Question

Use the following table for questions 11-13.A baseball manager believes a linear relationship exists between the number of Home Runs a player hits andthe number of times the player strikes out. The following data is for his eight starting players.9Home RunsStrikeouts66111 32 43 29 21 1674 162 238 159 125 101I7511. Write a regression equation for the data above. (MD1)a) y=4.6x + 28.26 b) y=28.26X+4.6

Answer 1

The regression equation of Y on X is given by the following formula:

`Y-\bar{Y}=b_(yx)(X-\bar{X})`

Where byx is given by the formula:

`b_(yx)=\frac{N\sum^{}_{}XY-\sum^{}_{}X\sum^{}_{}Y}{N\sum^{}_{}X^2-(\sum^{}_{}X)^2}`

Where N is the number of values (N=8). We need to find the sum of X values, the sum of Y values, the average of X, the average of Y, the sum of X*Y and the sum of X^2.

The table of values is:

The values we need to know are on the following table:

By replacing the known values in the formula we obtain:

`\begin{gathered} b_(yx)=(8\cdot26125-167\cdot995)/(8\cdot4649-(167)^2) \\ b_(yx)=(209000-166165)/(37192-27889) \\ b_(yx)=(42835)/(9303) \\ b_(yx)=4.6 \end{gathered}`

Now, the average of X and Y is the sum divided by N, then:

`\begin{gathered} \bar{X}=(167)/(8)=20.87 \\ \bar{Y}=(995)/(8)=124.37 \end{gathered}`

Replace these values in the formula and find the regression equation as follows:

`\begin{gathered} Y-124.37=4.6(X-20.87) \\ Y-124.37=4.6X-4.6\cdot20.87 \\ Y=4.6X-96.11+124.37 \\ Y=4.6X+28.26 \end{gathered}`

The answer is a) y=4.6x+28.26