Problem 34 Use the given data to find the m... [FREE SOLUTION] (2024)

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Chapter 7: Problem 34

Use the given data to find the minimum sample size required to estimate apopulation proportion or percentage. A sociologist plans to conduct a survey to estimate the percentage of adultswho believe in astrology. How many people must be surveyed if we want aconfidence 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 \(26 \%\) ofrespondents said that they believed in astrology.

Short Answer

Expert verified

a. 1037 people, b. 799 people

Step by step solution

01

Understand the Formula for Sample Size

The formula to calculate the sample size for estimating a population proportion is \[ n = \frac{Z^2 \times p \times (1 - p)}{E^2} \]Where: - \(n\) is the sample size- \(Z\) is the Z-value corresponding to the confidence level- \(p\) is the estimated proportion (if unknown, use 0.5)- \(E\) is the margin of error (in decimal form)

02

Identify the Z-value for 99% Confidence Level

For a 99% confidence level, the Z-value is approximately 2.576.

03

Step 3a: Calculate the Sample Size (Part a)

When nothing is known about the proportion, we use \(p = 0.5\). Plug the values into the formula: \[ n = \frac{(2.576)^2 \times 0.5 \times (1 - 0.5)}{0.04^2} \]First, calculate the numerator: \[ (2.576)^2 = 6.635 \]\[ 6.635 \times 0.5 \times 0.5 = 1.65875 \]Then, calculate the denominator: \[ 0.04^2 = 0.0016 \]Finally, \[ n = \frac{1.65875}{0.0016} = 1036.71875 \]Since the sample size must be a whole number, round up to 1037.

04

Step 3b: Calculate the Sample Size (Part b)

Using the previous survey result where \(p = 0.26\). Plug the values into the formula: \[ n = \frac{(2.576)^2 \times 0.26 \times (1 - 0.26)}{0.04^2} \]First, calculate the numerator: \[ (2.576)^2 = 6.635 \]\[ 6.635 \times 0.26 \times 0.74 = 1.277622 \]Then, calculate the denominator: \[ 0.04^2 = 0.0016 \]Finally, \[ n = \frac{1.277622}{0.0016} = 798.51375 \]Since the sample size must be a whole number, round up to 799.

Key Concepts

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

population proportion

Population proportion refers to the fraction of a population that has a particular attribute or characteristic. For example, if we want to find out how many adults believe in astrology, the population proportion will determine what percentage of the entire population believes in it. Knowing this proportion helps in making predictions and creating statistical estimates about the whole population based on a sample. When the proportion is unknown, a common practice is to assume it to be 0.5 (or 50%) for the worst-case scenario, which maximizes the required sample size to ensure a conservative estimate.

confidence level

The confidence level is a measure of how certain we are about our estimate. It is expressed as a percentage and typically chosen between 90%, 95%, and 99%. The higher the confidence level, the more certain we are that our estimate includes the true population parameter. For instance, with a 99% confidence level, we are 99% confident that the population proportion falls within the calculated margin of error. This high level of confidence translates into a higher Z-value, which impacts the sample size required.

margin of error

The margin of error represents the range within which we expect the true population parameter to lie. It is usually expressed as a percentage and corresponds to the maximum amount by which we believe our sample estimate may differ from the true population proportion. For instance, if we set our margin of error at 4 percentage points, it means our estimate could be plus or minus 4% off from the actual population proportion. A smaller margin of error indicates a more precise estimate and typically requires a larger sample size.

Z-value

The Z-value is a critical component in estimating sample size, derived from the standard normal distribution corresponding to the selected confidence level. For a confidence level of 99%, the Z-value is approximately 2.576. This value indicates how many standard deviations away from the mean we should be to achieve our desired confidence level. The Z-value is squared and multiplied by the estimated proportion and its complement in the sample size formula. Higher Z-values increase the sample size to ensure greater confidence in the estimate. The formula for sample size is: \[ n = \frac{Z^2 \times p \times (1 - p)}{E^2} \]Utilizing the correct Z-value ensures that your sample size aligns accurately with your confidence level requirements.

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Problem 34 Use the given data to find the m... [FREE SOLUTION] (3)

Most popular questions from this chapter

Find the sample size required to estimate the population mean. Data Set 1 "Body Data" in Appendix B includes pulse rates of 147 randomlyselected adult females, and those pulse rates vary from a low of 36 bpm to ahigh of 104 bpm. Find the minimum sample size required to estimate the meanpulse rate of adult females. Assume that we want \(99 \%\) confidence that thesample mean is within 2 bpm of the population mean. a. Find the sample size using the range rule of thumb to estimate \(\sigma .\) b. Assume that \(\sigma=12.5\) bpm, based on the value of \(s=12.5\) bpm for thesample of 147 female pulse rates. c. Compare the results from parts (a) and (b). Which result is likely to bebetter?Finding Critical Values and Confidence Intervals. In Exercises \(5-8,\) use thegiven information to find the number of degrees of freedom, the criticalvalues \(\mathcal{X}_{L}^{2}\) and \(\mathcal{X}_{R}^{2},\) and the confidenceinterval estimate of \(\boldsymbol{\sigma} .\) The samples are from Appendix\(\boldsymbol{B}\) and it is reasonable to assume that a simple random samplehas been selected from a population with a normal distribution. Heights of Men \(99 \%\) confidence; \(n=153, s=7.10 \mathrm{cm}\)Find the sample size required to estimate the population mean. The Wechsler IQ test is designed so that the mean is 100 and the standarddeviation is 15 for the population of normal adults. Find the sample sizenecessary to estimate the mean IQ score of college professors. We want to be99\% confident that our sample mean is within 4 IQ points of the true mean.The mean for this population is clearly greater than \(100 .\) The standarddeviation for this population is less than 15 because it is a group with lessvariation than a group randomly selected from the general population;therefore, if we use \(\sigma=15\) we are being conservative by using a valuethat will make the sample size at least as large as necessary. Assume thenthat \(\sigma=15\) and determine the required sample size. Does the sample sizeappear to be practical?Finding Critical Values and Confidence Intervals. In Exercises \(5-8,\) use thegiven information to find the number of degrees of freedom, the criticalvalues \(\mathcal{X}_{L}^{2}\) and \(\mathcal{X}_{R}^{2},\) and the confidenceinterval estimate of \(\boldsymbol{\sigma} .\) The samples are from Appendix\(\boldsymbol{B}\) and it is reasonable to assume that a simple random samplehas been selected from a population with a normal distribution. Platelet Counts of Women \(99 \%\) confidence; \(n=147, s=65.4\)
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Problem 34 Use the given data to find the m... [FREE SOLUTION] (2024)
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