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Chapter 3: Problem 13

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 \(s\). (e) Suppose the given data comprise the entire population of all \(x\) values. Compute the population variance \(\sigma^{2}\) and population standard deviation \(\sigma\).

Short Answer

Expert verified
Range: 15; Sample Variance and Standard Deviation: 37 and 6.08; Population Variance and Standard Deviation: 29.6 and 5.44.

Step by step solution

01

Calculate the Range

The range is the difference between the maximum and minimum values of the data set. First, identify the maximum and minimum values in the data set: Max = 30 and Min = 15. Then, compute the range:\[\text{Range} = \text{Max} - \text{Min} = 30 - 15 = 15\]
02

Verification of 危x

Calculate the sum of all data points and verify if it equals to 110:\[\Sigma x = 23 + 17 + 15 + 30 + 25 = 110\]Thus, \(\Sigma x = 110\) is verified.
03

Verification of 危x虏

Calculate the sum of the squares of each data point and verify if it equals to 2568:\[\Sigma x^2 = 23^2 + 17^2 + 15^2 + 30^2 + 25^2 = 529 + 289 + 225 + 900 + 625 = 2568\]Thus, \(\Sigma x^2 = 2568\) is verified.
04

Compute Sample Variance s虏 Using Computation Formulas

Using the formula for sample variance, \[s^2 = \frac{\Sigma x^2 - (\Sigma x)^2/n}{n-1}\], where \(n\) is the sample size, which is 5:\[s^2 = \frac{2568 - (110)^2/5}{5-1} = \frac{2568 - 2420}{4} = \frac{148}{4} = 37\]
05

Compute Sample Standard Deviation s

The sample standard deviation \(s\) is the square root of the sample variance \(s^2\):\[s = \sqrt{37} \approx 6.08\]
06

Compute Sample Variance s虏 Using Defining Formulas

Using defining formulas, first calculate the mean \(\bar{x}\):\[\bar{x} = \frac{\Sigma x}{n} = \frac{110}{5} = 22\]Then use \[s^2 = \frac{1}{n-1} \Sigma (x_i - \bar{x})^2\]:\[s^2 = \frac{(23-22)^2 + (17-22)^2 + (15-22)^2 + (30-22)^2 + (25-22)^2}{4} = \frac{1 + 25 + 49 + 64 + 9}{4} = \frac{148}{4} = 37\]
07

Compute Sample Standard Deviation s Using Defining Formulas

Again, the sample standard deviation \(s\) is:\[s = \sqrt{37} \approx 6.08\]
08

Compute the Population Variance 蟽虏

The population variance \(\sigma^2\) uses \(n\) instead of \(n-1\):\[\sigma^2 = \frac{\Sigma x^2 - (\Sigma x)^2/n}{n} = \frac{2568 - (110)^2/5}{5} = \frac{2568 - 2420}{5} = \frac{148}{5} = 29.6\]
09

Compute the Population Standard Deviation 蟽

The population standard deviation \(\sigma\) is the square root of the population variance \(\sigma^2\):\[\sigma = \sqrt{29.6} \approx 5.44\]

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Key Concepts

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

Range
The range in a data set is a simple yet insightful statistic that tells us about the spread of the data points. It is calculated as the difference between the maximum and minimum values. By understanding the range, we can quickly gauge how spread out the data points are. In this exercise, the maximum value is 30 and the minimum value is 15. To find the range, subtract the minimum from the maximum: 30 - 15 = 15.
This tells us that the spread of the data points is 15 units. While range provides a quick sense of variability, it's worth noting it only considers the most extreme values, so it does not reflect the distribution of data within that range.
Sample Variance
Sample variance is a measure of the variability or dispersion of sample data. It indicates how much the individual data points differ from the sample mean. The formula for sample variance is: \[ s^2 = \frac{\Sigma x^2 - (\Sigma x)^2/n}{n-1} \] where \( n \) is the number of data points. For our exercise, the sample size \( n \) is 5.
First, we calculated \( \Sigma x = 110 \) and \( \Sigma x^2 = 2568 \). Plugging these into our formula:- Substitute: \( s^2 = \frac{2568 - (110)^2/5}{5-1} = \frac{148}{4} = 37 \).
This shows that the average squared deviation from the mean is 37, which gives us a good measure of the data's spread.
Population Variance
Population variance, denoted as \( \sigma^2 \), measures the dispersion of the entire data set or population. The formula for population variance is:\[ \sigma^2 = \frac{\Sigma x^2 - (\Sigma x)^2/n}{n} \] This formula is typically used when you are analyzing an entire population, not just a sample. In our exercise, the same data set is assumed to represent the whole population.
Performed calculations show:- Population variance: \( \sigma^2 = \frac{2568 - (110)^2/5}{5} = 29.6 \).
Thus, the population variance is slightly lower than the sample variance because we divide by \( n \) instead of \( n-1 \), reflecting the use of population data rather than sample data.
Standard Deviation
Standard deviation is a crucial statistical tool that measures the amount of variation or dispersion in a set of values. It is the square root of variance and provides insights into how spread out the values are from the mean.
  • Sample standard deviation formula: \( s = \sqrt{s^2} \).
  • Population standard deviation formula: \( \sigma = \sqrt{\sigma^2} \).
In this context:- Sample standard deviation: \( s = \sqrt{37} \approx 6.08 \).- Population standard deviation: \( \sigma = \sqrt{29.6} \approx 5.44 \).
Both these metrics give us a tangible understanding of dispersion, making it clear how tightly or loosely the data values cluster around the mean. Standard deviation is widely used because it is expressed in the same units as the data, unlike variance, which is expressed in squared units.

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Most popular questions from this chapter

When computing the standard deviation, does it matter whether the data are sample data or data comprising the entire population? Explain.

Pax World Balanced is a highly respected, socially responsible mutual fund of stocks and bonds (see Viewpoint). Vanguard Balanced Index is another highly regarded fund that represents the entire U.S. stock and bond market (an index fund). The mean and standard deviation of annualized percent returns are shown below. The annualized mean and standard deviation are based on the years 1993 through 2002 (Source: Morningstar). Pax World Balanced: \(\bar{x}=9.58 \% ; s=14.05 \%\) Vanguard Balanced Index: \(\bar{x}=9.02 \% ; s=12.50 \%\) (a) Compute the coefficient of variation for each fund. If \(\bar{x}\) represents return and \(s\) represents risk, then explain why the coefficient of variation can be taken to represent risk per unit of return. From this point of view, which fund appears to be better? Explain. (b) Compute a \(75 \%\) Chebyshev interval around the mean for each fund. Use the intervals to compare the two funds. As usual, past performance does not guarantee future performance.

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.

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 \(\underline{s}\). (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?

In some reports, the mean and coefficient of variation are given. For instance, in Statistical Abstract of the United States, \(116 \mathrm{th}\) Edition, one report gives the average number of physician visits by males per year. The average reported is \(2.2\), and the reported coefficient of variation is \(1.5 \%\). Use this information to determine the standard deviation of the annual number of visits to physicians made by males.
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