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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 1 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 employec 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
Profits below $30k are low, profits above $40k are high compared to typical successful banks.

Step by step solution

01

Calculate Sample Mean \( \bar{x} 

The problem statement confirms that the sample mean \( \bar{x} = 36.0 \). We use a calculator or software to verify this. Add all data values and divide by the number of data points.
02

Calculate Standard Error \( SE 

The standard error (SE) is calculated using the formula \( SE = \frac{\sigma}{\sqrt{n}} \), where \( \sigma = 10.2 \) is the standard deviation and \( n = 44 \) is the sample size. Therefore, \( SE = \frac{10.2}{\sqrt{44}} \approx 1.54 \).
03

Find Critical Value for 75% Confidence Interval

For a 75% confidence interval, the critical value \( z_{\alpha/2} \approx 1.15 \). This value corresponds to the central 75% of the standard normal distribution.
04

Compute 75% Confidence Interval \( CI 

The 75% confidence interval is calculated using \( \bar{x} \pm z_{\alpha/2} \times SE \). Substituting the known values gives \( 36.0 \pm 1.15 \times 1.54 \). This yields the interval \( (34.2, 37.8) \).
05

Interpretation for Profits Below $30,000

With a 75% confidence interval of \( (34.2, 37.8) \), any profit less than \$30k per employee is lower than typical successful banks. This suggests it is somewhat low, as it falls outside this confidence range.
06

Interpretation for Profits Above $40,000

Profits more than \$40k per employee exceed the upper limit of the 75% confidence interval \( (34.2, 37.8) \). Thus, as a manager, you might feel better, as it indicates higher-than-average profitability.
07

Find Critical Value for 90% Confidence Interval

For a 90% confidence interval, the critical value \( z_{\alpha/2} \approx 1.645 \). This value corresponds to the central 90% of the standard normal distribution.
08

Compute 90% Confidence Interval

The 90% confidence interval is calculated using \( \bar{x} \pm z_{\alpha/2} \times SE \). Substituting the values gives \( 36.0 \pm 1.645 \times 1.54 \), resulting in the interval \( (33.47, 38.53) \).
09

Interpretation for Profits Below $30,000 (90% CI)

With a 90% confidence interval of \( (33.47, 38.53) \), any profit less than \$30k remains outside this range, confirming it is low compared to other banks.
10

Interpretation for Profits Above $40,000 (90% CI)

Profits above \$40k per employee exceed the 90% confidence interval \( (33.47, 38.53) \), indicating a comfortable position for a bank manager, as it is significantly above the average.

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Key Concepts

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

Standard Deviation
Standard deviation is a measure that indicates the amount of variation or dispersion in a set of data values.

In simpler terms, it tells us how much the individual data points usually differ from the average (mean) of those points.

In the context of the problem, the standard deviation is given as \( \sigma = 10.2 \) thousand dollars. This means, on average, each company's annual profit per employee is expected to deviate about $10,200 from the mean.

Standard deviation is crucial because it provides a tangible sense of how spread out the data is, giving an insight into the variability of the profits.
Sample Mean
The sample mean, represented by \( \bar{x} \), is essentially the average of all the data points in the sample.

To find it, you add up all the values and then divide by the number of values. In this problem, the sample mean is calculated to be 36.0.

This number serves as a central tendency measure, indicating what is typical or expected in the data set.
  • Provides a baseline to compare individual data points.
  • Helps calculate other statistical measures such as the standard error.
  • Establishes a central reference point for constructing confidence intervals.
Understanding the sample mean helps contextualize the rest of the statistical analysis, providing insight into what might be a typical annual profit for employees within this sector.
Standard Error
The standard error (SE) is a statistical measure that describes how much the sample mean is expected to vary from the true population mean. It accounts for the variability of the sample means if you took multiple samples from the population.

In this problem, the standard error is calculated using \( SE = \frac{\sigma}{\sqrt{n}} \), where \( \sigma = 10.2 \) is the standard deviation, and \( n = 44 \) is the sample size. Plugging in the numbers, we get: \( SE \approx 1.54 \).

The smaller the standard error, the closer the sample mean is likely to represent the population mean. Thus, SE is important for constructing confidence intervals, providing a basis for estimating how close your sample mean is to the actual population mean.
Critical Value
The critical value is a factor used to calculate confidence intervals and depends on the desired level of confidence.

For a 75% confidence interval, a critical value from the standard normal distribution is chosen.

In this example, the critical value is approximately 1.15.
  • It represents the "cut-off" point that separates the central region from the tail ends of the distribution in a standard normal distribution.
  • It gets multiplied by the standard error to produce the margin of error for the interval.
  • Subsequent construction of confidence intervals helps us assess if future samples lie within that interval.
Critical values change based on the confidence level — higher confidence levels (like 90%) typically have higher critical values. By applying this value in the calculation, we build confidence intervals that provide a range in which the true population mean likely falls.

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Most popular questions from this chapter

Lorraine was in a hurry when she computed a confidence interval for \(\mu\). Because \(\sigma\) was not known, she used a Student's \(t\) distribution. However, she accidentally used degrees of freedom \(n\) instead of \(n-1 .\) Was her confidence interval longer or shorter than one found using the correct degrees of freedom \(n-1 ?\) Explain.

For a binomial experiment with \(r\) successes out of \(n\) trials, what value do we use as a point estimate for the probability of success \(p\) on a single trial?

A random sample of 328 medical doctors showed that 171 have 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 bricf 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?

A requirement for using the normal distribution to approximate the \(\hat{p}\) distribution is that both \(n p > 5\) and \(n q > 5\) since we usually do not know \(p,\) we estimate \(p\) by \(\hat{p}\) and \(q\) by \(\hat{q}=1-\hat{p}\) Then we require that \(n \hat{p}>5\) and \(n \hat{q}>5 .\) Show that the conditions \(n \hat{p}>5\) and \(n \hat{q}>5\) are equivalent to the condition that out of \(n\) binomial trials, both the number of successes \(r\) and the number of failures \(n-r\) must exceed 5 Hint: In the inequality \(n \hat{p}>5,\) replace \(\hat{p}\) by \(r / n\) and solve for \(r .\) In the inequality \(n \hat{q}>5,\) replace \(\hat{q}\) by \((n-r) / n\) and solve for \(n-r\).

Assume that the population of \(x\) values has an approximately normal distribution. Wildlife: Mountain Lions How much do wild mountain lions weigh? The 77 th Annual Report of the New Mexico Department of Game and Fish, edited by Bill Montoya, gave the following information. Adult wild mountain lions (18 months or older) captured and released for the first time in the San Andres Mountains gave the following weights (pounds): \(68 \quad 104\) \(128 \quad 122 \quad 60\) (a) Use a calculator with mean and sample standard deviation keys to verify that \(\bar{x}=91.0\) pounds and \(s \approx 30.7\) pounds. (b) Find a \(75 \%\) confidence interval for the population average weight \(\mu\) of all adult mountain lions in the specified region. (c) Interpretation What does the confidence interval mean in the context of this problem?

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