/*! 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

Consider a \(90 \%\) confidence interval for \(\mu\). Assume \(\sigma\) is not known. For which sample size, \(n=10\) or \(n=20\), is the confidence interval longer?

A random sample of 328 medical doctors showed that 171 had a solo practice. (Source: Practice Patterns of General Internal Medicine, American Medical Association.) (a) Let \(p\) represent the proportion of all medical doctors who have a solo practice. Find a point estimate for \(p\). (b) Find a \(95 \%\) confidence interval for \(p .\) Give a brief explanation of the meaning of the interval. (c) As a news writer, how would you report the survey results regarding the percentage of medical doctors in solo practice? What is the margin of error based on a \(95 \%\) confidence interval?

If a \(90 \%\) confidence interval for the difference of proportions contains some positive and some negative values, what can we conclude about the relationship between \(p_{1}\) and \(p_{2}\) at the \(90 \%\) confidence level?

In the Focus Problem at the beginning of this chapter, a study was described comparing the hatch ratios of wood duck nesting boxes. Group I nesting boxes were well separated from each other and well hidden by available brush. There were a total of 474 eggs in group I boxes, of which a field count showed about 270 hatched. Group II nesting boxes were placed in highly visible locations and grouped closely together. There were a total of 805 eggs in group II boxes, of which a field count showed about 270 hatched. (a) Find a point estimate \(\hat{p}_{1}\) for \(p_{1}\), the proportion of eggs that hatch in group I nest box placements. Find a \(95 \%\) confidence interval for \(p_{1}\). (b) Find a point estimate \(\hat{p}_{2}\) for \(p_{2}\), the proportion of eggs that hatch in group II nest box placements. Find a \(95 \%\) confidence interval for \(p_{2}\). (c) Find a \(95 \%\) confidence interval for \(p_{1}-p_{2} .\) Does the interval indicate that the proportion of eggs hatched from group I nest boxes is higher than, lower than, or equal to the proportion of eggs hatched from group II nest boxes? (d) What conclusions about placement of nest boxes can be drawn? In the article discussed in the Focus Problem, additional concerns are raised about the higher cost of placing and maintaining group I nest box placements. Also at issue is the cost efficiency per successful wood duck hatch.

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.

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