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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?

Short Answer

Expert verified
a. 327, b. 209, c. The result using \text{σ} = 0.62°F is better.

Step by step solution

01

Determine the required sample size formula

The formula for the minimum sample size required to estimate the population mean is given by \[ n = \frac{(Z \times \frac{\text{standard deviation}}{\text{margin of error}})^2}{MOE^2} \]where Z is the Z-score for the confidence level, \text{standard deviation} is \text{σ}, and \text{margin of error} is \text{E}.
02

Find the Z-score for 98% Confidence Level

For a 98% confidence level, approximately 2% is outside the confidence interval. Since the normal distribution is symmetric, 1% is in each tail. So, we find the Z-score for 0.01 in the standard normal distribution table, which approximately equals 2.33.\[ Z = 2.33 \]
03

Estimate \text{σ} using the range rule of thumb

The range rule of thumb estimates the standard deviation as follows: \[ \text{σ} \text{≈} \frac{\text{Range}}{4} \]Given the range of body temperatures is 99.6°F - 96.5°F = 3.1°F, the standard deviation can be estimated as\[ \text{σ} \text{≈} \frac{3.1}{4} = 0.775°F \]
04

Calculate the sample size using the estimated \text{σ}

Using \text{σ} = 0.775°F and MOE = 0.1°F:\[ n = \frac{(2.33 \times 0.775)^2}{(0.1)^2} = \frac{(1.80775)^2}{0.01} ≈ \frac{3.268}{0.01} = 326.8 \]Rounding up, we get \[ n ≈ 327 \]
05

Calculate the sample size using the given \text{σ}

Given \text{σ} = 0.62°F:\[ n = \frac{(2.33 \times 0.62)^2}{(0.1)^2} = \frac{(1.4446)^2}{0.01} ≈ \frac{2.086}{0.01} = 208.6 \]Rounding up, we get \[ n ≈ 209 \]
06

Compare the results

The sample size estimated using the range rule of thumb (\( n ≈ 327 \)) is higher than the sample size calculated using the given standard deviation (\( n ≈ 209 \)). The result using the given standard deviation is likely to be better as it is based on actual data, which provides a more precise estimate.

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Key Concepts

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

Confidence Interval
Confidence intervals are ranges that estimate the population parameter, like the mean, with a specified probability—or confidence level. A 98% confidence interval means we are 98% sure the population mean lies within our interval. To find this interval, we use a Z-score, which depends on our desired confidence level. For a 98% confidence level, the Z-score is roughly 2.33. The formula for the interval combines this score with our data’s standard deviation and sample size.
Confidence intervals give us a range instead of a single estimate. This accounts for sample variability and provides more reliable results.
Standard Deviation
Standard deviation (σ) measures the amount of variation in a data set. It shows how much individual data points differ from the mean. In our exercise, the body temperature values range from 96.5°F to 99.6°F, with a given standard deviation of 0.62°F. Another way to estimate standard deviation is using the range rule of thumb, which suggests σ is approximately the range divided by 4. For our temperatures, that calculation gives us 0.775°F.
Lower standard deviations indicate that the data points are close to the mean, whereas higher standard deviations show more spread-out data.
Population Mean
The population mean is the average of a population’s values. Unlike sample means, which are averages of a subset, the population mean encompasses the entire group. To estimate it accurately, we need a good sample size. We increase our confidence in the estimate by using a sufficiently large sample, leading to more precise results. In our exercise, we use the body temperatures data to estimate the mean body temperature of all adults, ensuring a high confidence level by using calculated sample sizes.
Z-score
A Z-score indicates how many standard deviations a value is from the mean. For confidence intervals, Z-scores help define how confident we want to be. At a 98% confidence level, the Z-score is 2.33, meaning we are highly confident that the true population mean lies within our calculated range.
Z-scores are critical because they standardize scores, allowing comparison across different data sets or distributions.
Margin of Error
The margin of error (E) indicates the range within which we expect the true population mean to lie, given our sample data. It’s a critical component of calculating sample size. In our exercise, the margin of error was set to 0.1°F, meaning our sample mean should be no more than 0.1°F away from the true population mean.
Smaller margins of error require larger sample sizes to maintain the same confidence level, leading to more precise estimates.

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Most popular questions from this chapter

A study of 420,095 Danish cell phone users found that \(0.0321 \%\) of them developed cancer of the brain or nervous system. Prior to this study of cell phone use, the rate of such cancer was found to be \(0.0340 \%\) for those not using cell phones. The data are from the Journal of the National Cancer Institute. a. Use the sample data to construct a \(90 \%\) confidence interval estimate of the percentage of cell phone users who develop cancer of the brain or nervous system. b. Do cell phone users appear to have a rate of cancer of the brain or nervous system that is different from the rate of such cancer among those not using cell phones? 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.

Here is a sample of measured radiation emissions \((\mathrm{cW} / \mathrm{kg})\) for cell phones (based on data from the Environmental Working Group): \(38,55,86,145\). Here are ten bootstrapsamples: \(\\{38,145,55,86\\},\\{86,38,145,145\\},\\{145,86,55,55\\},\\{55,55,55,145\\}\), \(\\{86,86,55,55\\},\\{38,38,86,86\\},\\{145,38,86,55\\},\\{55,86,86,86\\},\\{145,86,55,86\\}\), \(\\{38,145,86,55\\}\) a. Using only the ten given bootstrap samples, construct an \(80 \%\) confidence interval estimate of the population mean. b. Using only the ten given bootstrap samples, construct an \(80 \%\) confidence interval estimate of the population standard deviation.

Data Set 3 "Body Temperatures" in Appendix B includes a sample of 106 body temperatures having a mean of \(98.20^{\circ} \mathrm{F}\) and a standard deviation of \(0.62^{\circ} \mathrm{F}\) (for day 2 at \(12 \mathrm{AM}\) ). Construct a \(95 \%\) confidence interval estimate of the standard deviation of the body temperatures for the entire population.

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?

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