Question

The stock market Some people think that the behavior of the stock market in January predicts its behavior for the rest of the year. Take the explanatory variable $x$to be the percent change in a stock market index in January and the response variable $y$to be the change in the index for the entire year. We expect a positive correlation between $x$and y because the change during January contributes to the full year’s change. Calculation from data for an 18-year period gives

$$\begin{gathered} \overline{x}=1.75\mathrm{\%}{s}_z=5.36\mathrm{\%}\overline{y}=9.07\mathrm{\%} \\ {s}_y=15.35\mathrm{\%}r=0.596 \end{gathered}$$

(a) What percent of the observed variation in yearly changes in the index is explained by a straight-line relationship with the change during January?

(b) For these data, $s=8.3$Explain what this value means

Short answer

(a) A straight-line association with the change in January explains $35.52$percent of the observed range in yearly changes in the index.$35.2\%$

(b) The value $s=8.3$shows too much prediction error.

Step-by-step solution

Part (a) Step 1: Given Information

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.

Part (a) Step 2:  Explanation

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

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

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

localid="1650002477681" $$\begin{gathered} a=\overline{Y}-b\overline{X} \\ =9.07-1.707\times 1.75\mathrm{g} \\ =6.083 \end{gathered}$$

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

Part (b) Step 1: Given Information 

Given in the question that $s=8.3$.

We have to explain what this value means.

Part (b) Step 2: Explanation 

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.