/*! 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 24 Jobs and productivity! How do re... [FREE SOLUTION] | 91Ó°ÊÓ

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Jobs and productivity! How do retail stores rate? One way to answer this question is to examine annual profits per employee. The following data give annual profits per employee (in units of 1 thousand dollars per employee) for companies in retail sales. (See reference in Problem \(23 .\) ) Companies such as Gap, Nordstrom, Dillards, JCPenney, Sears, Wal-Mart, Office Depot, and Toys " \(\mathrm{A}\) " Us are included. Assume \(\sigma \approx 3.8\) thousand dollars. $$\begin{array}{rrrrrrrrrrrr}4.4 & 6.5 & 4.2 & 8.9 & 8.7 & 8.1 & 6.1 & 6.0 & 2.6 & 2.9 & 8.1 & -1 . \\\11.9 & 8.2 & 6.4 & 4.7 & 5.5 & 4.8 & 3.0 & 4.3 & -6.0 & 1.5 & 2.9 & 4 . \\ -1.7 & 9.4 & 5.5 & 5.8 & 4.7 & 6.2 & 15.0 & 4.1 & 3.7 & 5.1 & 4.2 &\end{array}$$ (a) Use a calculator or appropriate computer software to verify that, for the preceding data, \(\bar{x} \approx 5.1 .\) (b) Let us say that the preceding data are representative of the entire sector of retail sales companies. Find an \(80 \%\) confidence interval for \(\mu,\) the average annual profit per employee for retail sales. (c) Interpretation Let us say that you are the manager of a retail store with a large number of employees. Suppose the annual profits per employee are less than 3 thousand dollars per employee. Do you think this might be low compared with other retail stores? Explain by referring to the confidence interval you computed in part (b). (d) Interpretation Suppose the annual profits are more than 6.5 thousand dollars per employee. As store manager, 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 \(95 \%\) confidence interval.

Short Answer

Expert verified
(b) 80% CI: [4.29, 5.91]; (c) Yes, 3 is low; (d) Yes, feel better; (e) 95% CI: [3.86, 6.34], same conclusions as before.

Step by step solution

01

Calculate the Sample Mean

Given the data set and provided values, we have the sample mean \( \bar{x} \approx 5.1 \). This was verified using a calculator or software.
02

Determine the Z-Score for 80% Confidence Interval

For an 80% confidence interval, find the z-score that corresponds to the tail area of 10% on each side of the normal distribution. From standard tables, \( z \approx 1.28 \).
03

Calculate the 80% Confidence Interval

Use the formula for the confidence interval: \[ \bar{x} \pm z \left( \frac{\sigma}{\sqrt{n}} \right) \] where \( \sigma = 3.8 \) and \( n = 36 \). Calculate:\[ 5.1 \pm 1.28 \left( \frac{3.8}{\sqrt{36}} \right) \] \[ 5.1 \pm 0.81 \] This gives the interval \([4.29, 5.91]\).
04

Interpretation for Part (c)

Comparing 3 thousand dollars with the confidence interval \([4.29, 5.91]\), it falls outside and below this range. Thus, 3 thousand is considered low compared to other retail stores.
05

Interpretation for Part (d)

If the profits are more than 6.5 thousand, it exceeds the upper bound of the interval \([4.29, 5.91]\). Thus, it is higher than average, and as a manager, you would feel better compared to the average store.
06

Determine the Z-Score for 95% Confidence Interval

For a 95% confidence interval, find the z-score that corresponds to the tail area of 2.5% on each side. From the z-table, \( z \approx 1.96 \).
07

Calculate the 95% Confidence Interval

Use the formula again: \[ \bar{x} \pm z \left( \frac{\sigma}{\sqrt{n}} \right) \] Calculate:\[ 5.1 \pm 1.96 \left( \frac{3.8}{\sqrt{36}} \right) \] \[ 5.1 \pm 1.24 \] This provides the interval \([3.86, 6.34]\).
08

Interpretation for Part (e) - 80% Interval Comparisons

Using the 95% confidence interval \([3.86, 6.34]\), 3 thousand is outside and below this range, still indicating it is low. On the other hand, profits over 6.5 thousand would now exceed this interval's upper limit, indicating a higher-than-average profit.

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Key Concepts

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

Retail Sales
Retail sales refer to the sale of goods and services to consumers through multiple channels of distribution. It includes everything from clothing, electronics, groceries, and more, to services like beauty salons and car repairs. Retail sales are a significant indicator of the economic health of a country.

This is because consumer spending drives a large portion of the economy. Analyzing retail sales data, like profits per employee, helps understand the productivity and efficiency of retail companies. Profits per employee are an important metric in this context, as it reflects how well a company converts its workforce efforts into profits.
Sample Mean
The sample mean is a crucial statistic representing the average of a data set. It is calculated by summing all observations in the sample and dividing by the number of observations. In our exercise, the sample mean was calculated as \( \bar{x} \approx 5.1 \).

This tells us that, on average, the companies in the data set make approximately \( 5.1 \) thousand dollars in profit per employee.
  • The sample mean provides a central point that summarizes the data set.
  • It helps make inferences about the larger population from which the sample is drawn.

In statistical analysis, the sample mean is often used as an estimator for the population mean, which is the average of the entire population.
Z-Score
A z-score is a measure of how many standard deviations an element is from the mean. It is crucial in statistics because it allows for the comparison between different data points or to determine the probability of a certain value occurring. For confidence intervals, z-scores correspond to how confident we are about the range of values where the true mean lies.

In the exercise, the z-score is determined by the confidence level. For an 80% confidence interval, the z-score is found to be approximately \(1.28\). This z-score allows us to calculate the range of values (confidence interval) that likely contains the true mean of the population based on our sample data. The higher the confidence level desired, the larger the z-score and hence the wider the confidence interval.
Standard Deviation
Standard deviation is a statistic that measures the dispersion or spread of a data set. It represents how much variation there is from the mean. In this exercise, the standard deviation is known to be approximately \(3.8\) thousand dollars.

This information helps us understand how spread out the profits per employee are for different companies. A smaller standard deviation would imply that most company profits per employee are around the mean. Conversely, a larger standard deviation indicates a wider range of profit values.

Standard deviation is a key component in calculating the confidence interval, as shown in the formula: \[ \bar{x} \pm z \left( \frac{\sigma}{\sqrt{n}} \right) \]. In this context, it helps adjust the range of values in the confidence interval to reflect the uncertainty and variability present in the data.

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Most popular questions from this chapter

S. C. Jett is a professor of geography at the University of California, Davis. He and a colleague, V. E. Spencer, are experts on modern Navajo culture and geography. The following information is taken from their book Navajo Architecture: Forms, History, Distributions (University of Arizona Press). On the Navajo Reservation, a random sample of 210 permanent dwellings in the Fort Defiance region showed that 65 were traditional Navajo hogans. In the Indian Wells region, a random sample of 152 permanent dwellings showed that 18 were traditional hogans. Let \(p_{1}\) be the population proportion of all traditional hogans in the Fort Defiance region, and let \(p_{2}\) be the population proportion of all traditional hogans in the Indian Wells region. (a) Can a normal distribution be used to approximate the \(\hat{p}_{1}-\hat{p}_{2}\) distribution? Explain. (b) Find a \(99 \%\) confidence interval for \(p_{1}-p_{2}\) (c) Examine the confidence interval and comment on its meaning. Does it include numbers that are all positive? all negative? mixed? What if it is hypothesized that Navajo who follow the traditional culture of their people tend to occupy hogans? Comment on the confidence interval for \(p_{1}-p_{2}\) in this context.

A random sample is drawn from a population with \(\sigma=12 .\) The sample mean is 30 (a) Compute a \(95 \%\) confidence interval for \(\mu\) based on a sample of size 49 What is the value of the margin of error? (b) Compute a \(95 \%\) confidence interval for \(\mu\) based on a sample of size \(100 .\) What is the value of the margin of error? (c) Compute a \(95 \%\) confidence interval for \(\mu\) based on a sample of size \(225 .\) What is the value of the margin of error? (d) Compare the margins of error for parts (a) through (c). As the sample size increases, does the margin of error decrease? (e) Critical Thinking Compare the lengths of the confidence intervals for parts (a) through (c). As the sample size increases, does the length of a \(90 \%\) confidence interval decrease?

In a marketing survey, a random sample of 1001 supermarket shoppers revealed that 273 always stock up on an item when they find that item at a real bargain price. (See reference in Problem 19.) (a) Let \(p\) represent the proportion of all supermarket shoppers who always stock up on an item when they find a real bargain. 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 on the percentage of supermarket shoppers who stock up on real-bargain items? What is the margin of error based on a \(95 \%\) confidence interval?

You want to conduct a survey to determine the proportion of people who favor a proposed tax policy. How does increasing the sample size affect the size of the margin of error?

Independent random samples of professional football and basketball players gave the following information (References: Sports Encyclopedia of Pro Football and Official NBA Basketball Encyclopedia ). Note: These data are also available for download at the Companion Sites for this text. Assume that the weight distributions are mound-shaped and symmetric. (a) Use a calculator with mean and standard deviation keys to verify that \(\bar{x}_{1}=259.6, s_{1}=12.1, \bar{x}_{2}=205.8,\) and \(s_{2}=12.9\) (b) Let \(\mu_{1}\) be the population mean for \(x_{1}\) and let \(\mu_{2}\) be the population mean for \(x_{2} .\) Find a \(99 \%\) confidence interval for \(\mu_{1}-\mu_{2}\) (c) 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 \(99 \%\) level of confidence, do professional football players tend to have a higher population mean weight than professional basketball players? (d) Which distribution (standard normal or Student's \(t\) ) did you use? Why?

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