/*! 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 16 Assume that the population of \(... [FREE SOLUTION] | 91Ó°ÊÓ

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Assume that the population of \(x\) values has an approximately normal distribution. Franchise: Candy Store Do you want to own your own candy store? 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 ). \(\begin{array}{lllllllll}95 & 173 & 129 & 95 & 75 & 94 & 116 & 100 & 85\end{array}\) (a) 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. (b) Find a \(90 \%\) confidence interval for the population average startup costs \(\mu\) for candy store franchises. (c) Interpretation What does the confidence interval mean in the context of this problem?

Short Answer

Expert verified
The 90% confidence interval for the average startup cost is $90,779 to $123,021.

Step by step solution

01

Calculate Mean and Sample Standard Deviation

To verify given values, first calculate the mean (\(\bar{x}\)) of the startup costs. Add all the sample values and divide by the number of samples (9): \(\bar{x} = \frac{95 + 173 + 129 + 95 + 75 + 94 + 116 + 100 + 85}{9} = 106.9\). Then, calculate the sample standard deviation (\(s\)), which measures the spread around the mean. Computing \(s\) manually or using a calculator confirms it is approximately 29.4.
02

Understand Confidence Interval Formula

A confidence interval estimates the range in which the population mean might lie. For a 90% confidence interval, the formula is: \(\bar{x} \pm z_{\alpha/2} \times \frac{s}{\sqrt{n}}\). Here, \(z_{\alpha/2}\) corresponds to the z-score for a 90% confidence level, \(s\) is the standard deviation, and \(n\) is the sample size.
03

Obtain the Z-Score for 90% Confidence

For a 90% confidence interval, \(z_{\alpha/2} = 1.645\). This is obtained from a standard normal distribution table or calculator.
04

Calculate the Confidence Interval

Plug values into the confidence interval formula: \(106.9 \pm 1.645 \times \frac{29.4}{3}\). Calculate the margin of error: \(1.645 \times 9.8 = 16.121\). Therefore, the confidence interval is \([106.9 - 16.121, 106.9 + 16.121]\) which is approximately \([90.779, 123.021]\).
05

Interpret the Confidence Interval

The 90% confidence interval suggests that if we repeated this sampling process many times, 90% of the calculated intervals would contain the true population mean startup cost. Here, we are 90% confident the true average startup cost is between $90,779 and $123,021.

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Key Concepts

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

Normal Distribution
The normal distribution is a key concept in statistics that refers to a bell-shaped, symmetrical curve. This distribution is important because many natural phenomena, including measurement errors and test scores, follow it. Its distinctive shape is defined such that most of the data in a set will fall near the mean (average), with fewer data points appearing as you move away from the center.
In the context of the candy store franchise example, we assume that the startup costs for candy stores are approximately normally distributed. This assumption allows us to apply statistical techniques, like confidence intervals, to infer about the population from a sample. This is crucial in practical scenarios where we cannot measure the whole population and need a reliable method to make predictions.
  • The curve is symmetric around the mean.
  • The mean, median, and mode of a normal distribution are equal.
  • About 68% of values lie within one standard deviation of the mean.
  • Approximately 95% are within two standard deviations, and 99.7% within three standard deviations.
Sample Standard Deviation
Sample standard deviation (\(s\)) is a measure that indicates how much individual data points in a sample deviate from the sample mean (\(\bar{x}\)). It's used to quantify the amount of variation or dispersion in a set of data points. When calculating it for a sample, rather than an entire population, it's known as the sample standard deviation.
This measure is particularly useful when making inferences or predictions based on just a part of the population.
In our case of startup costs for candy stores, the sample standard deviation is approximately 29.4 thousand dollars. This tells us about the spread or variability in the startup costs among the sampled stores. A higher value would indicate that costs are more spread out or varied, whereas a lower value suggests that they are more consistent.
Z-Score
A Z-score is a statistical measurement that describes a value's relation to the mean of a group of values. It is expressed as the number of standard deviations away from the mean a data point is. The formula to calculate a Z-score is:\[ Z = \frac{x - \bar{x}}{s} \]where \(x\) is the value being analyzed, \(\bar{x}\) is the mean, and \(s\) is the standard deviation.
In constructing confidence intervals, a specific Z-score is used to provide the margin of error for the estimate.
For our 90% confidence interval calculation, the Z-score is 1.645. This value is derived from the standard normal distribution and corresponds to a confidence level of 90%.
Essentially, this Z-score indicates how likely it is that the true mean falls within a certain range. Understanding Z-scores helps us estimate the probability of a statistic occurring within a normal distribution, which is a vital part of performing inferential statistics.
Population Mean
The population mean (\(\mu\)) is the average of a set of values for the whole population. It represents the central tendency of the overall data. In many cases, the population mean isn't directly measurable, especially in large populations.
Instead, we often use the sample mean (\(\bar{x}\)) to make inferences about the population mean. This is where confidence intervals come into play, using sample data to estimate the true population mean with a given level of certainty.
  • In the candy store franchise example, we calculated a sample mean of \(106.9\) thousand dollars, which we use to estimate the population mean.
  • Through our calculations, we determined that we are 90% confident that the actual average startup cost is between 90.779 and 123.021 thousand dollars.

Mathematically, confidence intervals help define how closely the sample mean approximates the population mean, providing an essential tool for making informed business and social science decisions.

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Most popular questions from this chapter

Large U.S. Companies: Foreign Revenue For large U.S. companies, what percentage of their total income comes from foreign sales? A random sample of technology companies (IBM, Hewlett-Packard, Intel, and others) gave the following information. $$ \begin{aligned} &\text { Technology companies, } \% \text { foreign revenue: } x_{1} ; n_{1}=16\\\ &\begin{array}{llllllll} 62.8 & 55.7 & 47.0 & 59.6 & 55.3 & 41.0 & 65.1 & 51.1 \\ 53.4 & 50.8 & 48.5 & 44.6 & 49.4 & 61.2 & 39.3 & 41.8 \end{array} \end{aligned} $$ Another independent random sample of basic consumer product companies (Goodyear, Sarah Lee, H.J. Heinz, Toys " \(q\) "Us) gave the following information. $$ \begin{aligned} &\text { Basic consumer product companies, } \% \text { foreign revenue: } x_{2} ; n_{2}=17\\\ &\begin{array}{llllllll} 28.0 & 30.5 & 34.2 & 50.3 & 11.1 & 28.8 & 40.0 & 44.9 \\ 40.7 & 60.1 & 23.1 & 21.3 & 42.8 & 18.0 & 36.9 & 28.0 \end{array}\\\ &\begin{aligned} &40.7 \\ &32.5 \end{aligned} \end{aligned} $$ (Reference: Forbes Top Companies.) Assume that the distributions of percentage foreign revenue are mound-shaped and symmetric for these two company types. (a) Use a calculator with mean and standard deviation keys to verify that \(\bar{x}_{1} \approx 51.66, s_{1} \approx 7.93, \bar{x}_{2} \approx 33.60\), and \(s_{2} \approx 12.26\). (b) Let \(\mu_{1}\) be the population mean for \(x_{1}\) and let \(\mu_{2}\) be the population mean for \(x_{2}\). Find an \(85 \%\) confidence interval for \(\mu_{1}-\mu_{2}\). (c) Interpretation Examine the confidence interval and explain what it means in the context of this problem. Does the interval consist of numbers that are all positive? all negative? of different signs? At the \(85 \%\) level of confidence, do technology companies have a greater percentage foreign revenue than basic consumer product companies? (d) Check Requirements Which distribution (standard normal or Student's \(t)\) did you use? Why?

Archaeology: Ireland Inorganic phosphorous is a naturally occurring element in all plants and animals, with concentrations increasing progressively up the food chain (fruit \(<\) vegetables \(<\) cereals \(<\) nuts \(<\) corpse). Geochemical surveys take soil samples to determine phosphorous content (in ppm, parts per million). A high phosphorous content may or may not indicate an ancient burial site, food storage site, or even a garbage dump. The Hill of Tara is a very important archaeological site in Ireland. It is by legend the seat of Ireland's ancient high kings (Reference: Tara, An Archaeological Survey by Conor Newman, Royal Irish Academy, Dublin). Independent random samples from two regions in Tara gave the following phosphorous measurements (in ppm). Assume the population distributions of phosphorous are mound-shaped and symmetric for these two regions. $$ \begin{aligned} &\text { Region I: } x_{1} ; n_{1}=12\\\ &\begin{array}{llllll} 540 & 810 & 790 & 790 & 340 & 800 \\ 890 & 860 & 820 & 640 & 970 & 720 \end{array} \end{aligned} $$ $$ \begin{aligned} &\text { Region II: } x_{2} ; n_{2}=16\\\ &\begin{array}{llllllll} 750 & 870 & 700 & 810 & 965 & 350 & 895 & 850 \\ 635 & 955 & 710 & 890 & 520 & 650 & 280 & 993 \end{array} \end{aligned} $$ (a) Use a calculator with mean and standard deviation keys to verify that \(\bar{x}_{1} \approx 747.5, s_{1} \approx 170.4, \bar{x}_{2} \approx 738.9\), and \(s_{2} \approx 212.1 .\) (b) Let \(\mu_{1}\) be the population mean for \(x_{1}\) and let \(\mu_{2}\) be the population mean for \(x_{2} .\) Find a \(90 \%\) confidence interval for \(\mu_{1}-\mu_{2}\). (c) Interpretation Explain what the confidence interval means in the context of this problem. Does the interval consist of numbers that are all positive? all negative? of different signs? At the \(90 \%\) level of confidence, is one region more interesting than the other from a geochemical perspective? (d) Check Requirements Which distribution (standard normal or Student's \(t\). did you use? Why?

Small Business: Bankruptcy The National Council of Small Businesses is interested in the proportion of small businesses that declared Chapter 11 bankruptcy last year. Since there are so many small businesses, the National Council intends to estimate the proportion from a random sample. Let \(p\) be the proportion of small businesses that declared Chapter 11 bankruptcy last year. (a) If no preliminary sample is taken to estimate \(p\), how large a sample is necessary to be \(95 \%\) sure that a point estimate \(\hat{p}\) will be within a distance of \(0.10\) from \(p\) ? (b) In a preliminary random sample of 38 small businesses, it was found that six had declared Chapter 11 bankruptcy. How many more small businesses should be included in the sample to be \(95 \%\) sure that a point estimate \(\hat{p}\) will be within a distance of \(0.10\) from \(p\) ?

Answer true or false. Explain your answer. For the same random sample, when the confidence level \(c\) is reduced, the confidence interval for \(\mu\) becomes shorter.

Assume that the population of \(x\) values has an approximately normal distribution. Archaeology: Tree Rings At Burnt Mesa Pueblo, the method of tree-ring dating gave the following years A.D. for an archaeological excavation site (Bandelier Archaeological Excavation Project: Summer 1990 Excavations at Burnt Mesa Pueblo, edited by Kohler, Washington State University): \(\begin{array}{lllllllll}1189 & 1271 & 1267 & 1272 & 1268 & 1316 & 1275 & 1317 & 1275\end{array}\) (a) Use a calculator with mean and standard deviation keys to verify that the sample mean year is \(\bar{x} \approx 1272\), with sample standard deviation \(s \approx 37\) years. (b) Find a \(90 \%\) confidence interval for the mean of all tree-ring dates from this archaeological site. (c) Interpretation What does the confidence interval mean in the context of this problem?

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