Question

the accompanying data are the number of wins and the earned run averages (mean number of earned runs allowed per nine innings pitched) for eight baseball pitchers in a recent season. find the equation of the regression line. then construct a scatter plot of the data and draw the regression line. then use the regression equation to predict the value of y for each of the given x - values, if meaningful. if the x - value is not meaningful to predict the value of y, explain why not. (a)x = 5 wins (b)x = 10 wins (c)x = 19 wins (d)x = 15 wins click the icon to view the table of numbers of wins and earned run average the equation of the regression line is \\(\hat{y}=\square x+\square\\) (round to two decimal places as needed.) wins and era wins, x earned run average, y 20 2.81 18 3.36 17 2.64 16 3.81 14 3.89 12 4.25 11 3.73 9 5.04

Explanation:

Step1: Calculate sums

Let \(n = 8\). Calculate \(\sum x\), \(\sum y\), \(\sum xy\), \(\sum x^{2}\).
\(\sum x=20 + 18+17+16+14+12+11+9=117\)
\(\sum y=2.81 + 3.36+2.64+3.81+3.89+4.25+3.73+5.04 = 29.53\)
\(\sum xy=20\times2.81+18\times3.36 + 17\times2.64+16\times3.81+14\times3.89+12\times4.25+11\times3.73+9\times5.04=409.91\)
\(\sum x^{2}=20^{2}+18^{2}+17^{2}+16^{2}+14^{2}+12^{2}+11^{2}+9^{2}=1887\)

Step2: Calculate slope \(b_1\)

The formula for the slope \(b_1\) of the regression - line is \(b_1=\frac{n\sum xy-\sum x\sum y}{n\sum x^{2}-(\sum x)^{2}}\)
\[

$$\begin{align*} b_1&=\frac{8\times409.91-117\times29.53}{8\times1887 - 117^{2}}\\ &=\frac{3279.28-3455.01}{15096-13689}\\ &=\frac{- 175.73}{1407}\\ &\approx - 0.13 \end{align*}$$

\]

Step3: Calculate intercept \(b_0\)

The formula for the intercept \(b_0\) is \(b_0=\bar{y}-b_1\bar{x}\), where \(\bar{x}=\frac{\sum x}{n}=\frac{117}{8}=14.625\) and \(\bar{y}=\frac{\sum y}{n}=\frac{29.53}{8}=3.69125\)
\[

$$\begin{align*} b_0&=3.69125-(-0.13)\times14.625\\ &=3.69125 + 1.90125\\ &=5.5925\approx5.59 \end{align*}$$

\]

The equation of the regression line is \(\hat{y}=-0.13x + 5.59\)

Step4: Predict values

(a) For \(x = 5\) wins

Since \(5\) is outside the range of the \(x - \)values (\(9\leq x\leq20\)) in the data set, it is not meaningful to predict \(y\) using the regression line.

(b) For \(x = 10\) wins

\(\hat{y}=-0.13\times10 + 5.59=-1.3 + 5.59 = 4.29\)

(c) For \(x = 19\) wins

\(\hat{y}=-0.13\times19+5.59=-2.47 + 5.59 = 3.12\)

(d) For \(x = 15\) wins

\(\hat{y}=-0.13\times15+5.59=-1.95 + 5.59 = 3.64\)

Answer:

The equation of the regression line is \(\hat{y}=-0.13x + 5.59\)
(a) Not meaningful as \(x = 5\) is outside the data - range.
(b) \(\hat{y}=4.29\)
(c) \(\hat{y}=3.12\)
(d) \(\hat{y}=3.64\)