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Chapter 3: Problem 10

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?

Short Answer

Expert verified
Adding a constant to each data value doesn't change the standard deviation.

Step by step solution

01

Compute the Mean of the Original Data Set

Find the mean of the data set \(5, 9, 10, 11, 15\). The mean is the sum of all data points divided by the number of data points. Thus, \( \bar{x} = \frac{5 + 9 + 10 + 11 + 15}{5} = \frac{50}{5} = 10 \).
02

Compute the Variance of the Original Data Set

The variance is calculated by finding the average of the squared differences from the mean. \[\text{Variance } (s^2) = \frac{(5-10)^2 + (9-10)^2 + (10-10)^2 + (11-10)^2 + (15-10)^2}{5-1}\]\[= \frac{25 + 1 + 0 + 1 + 25}{4} = \frac{52}{4} = 13.\]
03

Compute the Standard Deviation of the Original Data Set

The standard deviation is the square root of the variance. Thus, \( s = \sqrt{13} \approx 3.61 \).
04

Compute the Mean of the New Data Set

After adding 5 to each data value, the new data set is \(10, 14, 15, 16, 20\). Calculate the mean: \[\bar{x}_{\text{new}} = \frac{10 + 14 + 15 + 16 + 20}{5} = \frac{75}{5} = 15.\]
05

Compute the Variance for the New Data Set

Calculate the variance for the new data set: \[\text{Variance}_{\text{new}} (s^2_{\text{new}}) = \frac{(10-15)^2 + (14-15)^2 + (15-15)^2 + (16-15)^2 + (20-15)^2}{5-1}\]\[= \frac{25 + 1 + 0 + 1 + 25}{4} = \frac{52}{4} = 13.\]
06

Compute the Standard Deviation of the New Data Set

The standard deviation for the new data set is the square root of the variance. Thus, \( s_{\text{new}} = \sqrt{13} \approx 3.61 \).
07

Compare Both Standard Deviations

The standard deviation of both the original and the new data set remains the same at approximately 3.61. Thus, adding a constant to each data value does not change the standard deviation.

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Key Concepts

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

Variance
Variance is a measure of how spread out the values in a data set are. It helps us understand the degree of variation or dispersion from the mean (average). To calculate variance, we first find the squared differences between each data point and the mean. Then, we take the average of those squared differences.

In the original data set, \\( ext{Variance } (s^2)\) was calculated by:
  • Finding each data point's deviation from the mean, (e.g., \(5 - 10\) for the value 5).
  • Squaring each deviation (e.g., \( (5-10)^2 = 25\) ).
  • Taking the average: \[ \text{Variance} = \frac{25 + 1 + 0 + 1 + 25}{4} = 13.\]
Variance gives us insight into the variability of a data set. A higher variance means data points are more spread out. A lower variance indicates they are closer together.
Mean
The mean is simply the average of a data set. It represents a central value around which data points are distributed.

To compute the mean:
  • Add up all the data points.
  • Divide that sum by the number of data points.
For the original data set \(5, 9, 10, 11, 15\), the mean was calculated as:
\[ \bar{x} = \frac{5 + 9 + 10 + 11 + 15}{5} = 10. \]

The mean shifts when a constant is added to each data point, as seen when 5 was added:
\[ \bar{x}_{\text{new}} = \frac{10 + 14 + 15 + 16 + 20}{5} = 15. \]The mean shifted by the same constant, highlighting that the arithmetic operation on all values affects the mean directly.
Data Set
A data set is simply a collection of numbers or values that we want to study or analyze. It can be anything from temperatures of a city over a week to the weights of students in a class.

Consider the data set from the original exercise: \(5, 9, 10, 11, 15\). This set is our subject of analysis.

After adding 5 to each number, the new data set became \(10, 14, 15, 16, 20\), which still contains 5 values just like the original. Analyzing a data set involves understanding its characteristics by calculating statistical measures like mean, median, variance, etc.The importance of analyzing data sets lies in extracting meaningful information and spotting trends, anomalies, and regularities.
Effect of Constant Addition
When a constant is added to each value in a data set, some properties remain unchanged, while others shift. The mean changes, but interestingly, the standard deviation stays the same.

This is because:
  • Adding a constant alters data points equally in relation to their original positions.
  • The relative differences between the data points do not change.
  • Thus, measures of spread like variance and standard deviation do not get affected.
This can be surprising, as you might expect all properties of a data set to change. However, the constant addition is merely shifting the entire data set uniformly on the number line without altering how spread out the numbers are from each other.

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Most popular questions from this chapter

The Hill of Tara in Ireland is a place of great archaeological importance. This region has been occupied by people for more than 4000 years. Geomagnetic surveys detect subsurface anomalies in the earth's magnetic field. These surveys have led to many significant archaeological discoveries. After collecting data, the next step is to begin a statistical study. The following data measure magnetic susceptibility (centimeter-gram-second \(\times 10^{-6}\) ) on two of the main grids of the Hill of Tara (Reference: Tara: An Archaeological Survey by Conor Newman, Royal Irish Academy, Dublin). Grid \(\mathbf{E}: x\) variable $$\begin{array}{ccccccc} 13.20 & 5.60 & 19.80 & 15.05 & 21.40 & 17.25 & 27.45 \\ 16.95 & 23.90 & 32.40 & 40.75 & 5.10 & 17.75 & 28.35 \end{array}$$ Grid H: \(y\) variable $$\begin{array}{lllllll} 11.85 & 15.25 & 21.30 & 17.30 & 27.50 & 10.35 & 14.90 \\ 48.70 & 25.40 & 25.95 & 57.60 & 34.35 & 38.80 & 41.00 \\ 31.25 & & & & & \end{array}$$ (a) Compute \(\Sigma x, \Sigma x^{2}, \Sigma y,\) and \(\Sigma y^{2}\). (b) Use the results of part (a) to compute the sample mean, variance, and standard deviation for \(x\) and for \(y\). (c) Compute a \(75 \%\) Chebyshev interval around the mean for \(x\) values and also for \(y\) values. Use the intervals to compare the magnetic susceptibility on the two grids. Higher numbers indicate higher magnetic susceptibility. However, extreme values, high or low, could mean an anomaly and possible archaeological treasure. (d) Compute the sample coefficient of variation for each grid. Use the \(C V\) s to compare the two grids. If \(s\) represents variability in the signal (magnetic susceptibility) and \(\bar{x}\) represents the expected level of the signal, then \(s / \bar{x}\) can be thought of as a measure of the variability per unit of expected signal. Remember, a considerable variability in the signal (above or below average) might indicate buried artifacts. Why, in this case, would a large \(C V\) be better, or at least more exciting? Explain.

Basic Computation: Weighted Average Find the weighted average of a data set where 10 has a weight of \(5 ; 20\) has a weight of \(3 ; 30\) has a weight of 2

One standard for admission to Redfield College is that the student rank in the upper quartile of his or her graduating high school class. What is the minimal percentile rank of a successful applicant?

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.

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