/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 12 How much does a sleeping bag cos... [FREE SOLUTION] | 91Ó°ÊÓ

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How much does a sleeping bag cost? Let's say you want a sleeping bag that should keep you warm in temperatures from \(20^{\circ} \mathrm{F}\) to \(45^{\circ} \mathrm{F}\). A random sample of prices (\$) for sleeping bags in this temperature range was taken from Backpacker Magazine: Gear Guide (Vol. 25 , Issue 157, No. 2). Brand names include American Camper, Cabela's, Camp 7, Caribou, Cascade, and Coleman. \(\begin{array}{rrrrrrrrrr}80 & 90 & 100 & 120 & 75 & 37 & 30 & 23 & 100 & 110 \\ 105 & 95 & 105 & 60 & 110 & 120 & 95 & 90 & 60 & 70\end{array}\) (a) Use a calculator with mean and sample standard deviation keys to verify that \(\bar{x} \approx \$ 83.75\) and \(s \approx \$ 28.97\). (b) Using the given data as representative of the population of prices of all summer sleeping bags, find a \(90 \%\) confidence interval for the mean price \(\mu\) of all summer sleeping bags.

Short Answer

Expert verified
The 90% confidence interval for the mean price is \((72.56, 94.94)\).

Step by step solution

01

Understand the problem

We are given a sample of sleeping bag prices and are tasked with verifying the mean and standard deviation. Then, using this sample, we need to calculate a 90% confidence interval for the mean price of all summer sleeping bags.
02

Verify the mean and sample standard deviation

Calculate the sample mean \( \bar{x} \) and sample standard deviation \( s \) using the given data. The sample mean \( \bar{x} \) is calculated as the sum of all prices divided by the number of prices. The sample standard deviation \( s \) measures the spread of the data about the mean.
03

Compute the sample mean \( \bar{x} \)

Sum all the provided prices: \( 80 + 90 + 100 + 120 + 75 + 37 + 30 + 23 + 100 + 110 + 105 + 95 + 105 + 60 + 110 + 120 + 95 + 90 + 60 + 70 = 1675 \). Divide the sum by the number of prices (20) to find \( \bar{x} \): \( \bar{x} = \frac{1675}{20} = 83.75 \).
04

Calculate sample standard deviation \( s \)

To find \( s \), first compute the variance: find the difference between each price and the mean, square the differences, sum them up, and divide by \( n-1 \) (19). Then take the square root of the variance to get \( s \). The calculations confirm \( s \approx 28.97 \).
05

Determine the 90% confidence interval

The formula for a confidence interval is \( \bar{x} \pm t_{\alpha/2} \times \frac{s}{\sqrt{n}} \). Since we are using a 90% confidence level and \( n = 20 \), find the critical value \( t_{\alpha/2} \) in a t-distribution table corresponding to 19 degrees of freedom. Use this to calculate the margin of error and, consequently, the confidence interval.
06

Find the critical value and calculate margin of error

For 19 degrees of freedom, the critical value \( t_{\alpha/2} \approx 1.729 \). Calculate the margin of error: \( E = 1.729 \times \frac{28.97}{\sqrt{20}} \approx 11.19 \).
07

Calculate the confidence interval endpoints

The 90% confidence interval is \( (\bar{x} - E, \bar{x} + E) \), which computes to \( (83.75 - 11.19, 83.75 + 11.19) \), or \( (72.56, 94.94) \).

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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, often symbolized as \( \bar{x} \), is a fundamental statistic used to summarize a set of data. It's simply the arithmetic average of all observations in your dataset. In the context of the sleeping bag prices, the sample mean is calculated by adding up all the prices and dividing by the total number of prices. This gives us a measure of central tendency, or in simpler terms, a single value that attempts to describe the center of a dataset.
  • **Adding all prices together**: In our example, we sum up prices: \( 80 + 90 + 100 + \ldots + 70 = 1675 \)
  • **Dividing by the number of observations**: We have 20 sleeping bag prices in total. Thus, our sample mean \( \bar{x} = \frac{1675}{20} = 83.75 \)
This sample mean \( \$ 83.75 \) gives us a good idea of what the typical price is within our sample. It's important to note that this is just an estimate of the true population mean, which would be the average price if we considered all possible sleeping bags in this temperature range.
Sample Standard Deviation
The sample standard deviation, denoted as \( s \), gives us insight into how spread out our data is around the sample mean. It is critical for understanding the variability in your dataset. If the prices are relatively clustered around the mean, the standard deviation will be small. Conversely, if there is a wide range of prices, \( s \) will be larger.To calculate \( s \), follow these steps:
  • **Subtract the mean from each observation** to find deviations: E.g., for the first price, \( 80 - 83.75 = -3.75 \)
  • **Square each deviation** to eliminate negatives, and sum them up
  • **Divide by \( n-1 \)**, where \( n = 20 \), for an unbiased estimate. This calculation yields the variance: \( s^2 \)
  • **Take the square root** of the variance to return to the original units, resulting in \( s \approx 28.97 \)
This sample standard deviation tells us that, on average, the prices vary by about \( \$28.97 \) from the mean price. Understanding this concept helps us appreciate how spread out the prices of sleeping bags are compared to their mean value.
t-distribution
When constructing a confidence interval for the mean, particularly when dealing with small sample sizes (like our 20 sleeping bag prices), the t-distribution is a useful tool. Unlike the normal distribution, the t-distribution accommodates variability better with smaller sample sizes.Key aspects of the t-distribution:
  • **Shape and Spread**: The t-distribution is similar to the normal distribution but has heavier tails. This means it accounts for more dispersion of data around the mean, which is helpful when our sample size is small.
  • **Degrees of Freedom (df)**: Calculated as \( n-1 \), where \( n \) is the sample size. For our sleeping bag prices, \( df = 20 - 1 = 19 \).
  • **t-Critical Values**: Utilized to identify the margin of error in confidence intervals. For a 90% confidence level and \( df = 19 \), the t-critical value is approximately \( 1.729 \).
In our confidence interval calculation, the t-distribution ensures that our interval estimation captures the true population mean correctly. We multiplied the t-critical value by the standard error (\( \frac{28.97}{\sqrt{20}} \)) to determine the margin of error and finalized the confidence interval for the true mean price of summer sleeping bags. This method is reliable even when our sample isn't huge.

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Most popular questions from this chapter

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.

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.

If the original \(x\) distribution has a relatively small standard deviation, the confidence interval for \(\mu\) will be relatively short.

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 \(7.2 .\)

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