Chapter 3.2, Problem 50E

(a)

To determine

To explain: the observed variation in yearly changes in the index

Answer to Problem 50E

About 35.52% of the variation in Y is due to the straight line relationship between Y and X.

The other 64.48% is individual variation that is not explained by the linear relationship.

Explanation of Solution

Given:

	Mean	Standard deviation
Percent change in a stock market index in January $\left(X\right)$	$\overline{X}=1.75\%$	$S_X=5.36\%$
Percent change in a stock market index for the entire year $\left(Y\right)$	$\overline{Y}=9.07\%$	$S_Y=15.35\%$

  $r=0.596$

Calculation:

The least-squares regression line of percent change in a stock market index for the entire year on percent change in a stock market index for the January has slope is$\begin{array}{l}b=r\frac{S_Y}{S_X}\\ b=0.596\times \frac{15.35}{5.36}\\ =1.707\end{array}$

To find the intercept, we use the fact that the least-squares line passes through $\left(\overline{X},\overline{Y}\right)$ :

  $\begin{array}{l}a=\overline{Y}-b\overline{X}\\ =9.07-1.707\times 1.75\\ =6.083\end{array}$ g

So the equation of the least-squares line is:

  $\widehat{Y}=6.083+1.707X$

The coefficient of determination $r^2$ is the fraction of the variation in the values of Y that is accounted for by the least-squares regression line Y on X.

Here,$\begin{array}{l}r=0.596\\ r^2={\left(0.596\right)}^2\\ =0.3552\end{array}$

Then about 35.52% of the variation in Y is due to the straight line relationship between Y and X. The other 64.48% is individual variation that is not explained by the linear relationship.

Conclusion:

Hence, about 35.52% of the variation in Y is due to the straight line relationship between Y and X. The other 64.48% is individual variation that is not explained by the linear relationship.

(b)

To determine

To explain:the value of thestandard deviation $s=8.3$ .

Answer to Problem 50E

The standard deviationof the residuals $s$ measures the average size of the prediction errors (residuals) when using the regression line. The value $s=8.3$ shows too much prediction error.

Explanation of Solution

Explanation:

For these data the standard deviation of the residuals is $s=8.3$

The standard deviation of the residuals $s$ measures the average size of the prediction errors (residuals) when using the regression line.The value $s=8.3$ shows too much prediction error.