91Ó°ÊÓ

Given in the question that, calculation from data for an 18-year period

$x=1.75\mathrm{\%}$$s_z=5.36\mathrm{\%}$$y=9.07\mathrm{\%}$

$s_y=15.35\mathrm{\%}$$r=0.596$

We have to find out that What percent of the observed variation in yearly changes in the index is explained by a straight-line relationship with the change during January.

The slope of the least-squares regression line of percent change in a stock market index during the course of the year on percent change in a stock market index in January is

$$\begin{gathered} b=r\frac{S_Y}{S_X} \\ b=0.596×\frac{15.35}{5.36} \\ =1.707 \end{gathered}$$

To find the intercept, we use the fact that the least-squares line passes through $(\stackrel{¯}{X},\stackrel{¯}{Y})$

localid="1650002477681" $$\begin{gathered} a=\stackrel{¯}{Y}-b\stackrel{¯}{X} \\ =9.07−1.707×1.75\mathrm{g} \\ =6.083 \end{gathered}$$

So the equation of the least-squares line is:

$\hat{Y}=6.083+1.707X$

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

$r=0.596$

Here,

localid="1650002483859" $$\begin{gathered} r^2=(0.596{)}^2 \\ =0.3552 \end{gathered}$$

Given in the question that $s=8.3$.

We have to explain what this value means.

The standard deviation is a measurement of a set of values' variation or dispersion. The standard deviation of the residuals $s$measures the average size of the prediction errors (residuals) when using the regression line.