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Chapter 8: Problem 13

How much do wild mountain lions weigh? The 77 th Annual Report of the New Mexico Department of Game and Fish, edited by Bill Montoya, gave the following information. Adult wild mountain lions (18 months or older) captured and released for the first time in the San Andres Mountains gave the following weights (pounds): \(\begin{array}{llllll}68 & 104 & 128 & 122 & 60 & 64\end{array}\) (a) Use a calculator with mean and sample standard deviation keys to verify that \(\bar{x}=91.0\) pounds and \(s \approx 30.7\) pounds. (b) Find a \(75 \%\) confidence interval for the population average weight \(\mu\) of all adult mountain lions in the specified region.

Short Answer

Expert verified
The 75% confidence interval for the average weight is approximately (76.61, 105.39) pounds.

Step by step solution

01

Calculate the Mean

To find the mean (x) of the weights, sum all weights together and divide by the total number of weights.\[\bar{x} = \frac{68 + 104 + 128 + 122 + 60 + 64}{6} = \frac{546}{6} = 91.0\]
02

Calculate the Sample Standard Deviation

Use the sample standard deviation formula. First, find the deviations from the mean, square them, sum them up and then divide by one less than the number of observations. Finally, take the square root:\[s = \sqrt{\frac{1}{n-1} \sum_{i=1}^{n} (x_i - \bar{x})^2}\]Applying the values:\[s = \sqrt{\frac{(68 - 91)^2 + (104 - 91)^2 + (128 - 91)^2 + (122 - 91)^2 + (60 - 91)^2 + (64 - 91)^2}{5}}\]\[s = \sqrt{\frac{529 + 169 + 1369 + 961 + 961 + 729}{5}} \approx 30.7\]
03

Find the Z-Score for 75% Confidence Interval

A 75% confidence interval corresponds to a z-score of approximately 1.150. This can be found using statistical tables or a calculator.
04

Calculate the 75% Confidence Interval

Use the formula for the confidence interval: \[\bar{x} \pm z \left(\frac{s}{\sqrt{n}}\right)\]Substitute the known values:\[91 \pm 1.150 \left(\frac{30.7}{\sqrt{6}}\right)\]Calculate the margin of error:\[1.150 \times \frac{30.7}{2.45} \approx 14.393\]Thus, the confidence interval is:\[ 91 \pm 14.393 = (76.607, 105.393)\]

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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 is an average value of a set of data points, often depicted as \(\bar{x}\). Calculating the sample mean is a fundamental step in statistical analysis because it gives us a central point, or "center of mass," for our data. To find the sample mean, you simply add all the data values together, and then divide by the number of data points.
For the mountain lion weights example:
  • Add the weights: 68 + 104 + 128 + 122 + 60 + 64 = 546.
  • Count the number of data points: 6.
  • Divide the sum by the number of data points: \(546 / 6 = 91\).
This tells us that the average weight of the sampled mountain lions is 91 pounds. Simple and straightforward, the sample mean helps churn vast data into manageable insight.
Sample Standard Deviation
The sample standard deviation, symbolized as \(s\), measures how much the individual data points deviate from the mean. Unlike the mean, which shows central tendency, the standard deviation reflects variability or dispersion.
You calculate it with this process:
  • Find each data point's deviation from the mean: Subtract the mean from each data point's value.
  • Square each deviation: Convert all differences to positive values and exaggerate bigger deviations.
  • Calculate the average of these squared deviations, considering \(n-1\) where \(n\) is the number of data points: Account for the spread without biasing towards your sample.
  • Find the square root of this result: Turn the variance back into standard units.
Using this technique on our mountain lion weights calculates around 30.7 pounds, suggesting considerable spread from the average.
Z-Score
A Z-score is a statistical measurement that describes a data point's relationship to the mean of a group of points. It tells us how many standard deviations an element is from the mean. In the context of confidence intervals, the Z-score represents the point along a normal distribution that captures a certain percentage of the data.
For a 75% confidence interval, the Z-score is approximately 1.150.
This Z-score will help us compute the range within which we expect the true population mean to fall with a 75% probability. It's a critical piece in adjusting for variation and sampling error, giving us the margin of error that fine-tunes our mean estimate for reliability.
Population Mean
The population mean, denoted as \(\mu\), is the average of a set of values for the entire population. Unlike the sample mean, which averages a selection of data, the population mean considers every data point possible, providing a complete representation.
In research and statistics, often the goal of using a sample is to estimate this true population mean. We use methods like calculating confidence intervals to infer conclusions about \(\mu\) efficiently.
For the mountain lion weights example, the 75% confidence interval calculated estimates that the population mean weight of all adult mountain lions in the San Andres Mountains lies between approximately 76.607 and 105.393 pounds.
Understanding \(\mu\) expands the significance of a study's findings beyond just the sample, providing insights into the total subject group represented by the sample.

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Most popular questions from this chapter

Jobs and productivity! How do banks rate? One way to answer this question is to examine annual profits per employee. Forbes Top Companies, edited by J. T. Davis (John Wiley \& Sons), gave the following data about annual profits per employee (in units of one thousand dollars per employee) for representative companies in financial services. Companies such as Wells Fargo, First Bank System, and Key Banks were included. Assume \(\sigma \approx 10.2\) thousand dollars. $$ \begin{array}{llllllllll} 42.9 & 43.8 & 48.2 & 60.6 & 54.9 & 55.1 & 52.9 & 54.9 & 42.5 & 33.0 & 33.6 \\ 36.9 & 27.0 & 47.1 & 33.8 & 28.1 & 28.5 & 29.1 & 36.5 & 36.1 & 26.9 & 27.8 \\ 28.8 & 29.3 & 31.5 & 31.7 & 31.1 & 38.0 & 32.0 & 31.7 & 32.9 & 23.1 & 54.9 \\ 43.8 & 36.9 & 31.9 & 25.5 & 23.2 & 29.8 & 22.3 & 26.5 & 26.7 & & \end{array} $$ (a) Use a calculator or appropriate computer software to verify that, for the preceding data, \(\bar{x} \approx 36.0\). (b) Let us say that the preceding data are representative of the entire sector of (successful) financial services corporations. Find a \(75 \%\) confidence interval for \(\mu\), the average annual profit per employee for all successful banks. (c)Let us say that you are the manager of a local bank with a large number of employees. Suppose the annual profits per employee are less than 30 thousand dollars per employee. Do you think this might be somewhat low compared with other successful financial institutions? Explain by referring to the confidence interval you computed in part (b). (d) Suppose the annual profits are more than 40 thousand dollars per employee. As manager of the bank, would you feel somewhat better? Explain by referring to the confidence interval you computed in part (b). (e) Repeat parts (b), (c), and (d) for a \(90 \%\) confidence level.

How hot is the air in the top (crown) of a hot air balloon? Information from Ballooning: The Complete Guide to Riding the Winds, by Wirth and Young (Random House), claims that the air in the crown should be an average of \(100^{\circ} \mathrm{C}\) for a balloon to be in a state of equilibrium. However, the temperature does not need to be exactly \(100^{\circ} \mathrm{C}\). What is a reasonable and safe range of temperatures? This range may vary with the size and (decorative) shape of the balloon. All balloons have a temperature gauge in the crown. Suppose that 56 readings (for a balloon in equilibrium) gave a mean temperature of \(\bar{x}=97^{\circ} \mathrm{C}\). For this balloon, \(\sigma \approx 17^{\circ} \mathrm{C}\). (a) Compute a \(95 \%\) confidence interval for the average temperature at which this balloon will be in a steady-state equilibrium. (b) If the average temperature in the crown of the balloon goes above the high end of your confidence interval, do you expect that the balloon will go up or down? Explain.

The home run percentage is the number of home runs per 100 times at bat. A random sample of 43 professional baseball players gave the following data for home run percentages (Reference: The Baseball Encyclopedia, Macmillan). $$ \begin{array}{llllllllll} 1.6 & 2.4 & 1.2 & 6.6 & 2.3 & 0.0 & 1.8 & 2.5 & 6.5 & 1.8 \\ 2.7 & 2.0 & 1.9 & 1.3 & 2.7 & 1.7 & 1.3 & 2.1 & 2.8 & 1.4 \\ 3.8 & 2.1 & 3.4 & 1.3 & 1.5 & 2.9 & 2.6 & 0.0 & 4.1 & 2.9 \\ 1.9 & 2.4 & 0.0 & 1.8 & 3.1 & 3.8 & 3.2 & 1.6 & 4.2 & 0.0 \\ 1.2 & 1.8 & 2.4 & & & & & & & \end{array} $$ (a) Use a calculator with mean and standard deviation keys to verify that \(\bar{x} \approx 2.29\) and \(s \approx 1.40 .\) (b) Compute a \(90 \%\) confidence interval for the population mean \(\mu\) of home run percentages for all professional baseball players. Hint: If you use Table 6 of Appendix II, be sure to use the closest \(d\). \(f\). that is smaller. (c) Compute a \(99 \%\) confidence interval for the population mean \(\mu\) of home run percentages for all professional baseball players. (d) The home run percentages for three professional players are Tim Huelett, \(2.5 \quad\) Herb Hunter, \(2.0 \quad\) Jackie Jensen, \(3.8\) Examine your confidence intervals and describe how the home run percentages for these players compare to the population average. (e) In previous problems, we assumed the \(x\) distribution was normal or approximately normal. Do we need to make such an assumption in this problem? Why or why not? Hint: See the central limit theorem in Section \(7.2 .\)

Over the past several months, an adult patient has been treated for tetany (severe muscle spasms). This condition is associated with an average total calcium level below \(6 \mathrm{mg} / \mathrm{dl}(\) Reference: Manual of Laboratory and Diagnostic Tests, F. Fischbach). Recently, the patient's total calcium tests gave the following readings (in \(\mathrm{mg} / \mathrm{dl}\) ). \(9.3\) \(\begin{array}{llllll}8.8 & 10.1 & 8.9 & 9.4 & 9.8 & 10.0\end{array}\) \(\begin{array}{lll}9.9 & 11.2 & 12.1\end{array}\) (a) Use a calculator to verify that \(\bar{x}=9.95\) and \(s \approx 1.02\). (b) Find a \(99.9 \%\) confidence interval for the population mean of total calcium in this patient's blood. (c) Based on your results in part (b), do you think this patient still has a calcium deficiency? Explain.

(a) Suppose a \(95 \%\) confidence interval for the difference of means contains both positive and negative numbers. Will a \(99 \%\) confidence interval based on the same data necessarily contain both positive and negative numbers? Explain. What about a \(90 \%\) confidence interval? Explain. (b) Suppose a \(95 \%\) confidence interval for the difference of proportions contains all positive numbers. Will a \(99 \%\) confidence interval based on the same data necessarily contain all positive numbers as well? Explain. What about a \(90 \%\) confidence interval? Explain.
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