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Chapter 7: Problem 21

Construct the confidence interval estimate of the mean. Listed below are the amounts of net worth (in millions of dollars) of these ten wealthiest celebrities: Tom Cruise, Will Smith, Robert De Niro, Drew Carey, George Clooney, John Travolta, Samuel L. Jackson, Larry King. Demi Moore, and Bruce Willis. Construct a \(98 \%\) confidence interval. What does the result tell us about the population of all celebrities? Do the data appear to be from a normally distributed population as required? $$\begin{array}{rrrrrrrrrr} 250 & 200 & 185 & 165 & 160 & 160 & 150 & 150 & 150 & 150 \end{array}$$

Short Answer

Expert verified
The 98% confidence interval for the mean net worth is from 143.31 to 200.69 million dollars. This interval suggests where the true mean net worth of all celebrities lies. Data assessment for normality is needed.

Step by step solution

01

Organize the Data

List the given celebrity net worths: 250, 200, 185, 165, 160, 160, 150, 150, 150, 150 (in millions of dollars).
02

Calculate the Sample Mean

Add all the values and then divide by the number of samples. \[ \text{Sample Mean} (\bar{x}) = \frac{250 + 200 + 185 + 165 + 160 + 160 + 150 + 150 + 150 + 150}{10} = \frac{1720}{10} = 172 \text{ million dollars} \]
03

Calculate the Sample Standard Deviation

First, find the variance.1. Find each deviation from the mean: \((250-172), (200-172), (185-172), (165-172), (160-172), (160-172), (150-172), (150-172), (150-172), (150-172) = 78, 28, 13, -7, -12, -12, -22, -22, -22, -22\)2. Square each deviation: \((78^2, 28^2, 13^2, -7^2, -12^2, -12^2, -22^2, -22^2, -22^2, -22^2) = (6084, 784, 169, 49, 144, 144, 484, 484, 484, 484)\)3. Sum the squared deviations: \[ \text{Sum} = 6084 + 784 + 169 + 49 + 144 + 144 + 484 + 484 + 484 + 484 = 9320 \]4. Divide by the number of samples minus one (n-1): \[ \text{Variance} = \frac{9320}{9} = 1035.56\]5. Take the square root: \[ s = \sqrt{1035.56} \approx 32.18 \text{ million dollars} \]
04

Find the t-Score for 98% Confidence Interval

Look up the t-score for a 98% confidence interval with 9 degrees of freedom (n-1). This value is approximately 2.821.
05

Calculate the Margin of Error (E)

\[ E = t \left(\frac{s}{\sqrt{n}} \right) = 2.821 \left( \frac{32.18}{\sqrt{10}} \right) = 2.821 \times 10.18 \approx 28.69 \text{ million dollars} \]
06

Construct the Confidence Interval

The 98% confidence interval is calculated as:\[ \bar{x} - E \text{ to } \bar{x} + E \]Substitute the values:\[ 172 - 28.69 \text{ to } 172 + 28.69 = 143.31 \text{ to } 200.69 \text{ million dollars} \]
07

Interpret the Confidence Interval

This interval provides an estimate of the true mean net worth of all celebrities. We are 98% confident that the true mean net worth falls between 143.31 and 200.69 million dollars.
08

Assess Normality

To determine if the data are approximately normally distributed, examine the dataset and consider using graphical methods like histograms or Q-Q plots. Sample size is small but assumes a normal distribution based on t-test application.

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Key Concepts

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

Sample Mean
The sample mean, denoted as \( \bar{x} \), is the average value of a sample set of observations. In this exercise, we calculated the sample mean for the net worths of ten celebrities. To find the sample mean, sum all the net worth values and then divide by the total number of observations. For instance, \[ \bar{x} = \frac{250 + 200 + 185 + 165 + 160 + 160 + 150 + 150 + 150 + 150}{10} = 172 \text{ million dollars} \].
This sample mean represents the average net worth of these ten celebrities and serves as a point estimate of the population mean.
Sample Standard Deviation
The sample standard deviation, represented as s, measures the amount of variation or dispersion in a set of values. A lower standard deviation indicates that the values tend to be closer to the mean, while a higher standard deviation shows more variability.
To calculate the sample standard deviation:
  • First, find each deviation from the mean (the difference between each value and the sample mean).
  • Square each deviation to get rid of negative signs.
  • Sum all squared deviations.
  • Divide the sum by the number of sample points minus one (n-1). This gives you the variance.
  • Take the square root of the variance to get the standard deviation.
For example,
\[ s = \frac{\text{Sum of squared deviations}}{n-1} \], and then find the square root. In this case, \[ s \approx 32.18 \text{ million dollars} \].
Margin of Error
The margin of error, denoted as E, shows how much the sample mean is expected to differ from the true population mean. This incorporates the t-score and sample standard deviation.
You calculate the margin of error using the formula
\[ E = t \times \frac{s}{\text{sqrt}(n)} \],
where:\ n \ is the number of samples, \ s \ is the sample standard deviation, and \ t \ is the t-score corresponding to the desired confidence level and degrees of freedom.
For example, for a 98% confidence interval with 9 degrees of freedom, the t-score is approximately 2.821. So:
\[ E = 2.821 \times \frac{32.18}{\text{sqrt}(10)} \approx 28.69 \text{ million dollars} \].
t-Score
The t-score helps in understanding how far the sample mean is likely to be from the population mean considering the sample's standard deviation.
It's derived from the Student's t-distribution and depends on two main factors: the desired confidence level and the degrees of freedom (which is the sample size minus one).
For instance, for a 98% confidence interval and 9 degrees of freedom, the t-score can be looked up in statistical tables or calculated using software, resulting in approximately 2.821.
Normal Distribution
A normal distribution is a bell-shaped curve where most of the data points lie close to the mean, and fewer points exist as you move away. It's crucial in statistics to assume or determine if the data follows this pattern because various techniques, like confidence intervals, depend on it.
In this exercise, although the sample size is small, the method assumes normality. Graphical methods like histograms or Q-Q plots can help to assess this visually. If the data is roughly symmetrical and follows the bell curve, it can be considered approximately normal.
Approximations help ensure statistical methods provide meaningful and reliable results.

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Most popular questions from this chapter

Finding Confidence Intervals. In Exercises \(9-16,\) assume that each sample is a simple random sample obtained from a population with a normal distribution. Garlic for Reducing Cholesterol 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 \(\mathrm{mg} / \mathrm{dL}\) ) 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 standard deviation of the changes in LDL cholesterol after the garlic treatment. Does the result indicate whether the treatment is effective?

Find the sample size required to estimate the population mean. Data Set 1 "Body Data" in Appendix B includes pulse rates of 153 randomly selected adult males, and those pulse rates vary from a low of 40 bpm to a high of 104 bpm. Find the minimum sample size required to estimate the mean pulse rate of adult males. Assume that we want \(99 \%\) confidence that the sample 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 \(\sigma=11.3\) bpm, based on the value of \(s=11.3\) bpm for the sample of 153 male pulse rates. c. Compare the results from parts (a) and (b). Which result is likely to be better?

Use the data and confidence level to construct a confidence interval estimate of \(p,\) then address the given question. In clinical trials of the drug Lipitor (atorvastatin), 270 subjects were given a placebo and 7 of them had allergic reactions. Among 863 subjects treated with \(10 \mathrm{mg}\) of the drug. 8 experienced allergic reactions. Construct the two \(95 \%\) confidence interval estimates of the percentages of allergic reactions. Compare the results. What do you conclude?

Find the critical value \(z_{a / 2}\) that corresponds to the given confidence level. $$98 \%$$

Construct the confidence interval estimate of the mean. Listed below are amounts of arsenic \((\mu \mathrm{g},\) or micrograms, per serving) in samples of brown rice from California (based on data from the Food and Drug Administration). Use a \(90 \%\) confidence level. The Food and Drug Administration also measured amounts of arsenic in samples of brown rice from Arkansas. Can the confidence interval be used to describe arsenic levels in Arkansas? $$\begin{array}{cccccccccc} 5.4 & 5.6 & 8.4 & 7.3 & 4.5 & 7.5 & 1.5 & 5.5 & 9.1 & 8.7 \end{array}$$
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