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Chapter 7: Problem 9

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.

Short Answer

Expert verified
The 95% confidence interval for the population standard deviation is approximately (0.53, 0.74).

Step by step solution

01

Identify given values

Given data includes a sample size () of 106, a mean () of 98.20掳F, and a standard deviation () of 0.62掳F. We want to construct a 95% confidence interval for the population standard deviation.
02

Understand the formula

The confidence interval for the population standard deviation is based on the chi-square distribution. The formula for the confidence interval is
03

Find critical values

Using chi-square distribution tables, find the critical values and for a 95% confidence level with degrees of freedom ().
04

Calculate confidence interval

Plug the known values into the confidence interval formula and compute. The interval is given by . This provides the lower and upper bounds for the population standard deviation.

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Key Concepts

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

sample size
Mathematical computations aside, does the sample size matter? Indeed, sample size plays a crucial role in creating a confidence interval.
Sample size affects both the degrees of freedom (df) and the width of the confidence interval. In our problem, the value of 106 determines the degrees of freedom used in the chi-square distribution (df = 1) and variability in our dataset.
A larger sample size generally leads to a narrower confidence interval, indicating a more precise estimate of the population standard deviation.
Conversely, a larger sample size would result in more variation and a broader confidence interval. This underscores the importance of having an adequate sample to draw accurate conclusions about the population from which the sample was drawn.
Larger samples help reduce variability and increase the reliability of statistical estimates, thereby providing more confidence in our findings.

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Most popular questions from this chapter

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

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.

In a study of the accuracy of fast food drive-through orders, Burger King had 264 accurate orders and 54 that were not accurate (based on data from \(Q S R\) magazine). a. Construct a \(99 \%\) confidence interval estimate of the percentage of orders that are not accurate. b. Compare the result from part (a) to this \(99 \%\) confidence interval for the percentage of orders that are not accurate at Wendy's: \(6.2 \%

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?

In a test of the effectiveness of garlic for lowering cholesterol, 49 subjects were treated with raw garlic. Cholesterol levels were measured before and after the treatment. The changes (before minus after) in their levels of LDL cholesterol (in mg/dL) had a mean of \(0.4\) and a standard deviation of \(21.0\) (based on data from "Effect of Raw Garlic vs Commercial Garlic Supplements on Plasma Lipid Concentrations in Adults with Moderate Hypercholesterolemia," by Gardner et al., Archives of Internal Medicine, Vol. 167). Construct a \(98 \%\) confidence interval estimate of the mean net change in LDL cholesterol after the garlic treatment. What does the confidence interval suggest about the effectiveness of garlic in reducing LDL cholesterol?
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