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Profits: Banks 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.$$ \begin{array}{lllllllllll} 42.9 & 43.8 & 48.2 & 60.6 & 54.9 & 55.1 & 52.9 & 54.9 & 42.5 & 33.0 & 33.6 \\ 36.9 & 27.0 & 47.1 & 33.8 & 28.1 & 28.5 & 29.1 & 36.5 & 36.1 & 26.9 & 27.8 \\ 28.8 & 29.3 & 31.5 & 31.7 & 31.1 & 38.0 & 32.0 & 31.7 & 32.9 & 23.1 & 54.9 \\ 43.8 & 36.9 & 31.9 & 25.5 & 23.2 & 29.8 & 22.3 & 26.5 & 26.7 & & \end{array} $$ (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) Interpretation 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) Interpretation 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.

Short Answer

Expert verified
For a 75% confidence interval, employee profits should ideally be between $34.1K and $37.9K; $30K is low, $40K is high. For a 90% interval, $33.4K to $38.6K are the bounds, providing similar conclusions.

Step by step solution

01

Verify the Mean

Using a calculator or software, compute the mean of the given data set. Test to ensure \( \bar{x} \approx 36.0 \) by summing all data points and dividing by the number of data points. Confirm \( \bar{x} = 36.0 \).
02

Calculate 75% Confidence Interval

To find a 75% confidence interval for the given data, first determine the z-score for a 75% confidence level (z = 1.150). Use the formula: \[ \bar{x} \pm z \left( \frac{\sigma}{\sqrt{n}} \right) \]. Plug in \( \bar{x} = 36.0 \), \( \sigma = 10.2 \), and \( n = 40 \), resulting in an interval of approximately 34.1 to 37.9.
03

Interpret Results for Income Below 30

If your bank's annual profits per employee are less than $30,000, this is below the lower bound of the 75% confidence interval (34.1). Thus, your profits are indeed low relative to the average of successful institutions.
04

Interpret Results for Income Above 40

If your bank's profits are over $40,000, this is above the upper bound of the 75% confidence interval (37.9). Therefore, you can feel confident that your bank is performing well compared to the average successful financial institutions.
05

Calculate 90% Confidence Interval

For a 90% confidence interval, calculate the standard z-score (z = 1.645). Use the formula: \[ \bar{x} \pm z \left( \frac{\sigma}{\sqrt{n}} \right) \]. With the same values, calculate the interval, approximately 33.4 to 38.6.
06

New Interpretation for Income Below 30

With a 90% confidence interval of 33.4 to 38.6, having profits below $30,000 is again below the lower bound, confirming that your bank's employee earnings are low.
07

New Interpretation for Income Above 40

With the 90% confidence interval, profits above $40,000 remain higher than the upper bound of 38.6, suggesting your bank is performing well.

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Key Concepts

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

Mean Calculation
Calculating the mean is a fundamental step in understanding a data set and is especially important in statistics education. For the given exercise on financial institutions, the mean, or average, represents the central value of annual profits per employee for several banks.
In simple terms, to calculate the mean, add up all the numbers in the data set and then divide by how many numbers you have. This gives you the typical value each data point represents. For our data, \[ \bar{x} = \frac{\sum_{i=1}^{n} x_i}{n} \]where \( n \) is the number of companies in the data set and \( x_i \) is each individual profit value. By following this calculation, you get an approximate mean of 36 thousand dollars per employee, a vital step to compare individual performance with typical benchmarks.
Z-Score
The z-score is a statistical measurement that describes the number of standard deviations a data point is from the mean. In the context of confidence intervals, it helps in determining how spread out values are around the mean.
For the financial institution data, the z-score aids in calculating the confidence intervals. For instance, in step two of the solution, a 75% confidence interval utilizes a z-score of 1.150. This is because the z-score determines how far the values can deviate while still maintaining a specific confidence level.
  • A positive z-score indicates a value above the mean.
  • A negative z-score shows a value below the mean.
This method provides a powerful tool in comparing given data in the context of broader trends, enabling financial analysts and students to make informed decisions.
Financial Institutions
Financial institutions such as banks play a key role in evaluating economic health through metrics like annual profits per employee. Such metrics shed light on productivity and efficiency.
This exercise highlights the significance of comparing individual bank performance to industry standards. A company with profits per employee significantly lower than the mean may require strategy analysis for improvement. On the other hand, companies exceeding industry averages can serve as leaders and benchmarks.
By statistically analyzing profit data, banks and other financial services can develop strategies to maintain or improve their competitive edge. This exercise underscores how quantitative measures can guide managerial decisions within financial institutions, proving essential for strategic planning.
Statistics Education
Understanding and applying statistical concepts are critical components of statistics education, especially in fields like financial management. Learning how to calculate means, z-scores, and confidence intervals paves the way for more insightful data analysis.
Students are trained to interpret data effectively, which is crucial in real-world applications, such as evaluating financial institutions' performance. By comparing means and establishing confidence intervals, students learn to identify trends, opportunities, and potential challenges.
  • Grasping these concepts provides foundational knowledge for more advanced statistical methods.
  • Practicing with real-world data, like bank profits, helps bridge theoretical learning and practical application.
This way, statistics education equips students with the tools necessary for informed decision-making in diverse professional environments.

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Most popular questions from this chapter

Fishing: Barbless Hooks In a combined study of northern pike, cutthroat trout, rainbow trout, and lake trout, it was found that 26 out of 855 fish died when caught and released using barbless hooks on flies or lures. All hooks were removed from the fish (Source: A National Symposium on Catch and Release Fishing, Humboldt State University Press). (a) Let \(p\) represent the proportion of all pike and trout that die (i.e., \(p\) is the mortality rate) when caught and released using barbless hooks. Find a point estimate for \(p\). (b) Find a \(99 \%\) confidence interval for \(p\), and give a brief explanation of the meaning of the interval. (c) Check Requirements Is the normal approximation to the binomial justified in this problem? Explain.

Marketing: Customer Loyalty In a marketing survey, a random sample of 730 women shoppers revealed that 628 remained loyal to their favorite supermarket during the past year (i.e., did not switch stores) (Source: Trends in the United States: Consumer Attitudes and the Supermarket, The Research Department, Food Marketing Institute). (a) Let \(p\) represent the proportion of all women shoppers who remain loyal to their favorite supermarket. 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) Interpretation As a news writer, how would you report the survey results regarding the percentage of women supermarket shoppers who remained loyal to their favorite supermarket during the past year? What is the margin of error based on a \(95 \%\) confidence interval?

Expand Your Knowledge: Sample Size, Difference of Proportions What about the sample size \(n\) for confidence intervals for the difference of proportions \(p_{1}-p_{2}\) ? Let us make the following assumptions: equal sample sizes \(n=n_{1}=n_{2}\) and all four quantities \(n_{1} \hat{p}_{1}, n_{1} \hat{q}_{1}, n_{2} \hat{p}_{2}\), and \(n_{2} \hat{q}_{2}\) are greater than \(5 .\) Those readers familiar with algebra can use the procedure outlined in Problem 28 to show that if we have preliminary estimates \(\hat{p}_{1}\) and \(\hat{p}_{2}\) and a given maximal margin of error \(E\) for a specified confidence level \(c\), then the sample size \(n\) should be at least $$ n=\left(\frac{z_{c}}{E}\right)^{2}\left(\hat{p}_{1} \hat{q}_{1}+\hat{p}_{2} \hat{q}_{2}\right) $$ However, if we have no preliminary estimates for \(\hat{p}_{1}\) and \(\hat{p}_{2}\), then the theory similar to that used in this section tells us that the sample size \(n\) should be at least $$ n=\frac{1}{2}\left(\frac{z_{c}}{E}\right)^{2} $$ (a) In Problem 17 (Myers-Briggs personality type indicators in common for married couples), suppose we want to be \(99 \%\) confident that our estimate \(\hat{p}_{1}-\hat{p}_{2}\) for the difference \(p_{1}-p_{2}\) has a maximal margin of error \(E=0.04\). Use the preliminary estimates \(\hat{p}_{1}=289 / 375\) for the proportion of couples sharing two personality traits and \(\hat{p}_{2}=23 / 571\) for the proportion having no traits in common. How large should the sample size be (assuming equal sample size-i.e., \(n=n_{1}=n_{2}\) )? (b) Suppose that in Problem 17 we have no preliminary estimates for \(\hat{p}_{1}\) and \(\hat{p}_{2}\) and we want to be \(95 \%\) confident that our estimate \(\hat{p}_{1}-\hat{p}_{2}\) for the difference \(p_{1}-p_{2}\) has a maximal margin of error \(E=0.05 .\) How large should the sample size be (assuming equal sample size-i.e., \(n=n_{1}=n_{2}\) )?

Answer true or false. Explain your answer. The point estimate for the population mean \(\mu\) of an \(x\) distribution is \(\bar{x}\), computed from a random sample of the \(x\) distribution.

Campus Life: Coeds What percentage of your campus student body is female? Let \(p\) be the proportion of women students on your campus. (a) If no preliminary study is made to estimate \(p\), how large a sample is needed to be \(99 \%\) sure that a point estimate \(\hat{p}\) will be within a distance of \(0.05\) from \(p ?\) (b) The Statistical Abstract of the United States, 112 th Edition, indicates that approximately \(54 \%\) of college students are female. Answer part (a) using this estimate for \(p\).

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