/*! 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 14 Do you want to own your own cand... [FREE SOLUTION] | 91Ó°ÊÓ

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Do you want to own your own candy store? Wow! With some interest in running your own business and a decent credit rating, you can probably get a bank loan on startup costs for franchises such as Candy Express, The Fudge Company, Karmel Corn, and Rocky Mountain Chocolate Factory. Startup costs (in thousands of dollars) for a random sample of candy stores are given below (Source: Entrepreneur Magazine, Vol. 23, No. 10\()\). \(\begin{array}{lllllllll}95 & 173 & 129 & 95 & 75 & 94 & 116 & 100 & 85\end{array}\) Use a calculator with mean and sample standard deviation keys to verify that \(\bar{x} \approx 106.9\) thousand dollars and \(s \approx 29.4\) thousand dollars. Find a \(90 \%\) confidence interval for the population average startup costs \(\mu\) for candy store franchises.

Short Answer

Expert verified
The 90% confidence interval for the population mean startup cost is \((88.67, 125.13)\) thousand dollars.

Step by step solution

01

Identify the Sample Size

Count the number of data points in the given sample. In this case, we have the dataset: \(95, 173, 129, 95, 75, 94, 116, 100, 85\). The sample size \( n \) is 9.
02

Calculate the Sample Mean

The sample mean \( \bar{x} \) is provided as \( \bar{x} = 106.9 \) thousand dollars. This represents the average of the data points in the sample.
03

Calculate the Sample Standard Deviation

The sample standard deviation \( s \) is provided as \( s = 29.4 \) thousand dollars. This measures the dispersion of the sample data points from the mean.
04

Determine the t-Value

Since the sample size is small (\( n = 9 \)) and the population standard deviation is unknown, we use the t-distribution. For a 90% confidence interval and degrees of freedom \( \text{df} = n - 1 = 8 \), the t-value is approximately \( 1.860 \). This value can be found using a t-table or a calculator.
05

Calculate the Margin of Error

The margin of error \( E \) is calculated using the formula: \[ E = t \frac{s}{\sqrt{n}} \]Substitute the known values: \[ E = 1.860 \times \frac{29.4}{\sqrt{9}} = 1.860 \times \frac{29.4}{3} = 1.860 \times 9.8 \approx 18.23 \] thousand dollars.
06

Construct the Confidence Interval

The 90% confidence interval for the population mean \( \mu \) is given by the formula: \( \bar{x} \pm E \).Substituting the known values: \[ 106.9 \pm 18.23 \]This results in the interval: \( (88.67, 125.13) \) thousand dollars.

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Key Concepts

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

Sample Mean
In statistics, the sample mean is a crucial measure of central tendency when working with a subset of data from a larger population. In simpler terms, it tells you the average of all the numbers in your sample. Calculating the sample mean is straightforward: you sum up all the individual data points in your sample and then divide by the number of observations in that sample.

For instance, if we consider a dataset of startup costs for various candy stores:
  • 95
  • 173
  • 129
  • 95
  • 75
  • 94
  • 116
  • 100
  • 85
We find that there are 9 numbers in total. The sample mean is the sum of these numbers divided by 9, resulting in an average (or sample mean) of approximately 106.9. This figure is essential because it serves as the center point for calculating other important statistics, like the confidence interval.
T-Distribution
The t-distribution arises when we are working with small sample sizes and don't know the population standard deviation. In these cases, using the t-distribution provides a more accurate assessment compared to a normal distribution. This is especially true for data with less than 30 observations.

When constructing confidence intervals from small sample sizes, the shape of the t-distribution accounts for added variability by having heavier tails compared to the normal distribution. The "heavier tails" imply that the t-distribution assumes more uncertainty about the mean value as we are not working with a large dataset.

The t-distribution is characterized by "degrees of freedom" (df), calculated as the sample size minus 1. When using a t-table or calculator, these degrees of freedom help to find the t-value, which is used to calculate the margin of error for the confidence interval.
Sample Standard Deviation
The sample standard deviation is a measure that quantifies the amount of variation or dispersion in a dataset. It tells you, on average, how much each data point in the sample deviates from the sample mean. This measure is critical because it impacts the calculation of the confidence interval and helps us understand the spread of data in the sample.

To find the sample standard deviation, you would:
  • Calculate the mean ( \( ar{x} \) ) of the sample.
  • Subtract the mean from each data point and square the result.
  • Sum all the squared differences.
  • Divide this sum by the number of data points minus one (n - 1) to find the variance.
  • Take the square root of the variance to get the standard deviation.
In the candy store example, the sample standard deviation ( \( s \) ) was given as 29.4 thousand dollars. This figure is necessary for calculating the margin of error, which ultimately helps define the confidence interval.
Margin of Error
The margin of error is an essential concept when assessing the precision of statistical estimates, especially within confidence intervals. It provides a range that likely contains the true population mean. Essentially, the margin of error is the plus-or-minus figure in a confidence interval.

To calculate it, you multiply the t-value (from the t-distribution) by the standard error of the sample mean. The standard error is obtained by dividing the sample standard deviation ( \( s \) ) by the square root of the sample size ( \( n \) ). Mathematically, it is expressed as:\[ E = t \times \frac{s}{\sqrt{n}} \]

This means that with our example, given the t-value of 1.860, a standard deviation of 29.4, and a sample size of 9, the margin of error becomes approximately 18.23 thousand dollars. This margin helps in stating the interval within which the true mean is likely found, adding scientific credibility to the data interpretation.

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Most popular questions from this chapter

A random sample of medical files is used to estimate the proportion \(p\) of all people who have blood type \(B\). (a) If you have no preliminary estimate for \(p\), how many medical files should you include in a random sample in order to be \(85 \%\) sure that the point estimate \(\hat{p}\) will be within a distance of \(0.05\) from \(p ?\) (b) Answer part (a) if you use the preliminary estimate that about 8 out of 90 people have blood type B. (Reference: Manual of Laboratory and Diagnostic Tests, F. Fischbach.)

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.

(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.

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 \(7.2 .\)

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.

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