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What \(i s\) normal, when it comes to people's body temperatures? A random sample of 130 human body temperatures, provided by Allen Shoemaker \(^{11}\) in the Journal of Statistical Education, had a mean of \(98.25^{\circ}\) Fahrenheit and a standard deviation of \(0.73^{\circ}\) Fahrenheit. 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 \(98.6^{\circ}\) Fahrenheit, the usual average temperature cited by physicians and others? If not, what conclusions can you draw?

Short Answer

Expert verified
Answer: No, the computed 99% confidence interval does not contain the commonly cited average body temperature of 98.6 degrees Fahrenheit. The interval is approximately (98.085, 98.415) degrees Fahrenheit.

Step by step solution

01

Gather the given information

We are given the following data: Sample size (n) = 130 Sample mean (\(\bar{x}\)) = \(98.25^{\circ}\) Fahrenheit Sample standard deviation (s) = \(0.73^{\circ}\) Fahrenheit Confidence level = 99%
02

Find the critical value\(z^*\)

Using a standard normal probability table or a calculator, find the \(z^*\) value that corresponds to a 99% confidence level (leaving 0.5% in each tail). This value is approximately \(2.576\).
03

Calculate the confidence interval

Now, use the formula to calculate the confidence interval: CI = \(\bar{x} \pm z^* \frac{s}{\sqrt{n}}\) CI = \(98.25 \pm 2.576\left(\frac{0.73}{\sqrt{130}}\right)\) Calculate the margin of error: \(2.576\left(\frac{0.73}{\sqrt{130}}\right) \approx 0.165\) Now find the lower and upper bounds of the confidence interval: Lower bound: \(98.25 - 0.165 \approx 98.085^{\circ}\) Fahrenheit Upper bound: \(98.25 + 0.165 \approx 98.415^{\circ}\) Fahrenheit So, the 99% confidence interval for the average body temperature of healthy people is approximately \((98.085, 98.415)^{\circ}\) Fahrenheit.
04

Check if the confidence interval contains \(98.6^{\circ}\) Fahrenheit

Now, we must determine if the interval \((98.085, 98.415)^{\circ}\) Fahrenheit contains \(98.6^{\circ}\) Fahrenheit. Since \(98.6^{\circ}\) Fahrenheit is outside this interval, we can conclude that the confidence interval does not contain the commonly cited average body temperature of \(98.6^{\circ}\) Fahrenheit. This finding suggests that the usual average temperature cited by physicians and others may not be an accurate representation of the average body temperature for healthy people, at least based on this sample data.

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Key Concepts

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

Statistics
Statistics is a branch of mathematics dealing with the collection, analysis, interpretation, and presentation of masses of numerical data. It provides us with methods for organizing and summarizing data, ranging from simple data visualization to complex statistical models. In the context of our exercise, statistics helps us understand the concept of a confidence interval, which is a range of values, derived from the sample data, that is likely to contain the population parameter with a certain level of confidence. In practice, it gives us a way to express how confident we are that a parameter, such as the average body temperature of healthy people, lies within a certain range.
  • The mean (average) provides a measure of the central tendency of the data.
  • The standard deviation indicates the amount of variability within a set of measurements.
  • Confidence intervals incorporate both the variability of the data and the size of the sample to provide a range around the estimated average.
Normal Distribution
The normal distribution, often referred to as the bell curve, is a continuous probability distribution that is symmetrical about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean. In graph form, it takes the shape of a bell, hence the name. Most of the observations cluster around the central peak and the probabilities for values further away from the mean taper off equally in both directions. In our exercise, the assumption that the population of body temperatures is normally distributed is crucial because it allows us to use the properties of the normal distribution to estimate the confidence interval for the average body temperature.
  • It's important when working with confidence intervals to verify or assume that the data is approximately normally distributed.
  • If the population is known to be normal, then the sample means will also be normally distributed, regardless of the sample size.
Sample Standard Deviation
Sample standard deviation is a measure of the amount of variation or dispersion in a set of values. A low standard deviation means that most of the numbers are close to the sample mean, while a high standard deviation means that the values are spread out over a wider range. When constructing a confidence interval, the sample standard deviation is used to estimate the standard deviation of the population, which then helps to determine how spread out the confidence interval will be. The formula for confidence interval takes this value into account through its use in the margin of error calculation.

Importance in Confidence Intervals

  • The sample standard deviation is a critical component in the margin of error calculation.
  • The larger the sample standard deviation, the wider the confidence interval for a given sample size and confidence level.
  • Understanding its impact helps interpret the results, such as why the commonly cited average body temperature could be outside our confidence interval.

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Most popular questions from this chapter

Excedrin or Tylenol? In a study to compare the effects of two pain relievers it was found that of \(n_{1}=200\) randomly selected individuals who used the first pain reliever, \(93 \%\) indicated that it relieved their pain. Of \(n_{2}=450\) randomly selected individuals who used the second pain reliever, \(96 \%\) indicated that it relieved their pain. a. Find a \(99 \%\) confidence interval for the difference in the proportions experiencing relief from pain for these two pain relievers. b. Based on the confidence interval in part a, is there sufficient evidence to indicate a difference in the proportions experiencing relief for the two pain relievers? Explain.

Use the information given in Exercises 9-15 to find the necessary confidence interval for the population mean \(\mu .\) Interpret the interval that you have constructed. \(\alpha=.01, n=38, \bar{x}=34, s^{2}=12\)

Construct a \(95 \%\) confidence interval for the difference in the population means. Then find a point estimate for the difference in the population means and calculate the margin of error: Compare your results. Can you conclude that there is a difference in the two population means? $$\begin{aligned}&\text { } n_{1}=35, n_{2}=45, \bar{x}_{1}=36.8, \bar{x}_{2}=33.6, s_{1}=4.9\\\ &s_{2}=3.4\end{aligned}$$

Can you afford the price of an NBA ticket during the regular season? The website Www.answers.com indicates that the low prices are around \(\$ 10\) for the high up seats while the court-side seats are around \(\$ 2000\) to \(\$ 5000\) per game and the average price of a ticket is \(\$ 75.50\) a game. \(^{7}\) Suppose that we test this claim by selecting a random sample of \(n=50\) ticket purchases from a computer database and find that the average ticket price is \(\$ 82.50\) with a standard deviation of \(\$ 75.25 .\) a. Do you think that \(x\), the price of an individual regular season ticket, has a mound-shaped distribution? If not, what shape would you expect? b. If the distribution of the ticket prices is not normal, you can still use the standard normal distribution to construct a confidence interval for \(\mu,\) the average price of a ticket. Why? c. Construct a \(95 \%\) confidence interval for \(\mu,\) the average price of a ticket. Does this confidence interval cause you support or question the claimed average price of \(\$ 75.50 ?\) Explain.

Calculate the margin of error in estimating a binomial proportion \(p\) using samples of size \(n=100\) and the values of p given in Exercises \(15-19 .\) What value of p produces the largest margin of error? \(p=.1\)

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