Chapter 14, Problem 25P

Problem 9 examined the relationship between weight and income for a sample of n = 10 women. Weights were classified in five categories and had a mean of M = 3 with SS = 20. Income, measured in thousands, had a mean score of M = 66 with SS = 7430, and SP = −359.

• a. Find the regression equation for predicting income from weight. (Identify the income scores as X values and the weight scores as Y values.)
• b. What percentage of the variance in the income is accounted for by the regression equation? (Compute the correlation, r. then find r².)
• c. Does the regression equation account for a significant portion of the variance in income? Use α= .05 to evaluate the F-ratio.

Weight (X)	Income (Y)
1	125
2	78
4	49
3	63
5	35
2	84
5	38
3	51
1	93
4	44

To determine

The regression equation for predicting income from weight.

Answer to Problem 25P

The regression equation of the data is $Y=-17.95X+119.85$

Explanation of Solution

Given info: For 10 women weights are classified in five categories with mean of  $M=3$ and $SS=20$. The income measured in thousands with mean $M=66$ and and $SS=7430$. The given data are shown below,

Weight(X)	Income(Y)
1	125
2	78
4	49
3	63
5	35
2	84
5	38
3	51
1	93
4	44

Calculation:

Generally the linear regression equation is defined as

$Y=bX+a$

Formula to calculate $b$ is,

$b=\frac{SP}{S{S}_x}$

Formula to calculate the $a$ is,

$a=M_Y-b{M}_X$

Mean of the variable $X$ is,

$\begin{array}{c}{M}_X=\frac{1+2+4+3+5+2+5+3+1+4}{10}\\ =\frac{30}{10}\\ =3\end{array}$

Thus, the mean of the variable $X$ is 3.

Mean of the variable $Y$ is,

$\begin{array}{c}{M}_Y=\frac{125+78+49+63+35+84+38+51+93+44}{10}\\ =\frac{660}{10}\\ =66\end{array}$

Thus the mean of the variable y is 66.

For calculating $S{S}_x$ and $SP$ the table is shown below,

Sr.no	$X$	$Y$	$X-M_X$	$Y-M_Y$	${\left(X-M_X\right)}^2$	${\left(Y-M_Y\right)}^2$	$\left(X-M_X\right)\left(Y-M_Y\right)$
1	1	125	12	59	4	3481	-118
2	2	78	-1	12	1	144	-12
3	4	49	1	-17	1	289	-17
4	3	63	0	-3	0	9	0
5	5	35	2	-31	4	961	-62
6	2	84	-1	18	1	324	-18
7	5	38	2	-28	4	784	-56
8	3	51	0	-15	0	225	0
9	1	93	-2	27	4	729	-54
10	4	44	1	-22	1	484	-22
Total			0	0	20	7430	-359

Where, $M_Y$ is the mean of the variable $Y$ and $M_X$ is the mean of variable $X$. Substitute $-359$ for $SP$ and 20 for $S{S}_X$ in the equation for calculating b to get the slop coefficient.

$\begin{array}{c}b=\frac{-359}{20}\\ =-17.95\end{array}$

Substitute $-17.95$ for $b$, 3 for $M_X$ and 66 for $M_Y$ In the equation for calculating a to get the coefficient.

$\begin{array}{c}a=66-3\times -17.95\\ =66+53.85\\ =119.85\end{array}$

Substitute $-17.95$ for $b$ and 119.85 for $a$ in equation for linear regression to get the regression equation.

$Y=-17.95X+119.85$

Thus, the regression equation of the data is $\underset{\_}{Y=-17.95X+119.85}$

To determine

The Percentage of variance in the income accounted for the regression equation.

Answer to Problem 25P

The percentage of variance in the income accounted for the regression equation is 86.7%.

Explanation of Solution

Calculation:

Formula to calculate the Pearson correlation is,

$r=\frac{SP}{\sqrt{S{S}_{x}S{S}_Y}}$

Where covariability of the variable x and y is $SP$, $S{S}_X$ is the variability of the x variable and $S{S}_Y$ is the variability of the y variable.

As calculated in part a $SP$ is $-359$, $S{S}_X$ is 7430 and $S{S}_X$ is 20.

Substitute $-359$ for $SP$, 20 for $S{S}_Y$ and 7430 for $S{S}_Y$ in the equation for calculating the Pearson correlation coefficient to get the Pearson correlation of the given data.

$\begin{array}{c}r=\frac{-359}{\sqrt{20\times 7430}}\\ =\frac{-359}{\sqrt{148600}}\\ =\frac{-359}{385.48}\\ =-0.931\end{array}$

Thus, the Pearson correlation of the given data is $-0.931$ and $r^2=0.867$. And percentage will be 86.7%.

Thus, 86.7% of the variance in the income is accounted for by the regression equation.

To determine

The significance of the regression equation.

Answer to Problem 25P

The regression equation is significance at 5% level of significance.

Explanation of Solution

Calculation:

As calculated in part a $SP$ is $-359$, $S{S}_Y$ is 7430 and $S{S}_X$ is 20.

Set the null hypothesis

$H_0:b=0$

The alternative hypothesis

$H_1:b\ne 0$

Formula to calculate the $F$ ratio is,

$F=\frac{M{S}_{regression}}{M{S}_{residual}}\text{with}\text{df}=1,n-2$

Where, $M{S}_{regression}=\frac{S{S}_{regression}}{d{f}_{regression}}\text{with}df=1$ and $M{S}_{residual}=\frac{S{S}_{residual}}{d{f}_{residual}}\text{with}df=n-2$.

$S{S}_{regression}=r^2\times S{S}_Y$

$S{S}_{residual}=\left(1-r^2\right)S{S}_Y$

Substitute $-0.931$  for $r$ and 7430 for $S{S}_Y$ in the equation (3) to get the $S{S}_{regression}$,

$\begin{array}{c}S{S}_{regression}={(}^{-}\times 7430\\ =0.867\times 7430\\ =6441.81\end{array}$

Substitute $-0.931$ for $r$ and 7430 for $S{S}_Y$ in the equation (4) to get the $S{S}_{residual}$,

$\begin{array}{c}S{S}_{residual}=\left(1-{\left(-0.931\right)}^2\right)\times 7430\\ =\left(1-0.867\right)\times 7430\\ =0.133\times 7430\\ =988.19\end{array}$

Thus the   $S{S}_{residual}$ is 988.19.

$\begin{array}{c}M{S}_{regression}=\frac{S{S}_{regression}}{df}\\ =\frac{6441.81}{1}\\ =6441.81\end{array}$

Thus the   $M{S}_{regression}$ is 6441.81.

$\begin{array}{c}M{S}_{residual}=\frac{S{S}_{residual}}{d{f}_{residual}}\\ =\frac{988.19}{8}\\ =123.523\end{array}$

Thus the  is 123.523.

Substitute 6441.81 for $M{S}_{regression}$ and 123.523 for $M{S}_{residual}$ in the equation (2)

$\begin{array}{c}F=\frac{6441.81}{123.523}\text{with}\text{df}=1,10-2\\ =52.15\end{array}$

Thus the F-ratio is 76.437.

From the F-table, the value of $F_{tab}$ at 5% level of significance with $df$ 1 and $n-2=8$ is 5.32.

Decision rule:

If the calculated F-value is less than table F-value, then there is enough evidence to accept to reject the null hypothesis.

As $F_{cal}$ is greater than the $F_{tab}$ so reject the null hypothesis and conclude that the regression equation does account for a significance portion of the variance for the $Y$ score.

Thus, the regression equation is significance at 5% level of significance.