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What is normal, when it comes to people's body temperatures? A random sample of 130 human body temperatures, provided by Allen Shoemaker \(^{9}\) in the Journal of Statistical Education, had a mean of 98.25 degrees and a standard deviation of 0.73 degrees. a. Construct a \(99 \%\) confidence interval for the average body temperature of healthy people. b. Does the confidence interval constructed in part a contain the value 98.6 degrees, the usual average temperature cited by physicians and others? If not, what conclusions can you draw?

Short Answer

Expert verified
Answer: No, the 99% confidence interval for the average body temperature of healthy people is (98.085, 98.415) and does not contain the value 98.6 degrees.

Step by step solution

01

a. Constructing a 99% confidence interval for the average body temperature of healthy people

To construct a 99% confidence interval, we first need to determine the critical z-value, \(z_{\frac{\alpha}{2}}\), for a 99% confidence level. We can find this using a z-table or from a calculator. Since this is a two-tailed confidence interval, we want to find the z-score that corresponds to the area between the z-score and -z-score, which is 0.995 (as half of the remaining area, i.e., 0.01, is split between the two tails). Using a z-table, we find the closest value to 0.995, which is 2.576. Now we can plug all the information into the formula for the confidence interval: \(\bar{x} \pm z_{\frac{\alpha}{2}} \times \frac{s}{\sqrt{n}}\) Where: \(\bar{x} = 98.25 \) (the sample mean) \(s = 0.73 \) (the sample standard deviation) \(n = 130 \) (the sample size) \(z_{\frac{\alpha}{2}} = 2.576 \) Calculating the confidence interval, we get: \(98.25 \pm 2.576 \times \frac{0.73}{\sqrt{130}}\) Calculate the margin of error: \(2.576 \times \frac{0.73}{\sqrt{130}} = 0.165\) So the 99% confidence interval for the average body temperature of healthy people is: \((98.25 - 0.165, 98.25 + 0.165) = (98.085, 98.415)\)
02

b. Checking if the confidence interval contains the value 98.6 degrees

Now we need to check if the confidence interval we calculated in part a contains the value 98.6 degrees. The confidence interval we calculated is \((98.085, 98.415)\). Since 98.6 is not within this interval, we can conclude that based on this sample, the usual average temperature cited by physicians and others (98.6 degrees) is not within the 99% confidence interval of the average body temperature for healthy people. This means that there is significant evidence at the 1% level of significance to reject the hypothesis that the true average body temperature is 98.6 degrees.

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Key Concepts

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

Standard Deviation
Standard deviation is a statistic that measures the amount of variation or dispersion in a set of values. It is often denoted by the letter 's' for a sample or 'σ' (sigma) for a population. In simple terms, a low standard deviation indicates that the values tend to be close to the mean (average) of the set, while a high standard deviation indicates that the values are spread out over a wider range.

For example, consider a random sample of human body temperatures with a standard deviation of 0.73 degrees. This indicates that most of the temperatures in our sample are within 0.73 degrees of the average temperature. Understanding standard deviation is crucial for interpreting the spread of data points and for calculating confidence intervals, which reflect the reliability of an estimate.
Sample Mean
The sample mean, often denoted as \(\bar{x}\), is the average of the values in a sample. It is calculated by summing all the values and then dividing by the number of values. The sample mean is a commonly used estimate of the population mean, but since it is based on a sample, it comes with uncertainty. This uncertainty can be quantified using a confidence interval.

Continuing with our example from the introduction, the sample mean of the body temperatures is 98.25 degrees. This is our best estimate for the average body temperature of healthy people based on the sample data. However, to make inferences about the average temperature for all healthy people, we need to calculate a confidence interval that will give us a range of plausible values for the population mean.
Z-Score
The z-score is a statistical measurement that describes a value's relationship to the mean of a group of values, all measured in terms of standard deviations. It indicates how many standard deviations an element is from the mean. A z-score can be positive or negative, with a positive value indicating that the score is above the mean, and a negative score indicating it's below the mean.

In the context of confidence intervals, the critical z-value (denoted as \(z_{\frac{\alpha}{2}}\)) corresponds to the desired confidence level. This z-score is used to determine the range within which a certain percentage of the data is expected to fall. For a 99% confidence level, the z-score is high, reflecting the fact that we expect 99% of all possible sample means to fall within the interval defined by this z-score.
Normal Distribution
The normal distribution, also known as the Gaussian distribution, is a probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean. In graph form, this distribution is represented by a bell-shaped curve, known as the 'bell curve.'

The normal distribution is extremely important in statistics because of its widespread appearance in natural phenomena. Many statistical tests assume that the data follows a normal distribution. For confidence interval calculations, if we assume that the population distribution is normal, then the sampling distribution of the sample mean is also normal. This property is what allows us to use z-scores and the standard deviation of the sample mean to calculate confidence intervals for population parameters like the average body temperature.

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Most popular questions from this chapter

A sample survey is designed to estimate the proportion of sports utility vehicles being driven in the state of California. A random sample of 500 registrations are selected from a Department of Motor Vehicles database, and 68 are classified as sports utility vehicles. a. Use a \(95 \%\) confidence interval to estimate the proportion of sports utility vehicles in California b. How can you estimate the proportion of sports utility vehicles in California with a higher degree of accuracy? (HINT: There are two answers.)

A researcher classified his subjects as innately right-handed or lefthanded by comparing thumbnail widths. He took a sample of 400 men and found that 80 men could be classified as left-handed according to his criterion. Estimate the proportion of all males in the population who would test to be left-handed using a \(95 \%\) confidence interval.

Due to a variation in laboratory techniques, impurities in materials, and other unknown factors, the results of an experiment in a chemistry laboratory will not always yield the same numerical answer. In an electrolysis experiment, a class measured the amount of copper precipitated from a saturated solution of copper sulfate over a 30 -minute period. The \(n=30\) students calculated a sample mean and standard deviation equal to .145 and .0051 mole, respectively. Find a \(90 \%\) confidence interval for the mean amount of copper precipitated from the solution over a 30 -minute period.

Refer to Exercise \(8.18 .\) The means and standard deviations for 50 billing statements from each of the computer databases of each of the three hotel chains are given in the table: $$\begin{array}{lccc} & \text { Marriott } & \text { Radisson } & \text { Wyndham } \\\\\hline \text { Sample Average } & \$ 170 & \$ 145 & \$ 160 \\\\\text { Sample Standard Deviation } & 17.5 & 10 & 16.5\end{array}$$ a. Find a \(95 \%\) confidence interval for the difference in the average room rates for the Marriott and the Wyndham hotel chains. b. Find a \(99 \%\) confidence interval for the difference in the average room rates for the Radisson and the Wyndham hotel chains. c. Do the intervals in parts a and b contain the value \(\left(\mu_{1}-\mu_{2}\right)=0 ?\) Why is this of interest to the researcher? d. Do the data indicate a difference in the average room rates between the Marriott and the Wyndham chains? Between the Radisson and the Wyndham chains?

A random sample of \(n=50\) observations from a quantitative population produced \(\bar{x}=56.4\) and \(s^{2}=2.6 .\) Give the best point estimate for the population mean \(\mu,\) and calculate the margin of error.

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