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BBUNDERSTAT12 7.1.023

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One way to examine how banks rate is to look at annual profits per employee. The following data represents annual profits per employee (in units of 1 thousand dollars per employee) for various companies in financial services. Assume [tex]$\sigma=9.8$[/tex] thousand dollars.

\begin{tabular}{lllllllllll}
46.0 & 50.9 & 33.4 & 45.3 & 30.6 & 33.4 & 32.0 & 34.8 & 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{tabular}

(a) Use a calculator or appropriate computer software to find [tex]$\bar{x}$[/tex] for the preceding data. (Round your answer to two decimal places.)
[tex]$\square$[/tex] thousand dollars

(b) Let us say that the preceding data are representative of the entire sector of (successful) financial services corporations. Find a [tex]$75 \%$[/tex] confidence interval for [tex]$\mu$[/tex], the average annual profit per employee for all successful banks. (Round your answers to two decimal places.)
Lower limit: [tex]$\square$[/tex] thousand dollars
Upper limit: [tex]$\square$[/tex] thousand dollars

(c) 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...



Answer :

GinnyAnswer
Sure, let's break down the solution to this problem step-by-step.

### Part (a): Finding [tex]\(\bar{x}\)[/tex]

First, let's calculate the mean of the data set. The data points in units of thousands of dollars per employee are:

[tex]\[ 46.0, 50.9, 33.4, 45.3, 30.6, 33.4, 32.0, 34.8, 42.5, 33.0, 33.6, 36.9, 27.0, 47.1, 33.8, 28.1, 28.5, 29.1, 36.5, 36.1, 26.9, 27.8, 28.8, 29.3, 31.5, 31.7, 31.1, 38.0, 32.0, 31.7, 32.9, 23.1, 54.9, 43.8, 36.9, 31.9, 25.5, 23.2, 29.8, 22.3, 26.5, 26.7 \][/tex]

When you find the average of these values, it results in:
[tex]\[ \bar{x} = 33.45 \, \text{thousand dollars} \][/tex]
This is the mean annual profit per employee.

### Part (b): 75% Confidence Interval for [tex]\(\mu\)[/tex]

To find the 75% confidence interval for the mean, we follow these steps:

1. Mean ([tex]\(\bar{x}\)[/tex]): From part (a), we know [tex]\(\bar{x} = 33.45\)[/tex] thousand dollars.
2. Standard deviation ([tex]\(\sigma\)[/tex]): Given as 9.8 thousand dollars.
3. Sample size (n): The total number of data points is 41.

To find the standard error ([tex]\(SE\)[/tex]), we use the formula:
[tex]\[ SE = \frac{\sigma}{\sqrt{n}} \][/tex]

Now, we need the z-value corresponding to a 75% confidence level. For a 75% confidence interval, we find:
[tex]\[ z = 1.15 \][/tex] (approximately, since we're using the standard normal distribution)

Next, calculate the margin of error:
[tex]\[ \text{Margin of Error} = z \times SE \][/tex]

Finally, calculate the lower and upper bounds of the confidence interval:
[tex]\[ \text{Lower Limit} = \bar{x} - \text{Margin of Error} \][/tex]
[tex]\[ \text{Upper Limit} = \bar{x} + \text{Margin of Error} \][/tex]

Plugging in the values, the confidence interval is:
[tex]\[ \text{Lower Limit} = 31.71 \, \text{thousand dollars} \][/tex]
[tex]\[ \text{Upper Limit} = 35.19 \, \text{thousand dollars} \][/tex]

### Summary

1. Mean [tex]\(\bar{x}\)[/tex]:
[tex]\[ \bar{x} = 33.45 \, \text{thousand dollars} \][/tex]

2. 75% Confidence Interval:
- Lower Limit:
[tex]\[ 31.71 \, \text{thousand dollars} \][/tex]
- Upper Limit:
[tex]\[ 35.19 \, \text{thousand dollars} \][/tex]

I hope this step-by-step explanation clarifies how we arrived at the mean and the 75% confidence interval for the data!

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