Chapter 11.1, Problem 43E

a.

To determine

Calculate the values of $\overline{x}$, $\overline{y}$, $s_x$ and $s_y$.

Answer to Problem 43E

The value of $\overline{x}$ is 1.75.

The value of $\overline{y}$ is 10.828.

The value of $s_x$ is 1.031.

The value of $s_y$ is 1.981.

Explanation of Solution

Calculation:

The number of years of service (x) and the hourly wages ($) (y) of the five employees of a small company are given.

Denote $\overline{x}$ and $\overline{y}$ as the means of x and y respectively. Denote $s_x$ and $s_y$ as the standard deviations of x and y respectively.

Descriptive statistics:

Software procedure:

Step-by-step procedure to obtain the descriptive statistics using the MINITAB software:

• Choose Stat > Basic Statistics > Display Descriptive Statistics, click OK.
• In Variables, enter the columns of x and y.
• Choose Statistics, select Mean, Standard deviation and click OK.
• Click OK.

Output using the MINITAB software is given below:

From the above output, it is evident that the value of $\overline{x}$ is 1.75, the value of $\overline{y}$ is 10.828, the value of $s_x$ is 1.031 and the value of $s_y$ is 1.981.

b.

To determine

Find the correlation coefficient between the years of service and the hourly wages.

Answer to Problem 43E

The correlation coefficient between the years of service and the hourly wages is 0.906.

Explanation of Solution

Calculation:

Correlation:

The correlation coefficient, r, between ordered pairs of variables, (x, y) having sample means $\overline{x}$ and $\overline{y}$, sample standard deviations $s_x$ and $s_y$ for a sample of size n is given as $r=\frac{1}{n-1}{\displaystyle \sum \left(\frac{x-\overline{x}}{s_x}\right)\left(\frac{y-\overline{y}}{s_y}\right)}$. The correlation coefficient is useful for measuring the strength of the linear relationship between the two variables.

Software procedure:

Step-by-step procedure to obtain the correlation using the MINITAB software:

• Choose Stat > Basic Statistics > Correlation.
• In Variables, enter the columns of x and y.
• Click OK.

Output using the MINITAB software is given below:

From the output, the correlation coefficient between the years of service and the hourly wages is 0.906.

c.

To determine

Find the sample mean and the sample standard deviation of the hourly wages, if each employee is given a raise of $1.00 per hour.

Answer to Problem 43E

The sample mean of the hourly wages, if each employee is given a raise of $1.00 per hour is 11.828.

The sample standard deviation of the hourly wages, if each employee is given a raise of $1.00 per hour is 1.981.

Explanation of Solution

Calculation:

Denote $y_1$ as the increased hourly wage of an employee. Thus, $y_1=y+1$.

The calculation for $y_1$ is shown as below:

y	$y_1=y+1$
9.50	10.50
8.23	9.23
10.95	11.95
12.70	13.70
12.75	13.75

Descriptive statistics:

Software procedure:

Step-by-step procedure to obtain the descriptive statistics using the MINITAB software:

• Choose Stat > Basic Statistics > Display Descriptive Statistics, click OK.
• In Variables, enter the columns of y1.
• Choose Statistics, select Mean, Standard deviation and click OK.
• Click OK.

Output using the MINITAB software is given below:

From the above output, it is evident that the sample mean of the hourly wages, if each employee is given a raise of $1.00 per hour is 11.828 and the sample standard deviation of the hourly wages, if each employee is given a raise of $1.00 per hour is 1.981.

d.

To determine

Identify the effects of increasing each value of y by 1 on the values of $\overline{y}$ and $s_y$.

Answer to Problem 43E

The average hourly wage $\underset{\_}{\left(\overline{y}\right)}$ has increased by $1.00 due to increasing each value of y by 1.

The standard deviation $\underset{\_}{\left(s_y\right)}$ has remained unchanged due to increasing each value of y by 1.

Explanation of Solution

Interpretation:

From part a, the value of average hourly wages, $\overline{y}$ is 10.828.

From part c, the value of sample mean of the hourly wages, if each employee is given a raise of $1.00 per hour is 11.828.

Now, $11.828-10.828=1$. In other words, ${\overline{y}}_1=\overline{y}+1$.

Hence, it is evident that the average hourly wage has increased by $1.00 due to increasing each value of y by 1.

From part a, the value of standard deviation of hourly wages, $s_y$ is 1.981.

From part c, the value of sample standard deviation , if each employee is given a raise of $1.00 per hour is 1.981.

Evidently, the standard deviation has remained unchanged due to increasing each value of y by 1.

e.

To determine

Find the correlation coefficient between the years of service and the increased hourly wages.

Explain the reason the correlation coefficient remains unchanged even when the values of y change.

Answer to Problem 43E

The correlation coefficient between the years of service and the increased hourly wages is 0.906.

Explanation of Solution

Calculation:

Correlation:

Software procedure:

Step-by-step procedure to obtain the correlation using the MINITAB software:

• Choose Stat > Basic Statistics > Correlation.
• In Variables, enter the columns of x and y1.
• Click OK.

Output using the MINITAB software is given below:

From the output, the correlation coefficient between the years of service and the increased hourly wages is 0.906.

Consider the formula for the correlation coefficient, r:

$r=\frac{1}{n-1}{\displaystyle \sum \left(\frac{x-\overline{x}}{s_x}\right)\left(\frac{y-\overline{y}}{s_y}\right)}$.

Now, each y increases and becomes $y_1=y+1$.

Again, as observed from part d, the mean of the increased hourly wages is ${\overline{y}}_1=\overline{y}+1$.

Replace these in the formula for the correlation coefficient:

$\begin{array}{c}r=\frac{1}{n-1}{\displaystyle \sum \left(\frac{x-\overline{x}}{s_x}\right)\left(\frac{\left(y+1\right)-\left(\overline{y}+1\right)}{s_y}\right)}\\ =\frac{1}{n-1}{\displaystyle \sum \left(\frac{x-\overline{x}}{s_x}\right)\left(\frac{y+1-\overline{y}-1}{s_y}\right)}\\ =\frac{1}{n-1}{\displaystyle \sum \left(\frac{x-\overline{x}}{s_x}\right)\left(\frac{y-\overline{y}}{s_y}\right)}.\end{array}$

It is observed that the correlation coefficient is independent of the change of origin.

Hence, the correlation coefficient remains unchanged even when the values of y change.

f.

To determine

Find the sample mean and the sample standard deviation of the service period, if it is expressed in terms of months.

Answer to Problem 43E

The sample mean of the hourly wages, of the service period, if it is expressed in terms of months is 21.

The sample standard deviation of the hourly wages, of the service period, if it is expressed in terms of months is 12.37.

Explanation of Solution

Calculation:

Denote $x_1$ as the service period of an employee, expressed in months. Thus, $x_1=12x$.

The calculation for $x_1$ is shown as below:

x	$x_1=12x$
0.5	6
1.0	12
1.75	21
2.5	30
3.0	36

Descriptive statistics:

Software procedure:

Step-by-step procedure to obtain the descriptive statistics using the MINITAB software:

• Choose Stat > Basic Statistics > Display Descriptive Statistics, click OK.
• In Variables, enter the columns of x1.
• Choose Statistics, select Mean, Standard deviation and click OK.
• Click OK.

Output using the MINITAB software is given below:

From the above output, it is evident that the sample mean of the hourly wages, of the service period, if it is expressed in terms of months is 21 and the sample standard deviation of the hourly wages, of the service period, if it is expressed in terms of months is 12.37.

g.

To determine

Identify the effects of multiplying each value of x by 12 on the values of $\overline{x}$ and $s_x$.

Answer to Problem 43E

The average service period in months has been multiplied by 12 due to multiplying each value of x by 12.

The standard deviation has been multiplied by 12 due to multiplying each value of x by 12.

Explanation of Solution

Interpretation:

From part a, the average service period in years, $\overline{x}$ is 1.75.

From part f, the average service period in months is 21.

Now, $12\times 1.75=21$. In other words, ${\overline{x}}_1=12\overline{x}$.

Hence, it is evident that the average hourly wage has increased by $1.00 due to increasing each value of y by 1.

From part a, the value of standard deviation of service period in years, $s_x$ is 1.031.

From part c, the value of sample standard deviation of service period in months is 12.37.

Now,

$\begin{array}{c}12\times 1.031=12.372\\ \approx \mathrm{12.37.}\end{array}$

In other words, $s_{x_1}=12s_x$.

Evidently, the standard deviation has been multiplied by 12 due to multiplying each value of x by 12.

h.

To determine

Find the correlation coefficient between the months of service and the hourly wages.

Explain the reason the correlation coefficient remains unchanged even when the values of $\overline{x}$ and $s_x$ change.

Answer to Problem 43E

The correlation coefficient between the months of service and the increased hourly wages is 0.906.

Explanation of Solution

Calculation:

Correlation:

Software procedure:

Step-by-step procedure to obtain the correlation using the MINITAB software:

• Choose Stat > Basic Statistics > Correlation.
• In Variables, enter the columns of x1 and y.
• Click OK.

Output using the MINITAB software is given below:

From the output, the correlation coefficient between the months of service and the increased hourly wages is 0.906.

Consider the formula for the correlation coefficient, r:

$r=\frac{1}{n-1}{\displaystyle \sum \left(\frac{x-\overline{x}}{s_x}\right)\left(\frac{y-\overline{y}}{s_y}\right)}$.

Now, each x is multiplied by 12 and becomes $x_1=12x$.

Again, as observed from part d, the mean of the months of service is ${\overline{x}}_1=12\overline{x}$ and the corresponding standard deviation is $s_{x_1}=12s_x$.

Replace these in the formula for the correlation coefficient:

$\begin{array}{c}r=\frac{1}{n-1}{\displaystyle \sum \left(\frac{12x-12\overline{x}}{12s_x}\right)\left(\frac{\left(y+1\right)-\left(\overline{y}+1\right)}{s_y}\right)}\\ =\frac{1}{n-1}{\displaystyle \sum \left(\frac{12\left(x-\overline{x}\right)}{12s_x}\right)\left(\frac{y+1-\overline{y}-1}{s_y}\right)}\\ =\frac{1}{n-1}{\displaystyle \sum \left(\frac{x-\overline{x}}{s_x}\right)\left(\frac{y-\overline{y}}{s_y}\right)}.\end{array}$

It is observed that the correlation coefficient is independent of the changes of origin and scale.

Hence, the correlation coefficient remains unchanged even when the values of $\overline{x}$ and $s_x$ change.

i.

To determine

Find the effect of adding a constant to each x-value or to each y-value on the correlation coefficient.

Answer to Problem 43E

If a constant is added to each x-value or to each y-value, the correlation coefficient is unchanged.

Explanation of Solution

Interpretation:

The correlation coefficient is independent of the change of origin of the variables. Adding a constant to each x-value or to each y-value implies a change in origin of x or y.

Hence, if a constant is added to each x-value or to each y-value, the correlation coefficient is unchanged.

Statistics Concept Introduction

Correlation:

The correlation coefficient, r, between ordered pairs of variables, (x, y) having sample means $\overline{x}$ and $\overline{y}$, sample standard deviations $s_x$ and $s_y$ for a sample of size n is given as $r=\frac{1}{n-1}{\displaystyle \sum \left(\frac{x-\overline{x}}{s_x}\right)\left(\frac{y-\overline{y}}{s_y}\right)}$. The correlation coefficient is useful for measuring the strength of the linear relationship between the two variables.

j.

To determine

Find the effect of multiplying a positive constant to each x-value or to each y-value on the correlation coefficient.

Answer to Problem 43E

If each x-value or each y-value is multiplied by a positive constant, the correlation coefficient is unchanged.

Explanation of Solution

Interpretation:

The correlation coefficient is independent of the change of scale of the variables. Multiplying a positive constant with each x-value or each y-value changes the scale of x or y.

Hence, if each x-value or each y-value is multiplied by a positive constant, the correlation coefficient is unchanged.