/*! 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 19 What are the big corporations do... [FREE SOLUTION] | 91Ó°ÊÓ

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What are the big corporations doing with their wealth? One way to answer this question is to examine profits as percentage of assets. A random sample of 50 Fortune 500 companies gave the following information. (Source: Based on information from Fortune 500 , Vol. 135 , No. 8.) \begin{tabular}{l|rrrrr} \hline Profit as percentage of assets & \(8.6-12.5\) & \(12.6-16.5\) & \(16.6-20.5\) & \(20.6-24.5\) & \(24.6-28.5\) \\ \hline Number of companies & 15 & 20 & 5 & 7 & 3 \\ \hline \end{tabular} Estimate the sample mean, sample variance, and sample standard deviation for profit as percentage of assets.

Short Answer

Expert verified
Sample mean = 15.59, sample variance = 23.467, sample standard deviation = 4.844.

Step by step solution

01

Identify the Midpoints

To estimate the sample mean, we first need the midpoints of each class interval. These are given by the average of the lower and upper bounds for each interval. Thus, the midpoints are calculated as:- For interval 8.6-12.5, midpoint = \( \frac{8.6 + 12.5}{2} = 10.55 \)- For interval 12.6-16.5, midpoint = \( \frac{12.6 + 16.5}{2} = 14.55 \)- For interval 16.6-20.5, midpoint = \( \frac{16.6 + 20.5}{2} = 18.55 \)- For interval 20.6-24.5, midpoint = \( \frac{20.6 + 24.5}{2} = 22.55 \)- For interval 24.6-28.5, midpoint = \( \frac{24.6 + 28.5}{2} = 26.55 \)
02

Compute the Sample Mean

The sample mean profit is calculated using the formula:\[ \bar{x} = \frac{\sum (f_i \cdot x_i)}{n} \]where \(f_i\) are the frequencies and \(x_i\) are the midpoints. Calculating each product:- \(15 \times 10.55 = 158.25\)- \(20 \times 14.55 = 291.00\)- \(5 \times 18.55 = 92.75\)- \(7 \times 22.55 = 157.85\)- \(3 \times 26.55 = 79.65\)Summing these gives \(779.5\) and since \(n = 50\), the mean is:\[ \bar{x} = \frac{779.5}{50} = 15.59\]
03

Calculate the Sample Variance

The sample variance is given by:\[ s^2 = \frac{\sum f_i (x_i - \bar{x})^2}{n-1} \]Firstly, calculate \((x_i - \bar{x})^2 \):- \((10.55 - 15.59)^2 = 25.6001\)- \((14.55 - 15.59)^2 = 1.081\)- \((18.55 - 15.59)^2 = 8.822\)- \((22.55 - 15.59)^2 = 48.4321\)- \((26.55 - 15.59)^2 = 120.4321\)Then calculate \(f_i (x_i - \bar{x})^2\) and sum:- \(15 \times 25.6001 = 384.0015\)- \(20 \times 1.081 = 21.62\)- \(5 \times 8.822 = 44.11\)- \(7 \times 48.4321 = 338.4247\)- \(3 \times 120.4321 = 361.2963\)Summing yields \(1149.4525\), and so:\[ s^2 = \frac{1149.4525}{49} = 23.467\]
04

Compute the Sample Standard Deviation

The sample standard deviation \(s\) is the square root of the variance:\[ s = \sqrt{s^2} = \sqrt{23.467} = 4.844\]

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Key Concepts

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

Sample Mean
The sample mean is a measure of the central tendency of a dataset. It provides an average value, offering insight into where the data roughly clusters. To compute the sample mean, especially with frequency distribution, follow these steps:

First, determine the midpoints for each range in your data. This is done by taking the average of the upper and lower interval bounds. For example, the midpoint for the interval 8.6-12.5 is \( \frac{8.6 + 12.5}{2} = 10.55 \). Repeat for all intervals: 12.6-16.5, 16.6-20.5, 20.6-24.5, and 24.6-28.5.

Next, multiply each midpoint by its respective frequency (number of companies falling in that range). Sum these products together to get a total weighted value.

Finally, divide this total by the total number of companies in your sample (in this case, 50) to find the sample mean. This shows the average profit as a percentage of assets for the companies in the sample.
Sample Variance
Sample variance helps measure the dispersion or spread of data points around the mean. High variance suggests a wide spread, while low variance indicates that the data points tend to be close to the mean.

To calculate it, start by finding the difference between each midpoint and the sample mean. Square this difference to remove negative values, which results in \((x_i - \bar{x})^2\) for each midpoint. This step ensures that all deviations contribute positively to the variance.

Then, multiply these squared differences by their respective frequencies to get the sum of squared deviations for each class interval.
  • For instance: for a class 10.55, deviation is \((10.55 - 15.59)^2 = 25.6001\) and for frequency 15, it becomes \(15 \times 25.6001 = 384.0015\).
Sum all these values and divide by \(n-1\) (where \(n\) is the sample size) to adjust for sample size, yielding the sample variance.
Frequency Distribution
Frequency distribution shows how data points are distributed across different intervals. It allows you to easily see how often different values or ranges of values occur in a dataset.

In the given exercise, data is grouped into five intervals (or "bins") representing the profit as a percentage of assets across 50 companies. Each interval is accompanied by a frequency, showing how many companies fall within that profit range. For instance:
  • The interval 8.6-12.5 has a frequency of 15.
  • The interval 12.6-16.5 has a frequency of 20, the highest in this case.
  • The lower intervals, such as 24.6-28.5, have fewer frequencies.
This distribution helps identify patterns in the data, such as which range is more common or less common among the companies. It's a crucial step before calculating measures like mean and variance, as it provides detailed insights into how data is grouped.
Standard Deviation
Standard deviation gives a clear picture of how spread out the values in a data set are. It is the square root of the variance, making it an average distance from the mean.

Since variance is in squared units, standard deviation brings it back to the original unit, making it easier to understand and interpret. For instance, if our variance within the context of companies' profit is \(23.467\), then the standard deviation will be \(\sqrt{23.467} \approx 4.844\).

The standard deviation tells us that, on average, the profit percentages deviate about 4.844% from the mean. This information can be essential for understanding the impression or variability of profits in these Fortune 500 companies. It helps companies assess risk and make more informed financial decisions.

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Most popular questions from this chapter

Given the sample data \(\begin{array}{llllll}x: & 23 & 17 & 15 & 30 & 25\end{array}\) (a) Find the range. (b) Verify that \(\Sigma x=110\) and \(\Sigma x^{2}=2568\). (c) Use the results of part (b) and appropriate computation formulas to compute the sample variance \(s^{2}\) and sample standard deviation \(s\). (d) Use the defining formulas to compute the sample variance \(s^{2}\) and sample standard deviation \(s\). (c) Suppose the given data comprise the entire population of all \(x\) values. Compute the population variance \(\sigma^{2}\) and population standard deviation \(\sigma\).

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Clayton and Timothy took different sections of Introduction to Economics. Each section had a different final exam. Timothy scored 83 out of 100 and had a percentile rank in his class of \(72 .\) Clayton scored 85 out of 100 but his percentile rank in his class was 70 . Who performed better with respect to the rest of the students in the class, Clayton or Timothy? Explain your answer.

Critical Thinking: Data Transformation In this problem, we explore the effect on the standard deviation of adding the same constant to each data value in a data set. Consider the data set \(5,9,10,11,15\). (a) Use the defining formula, the computation formula, or a calculator to compute \(s\). (b) Add 5 to each data value to get the new data set \(10,14,15,16,20 .\) Compute \(s\). (c) Compare the results of parts (a) and (b). In general, how do you think the standard deviation of a data set changes if the same constant is added to each data value?

Kevlar epoxy is a material used on the NASA Space Shuttle. Strands of this epoxy were tested at the \(90 \%\) breaking strength. The following data represent time to failure (in hours) for a random sample of 50 epoxy strands (Reference: \(\mathrm{R} . \mathrm{E} .\) Barlow, University of California, Berkeley). Let \(x\) be a random variable representing time to failure (in hours) at \(90 \%\) breaking strength. Note: These data are also available for download online in HM StatSPACETM. \(\begin{array}{llllllllll}0.54 & 1.80 & 1.52 & 2.05 & 1.03 & 1.18 & 0.80 & 1.33 & 1.29 & 1.11 \\ 3.34 & 1.54 & 0.08 & 0.12 & 0.60 & 0.72 & 0.92 & 1.05 & 1.43 & 3.03 \\ 1.81 & 2.17 & 0.63 & 0.56 & 0.03 & 0.09 & 0.18 & 0.34 & 1.51 & 1.45 \\ 1.52 & 0.19 & 1.55 & 0.02 & 0.07 & 0.65 & 0.40 & 0.24 & 1.51 & 1.45 \\ 1.60 & 1.80 & 4.69 & 0.08 & 7.89 & 1.58 & 1.64 & 0.03 & 0.23 & 0.72\end{array}\) (a) Find the range. (b) Use a calculator to verify that \(\Sigma x=62.11\) and \(\Sigma x^{2}=164.23\). (c) Use the results of part (b) to compute the sample mean, variance, and standard deviation for the time to failure. (d) Use the results of part (c) to compute the coefficient of variation. What does this number say about time to failure? Why does a small CV indicate more consistent data, whereas a larger CV indicates less consistent data? Explain.

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