrm{~g}\). First construct a \(95 \%\) confidence interval estimate of \(\mu\), assuming that the population is large; then construct a \(95 \%\) confidence interval estimate of the mean weight of M\&Ms in the full bag from which the sample was taken. The full bag has 465 M\&Ms. Compare the results.

What is different about the normality requirement for a confidence interval estimate of \(\sigma\) and the normality requirement for a confidence interval estimate of \(\mu\) ?

Comparing Waiting Lines a. The values listed below are waiting times (in minutes) of customers at the Jefferson Valley Bank, where customers enter a single waiting line that feeds three teller windows. Construct a \(95 \%\) confidence interval for the population standard deviation \(\sigma\). $$ \begin{array}{llllllllll} 6.5 & 6.6 & 6.7 & 6.8 & 7.1 & 7.3 & 7.4 & 7.7 & 7.7 & 7.7 \end{array} $$ b. The values listed below are waiting times (in minutes) of customers at the Bank of Providence, where customers may enter any one of three different lines that have formed at three teller windows. Construct a \(95 \%\) confidence interval for the population standard deviation \(\sigma\). $$ \begin{array}{llllllllll} 4.2 & 5.4 & 5.8 & 6.2 & 6.7 & 7.7 & 7.7 & 8.5 & 9.3 & 10.0 \end{array} $$ c. Interpret the results found in parts (a) and (b). Do the confidence intervals suggest a difference in the variation among waiting times? Which arrangement seems better: the single-line system or the multiple-line system?

Data Set 3 "Body Temperatures" in Appendix B includes 106 body temperatures of adults for Day 2 at \(12 \mathrm{AM}\), and they vary from a low of \(96.5^{\circ} \mathrm{F}\) to a high of \(99.6^{\circ} \mathrm{F}\). Find the minimum sample size required to estimate the mean body temperature of all adults. Assume that we want \(98 \%\) confidence that the sample mean is within \(0.1^{\circ} \mathrm{F}\) of the population mean. a. Find the sample size using the range rule of thumb to estimate \(\sigma\). b. Assume that \(\sigma=0.62^{\circ} \mathrm{F}\), based on the value of \(s=0.62^{\circ} \mathrm{F}\) for the sample of 106 body temperatures. c. Compare the results from parts (a) and (b). Which result is likely to be better?
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