Chapter 4.1, Problem 43E

Changing means and standard deviations: A small company has five employees. The following table presents the number of years each has been employed (x) and the wage dollars (y).

x (years) 0.5 1.0 1.75 2.5 3.0

y (dollars) 9.51 8.23 10.95 12.70 12.75

1. Compute $\bar{x},\bar{y},s_x$ and $s_y$ .
2. Compute the correlation coefficient between years of service and hourly wage.
3. Each employee is given a raise of $1.00 per hour, so each y-value is increased by 1. Using these newy-values, compute the sample mean and the sample standard deviation $s_y$ .
4. In part (c), each y-value was increased by 1. What was the effect on $\bar{y}$ ? What was the effect on $s_y$?
5. Compute the correlation coefficient between years of service and the increased hourly wage. Explain why the correlation coefficient is unchanged though the y -values have changed.
6. Convert x to months by multiplying each x-value by 12. So the new x-values are 6, 12, 21, 30, and 36. Compute the sample mean $\bar{x}$ and $s_x$ the sample standard deviation $s_x.$
7. In part (f), each x-value was multiplied by 12. What was the effect on $\bar{x}$ and $s_x$?
8. Compute the correlation coefficient between months of service and hourly wage in dollars. Explain why the correlation coefficient is unchanged even though the x-values, $\bar{x}$ and $s_x$ have changed.
9. Use the results of parts (a)-(h) to fill in the blank: If a constant is added to each x-value or to each y-value, the correlation coefficient is _____.
10. Use the results of parts (a)-(h) to fill in the blank: If each x-value or each y-value is multiplied by a positive constant, the correlation coefficient is____.

To determine

The values of $\overline{x},\overline{y},s_x,\text{ and }{s}_y$ .

Answer to Problem 43E

  $\overline{x}=1.75,\overline{y}=10.828,s_x=0.922,s_y=1.772$

Explanation of Solution

Given information:

Given, a small company has five employees and the following table presents the number of years each has been employed $x$ and the hourly wage in dollars $y$:

  

Concepts used:

The correlation coefficient of a data is given by:

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

Where,

  $\overline{x},\overline{y}$ represent the mean of $x$ and $y$ respectively. $s_x,s_y$ represent the standard deviations of $x$and $y$, $n$ represents the number of terms.

The standard deviations are given by:

  $s_x=\sqrt{\frac{{\displaystyle \sum {\left(x-\overline{x}\right)}^2}}{n}},s_y=\sqrt{\frac{{\displaystyle \sum {\left(y-\overline{y}\right)}^2}}{n}}$

Calculation:

The mean of $x$is given by:

  $\overline{x}=\frac{{\displaystyle \sum x}}{n}=\frac{0.5+1.0+1.75+2.5+3.0}{5}=\frac{8.75}{5}=1.75$

The mean of $y$is given by:

  $\overline{y}=\frac{{\displaystyle \sum y}}{n}=\frac{9.51+8.23+10.95+12.70+12.75}{5}=\frac{54.14}{5}=10.828$

The data can be represented in tabular form as:

  

Hence, the standard deviation is given by:

  $\begin{array}{l}{s}_x=\sqrt{\frac{{\displaystyle \sum {\left(x-\overline{x}\right)}^2}}{n}}\\ s_x=\sqrt{\frac{4.25}{5}}\\ s_x=\sqrt{0.85}\\ s_x=\mathrm{0.921954...}\\ s_x\approx 0.922\end{array}$

And,

  $\begin{array}{l}{s}_y=\sqrt{\frac{{\displaystyle \sum {\left(y-\overline{y}\right)}^2}}{n}}\\ s_y=\sqrt{\frac{15.70008}{5}}\\ s_y=\sqrt{3.140016}\\ s_y=\mathrm{1.772009...}\\ s_y\approx 1.772\end{array}$

Therefore, $\overline{x}=1.75,\overline{y}=10.828,s_x=0.922,s_y=1.772$

To determine

The correlation coefficient between years of service and hourly wage.

Answer to Problem 43E

  $0.906$ .

Explanation of Solution

Given information:

Given, a small company has five employees and the following table presents the number of years each has been employed $x$ and the hourly wage in dollars $y$ :

  

  $\overline{x}=1.75,\overline{y}=10.828,s_x=0.922,s_y=1.772$

Concepts used:

The correlation coefficient of a data is given by:

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

Where,

  $\overline{x},\overline{y}$ represent the mean of $x$ and $y$ respectively. $s_x,s_y$ represent the standard deviations of $x$and $y$, $n$ represents the number of terms.

The standard deviations are given by:

  $s_x=\sqrt{\frac{{\displaystyle \sum {\left(x-\overline{x}\right)}^2}}{n}},s_y=\sqrt{\frac{{\displaystyle \sum {\left(y-\overline{y}\right)}^2}}{n}}$

Calculation:

Consider,

Hence, the table for calculating coefficient of correlation is given by:

  

Putting the values in the formula,

  $r=\frac{1}{5}\cdot \frac{\left(7.4025\right)}{\left(0.922\right)\cdot \left(1.772\right)}=\mathrm{0.90621...}\approx 0.906$

Therefore, the correlation coefficient for the given data is $0.906$ .

To determine

  $\overline{y}\text{ and }{s}_y$ are to be calculated using new $y$ values.

Answer to Problem 43E

  $\overline{y}=11.828,s_y=1.772$

Explanation of Solution

Given information:

Given, a small company has five employees and the following table presents the number of years each has been employed $x$ and the hourly wage in dollars $y$:

  

Each employee is given a raise of $\$1.00$ per hour, so each $y-$ value is increased by $1$ .

Calculation:

When $y$ is increased by $1$ , the new $y$ values are given by $10.51,9.23,11.95,13.70,13.75$ respectively.

The standard deviation is given by:

  $s_y=\sqrt{\frac{{\displaystyle \sum {\left(y-\overline{y}\right)}^2}}{n}}$

The mean of $y$ is given by:

  $\overline{y}=\frac{{\displaystyle \sum y}}{n}=\frac{10.51+9.23+11.95+13.70+13.75}{5}=\frac{59.14}{5}=11.828$

The data can be represented in tabular form as:

  

Hence, the standard deviation is given by:

  $\begin{array}{l}{s}_y=\sqrt{\frac{{\displaystyle \sum {\left(y-\overline{y}\right)}^2}}{n}}\\ s_y=\sqrt{\frac{15.70008}{5}}\\ s_y=\sqrt{3.140016}\\ s_y=\mathrm{1.772009...}\\ s_y\approx 1.772\end{array}$

Hence, $\overline{y}=11.828,s_y=1.772$

To determine

The effect of increasing each $y-$ value by $1$ on $\overline{y}\text{ and }{s}_y$ .

Answer to Problem 43E

  $\overline{y}$ increased by $1$ and $s_y$ remained unchanged.

Explanation of Solution

Given information:

Given, a small company has five employees and the following table presents the number of years each has been employed $x$ and the hourly wage in dollars $y$ :

  

Each employee is given a raise of $\$1.00$ per hour, so each $y-$ value is increased by $1$ .

Calculation:

From part $a$ of this exercise, original values are given as $\overline{y}=10.828,s_y=1.772$

From part $c$ of this exercise, the new values are given by $\overline{y}=11.828,s_y=1.772$

From the given data, it is observed that when each $y-$ value is increased by $1$ , the mean $\overline{y}$ has increased by $1$ and the standard deviation $s_y$ remained unchanged.

To determine

The correlation coefficient between years of service and the increased hourly wage. And to explain why the correlation coefficient is unchanged.

Answer to Problem 43E

  $0.906$

Since the change did not affect he standard deviation of y, the correlation coefficient has not changed.

Explanation of Solution

Given information:

Given, a small company has five employees and the following table presents the number of years each has been employed $x$ and the hourly wage in dollars $y$:

  

Each employee is given a raise of $\$1.00$ per hour, so each $y-$ value is increased by $1$ .

Calculation:

From part $a$ of this exercise, original values are given as $\overline{x}=1.75,\overline{y}=10.828,s_x=0.922,s_y=1.772$

From part $c$ of this exercise, the new values are given by $\overline{y}=11.828,s_y=1.772$

From the given data, it is observed that when each $y-$ value is increased by $1$ , the mean $\overline{y}$ has increased by $1$ and the standard deviation $s_y$ remained unchanged.

Each employee is given a raise of $1.00 per hour, so each y-value is increased by 1.

Consider,

Hence, the table for calculating coefficient of correlation is given by:

  

Putting the values in the formula,

  $r=\frac{1}{5}\cdot \frac{\left(7.4025\right)}{\left(0.922\right)\cdot \left(1.772\right)}=\mathrm{0.90621...}\approx 0.906$

Therefore, the correlation coefficient for the given data is $0.906$ .

Since the change did not affect he standard deviation of $y$ , the correlation coefficient has not changed.

To determine

  $\overline{x}\text{ and }{s}_x$ are to be calculated using new $x$ values.

Answer to Problem 43E

  $\overline{x}=21,s_x=11.063$

Explanation of Solution

Given information:

Given, a small company has five employees and the following table presents the number of years each has been employed $x$ and the hourly wage in dollars $y$:

  

  $x$ is converted to months by multiplying each $x-$ value by $12$ .

Calculation:

  $x$ is converted to months by multiplying each $x-$ value by $12$ . So the new $x-$ values are

  $6,12,21,30\text{ and 36}$ .

The data can be represented in tabular form as:

  

Hence, the standard deviation is given by:

  $\begin{array}{l}{s}_x=\sqrt{\frac{{\displaystyle \sum {\left(x-\overline{x}\right)}^2}}{n}}\\ s_x=\sqrt{\frac{612}{5}}\\ s_x=\sqrt{122.4}\\ s_x=\mathrm{11.06345...}\\ s_x\approx 11.063\end{array}$

Therefore, $\overline{x}=21,s_x=11.063$

To determine

The effect of multiplying each $x-$ value by $12$ on $\overline{x}\text{ and }{s}_x$ is to be determined.

Answer to Problem 43E

  $\overline{x}$ and $s_x$ are multiplied by 12.

Explanation of Solution

Given information:

Given, a small company has five employees and the following table presents the number of years each has been employed $x$ and the hourly wage in dollars $y$:

  $x$ is converted to months by multiplying each $x-$ value by $12$ .

Calculation:

From part $a$ of this exercise, original values are given as $\overline{x}=1.75,s_x=0.922$

From part $f$ of this exercise, the new values are given by $\overline{x}=21,s_x=11.063$

From the given data, it is observed that when each x-value is multiplied by 12, the mean $\overline{x}$ and the standard deviation $s_x$ are also multiplied by 12.

To determine

The correlation coefficient between months of service and hourly wage. And to explain why the correlation coefficient is unchanged.

Answer to Problem 43E

  $0.906$

Both $x-\overline{x}\text{ and }{s}_x$ have been multiplied by $12$ which are present in numerator and denominator.

Explanation of Solution

Given information:

Given, a small company has five employees and the following table presents the number of years each has been employed $x$ and the hourly wage in dollars $y$:

  $x$ is converted to months by multiplying each $x-$ value by $12$ .

Calculation:

Original values:

  $\overline{x}=1.75,\overline{y}=10.828,s_x=0.922,s_y=1.772$

x is converted to months by multiplying each x-value by 12. So the new x-values are 6, 12, 21, 30 and 36.

So the new values are given by $\overline{x}=21,s_x=11.063$

Consider,

Hence, the table for calculating coefficient of correlation is given by:

  

Putting the values in the formula,

  $r=\frac{1}{5}\cdot \frac{\left(88.83\right)}{\left(11.063\right)\cdot \left(1.772\right)}=\mathrm{0.90621...}\approx 0.906$

Therefore, the correlation coefficient for the given data is $0.906$ .

Even though the values of $x$ , $\overline{x}\text{ and }{s}_x$ are changed, both $x-\overline{x}\text{ and }{s}_x$ have been multiplied by $12$ which are present in numerator and denominator thus nullifying the increase.

To determine

Using the results in the previous parts, the statement is to be completed by filling the blank.

Answer to Problem 43E

Unchanged.

Explanation of Solution

Given information:

Given statement is: If a constant is added to each x-value or to each y-value, the correlation coefficient is _____.

Calculation:

Given, a small company has five employees and the following table presents the number of years each has been employed $x$ and the hourly wage in dollars $y$ :

  

And $r=0.906$

Each employee is given a raise of $\$1.00$ per hour, so each $y-$ value is increased by $1$ .

And the new correlation coefficient is given by $r=0.906$

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

To determine

Using the results in the previous parts, the statement is to be completed by filling the blank.

Answer to Problem 43E

Unchanged.

Explanation of Solution

Given information:

Given statement is: If each x-value or each y-value is multiplied by a positive constant, the correlation coefficient is _____.

Calculation:

From the previous results:

Given, a small company has five employees and the following table presents the number of years each has been employed $x$ and the hourly wage in dollars $y$ :

  

And $r=0.906$

  $x$ is converted to months by multiplying each $x-$ value by $12$ .

And the new correlation coefficient is given by $r=0.906$

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