Question

Please explain thank you.

6. Confidence and prediction Interval estimates Aa Aa You are a starting pitcher in the major leagues. It's January 2008, and you are in the process of negotiating your salary for the 2008 season. You hire a statisticlan to help you with your negotiations. She specifies the folowing simple linear regression model: where y2008 salary (in millions of dollars), and x performance during the regular 2007 season Then she selects a random sample of 50 major league starting pitchers with signed salary agreements for the 2008 season. She collects data on their 2007 performances and their 2008 salaries and estimates her model several times, each time with a different performance varlable. The performance variable that provides the highest r2 is innings pitched (IP). You are not surprised at this result, because in major league baseball, the better a starting pitcher performs, the longer he remains in the game The sample data for 2008 salary and 2007 1P are located in the Pitchers data set shown in the following DataView tool. [TP data source: Canadian Business Online; Salary data source: USATODAY.com.) lo Sample Variables 2 Observations 50 Sample of Major League Starting Pitchers Oata Sources Canadian Business Online tinnings Pitched) and USATODAY.com ISalary Observstions Variable Type v Form v ValuesMissing 2008 Salary Quantitative Numeric Quantitative 50 Innings Pitched in 2007 Numeric 50 Variable Variable Correlation Correlation On the Data Set panel in the tool, click on the observations button to view the Pitchers data set. Then click on one of the Correlation panels. Select 2008 Salary in the dropdown menu for Dependent Variable, and select Innings Pitched in 2007 in the Session On the Data Set panet in the tool, cilck on the Observations button to view the Pitchers data set. Then click on one of the Correlation panels. Select 2008 Salary in the dropdown menu for Dependent Varlable, and select Innings Pitched in 2007 in the dropdown menu for Independent Varlabie. Click on the Scatter button to view the scatter diagram. Mouse over the ieast-squares line (in red) to obtain its equation from the label. Cick on the Linear Regresslon button to view the regression statistics. The estimated regression equation is also shown on this page You pitched 180 innings during the 2007 regular season. During your salary negotiations, you tell the team's owners that the mean 2008 salary of all starting pitchers who pitched 180 innings in the 2007 regular season is unknown, but its point estimate Is $5.60 million. You are given that the standard error of the estimate, s-451685, when x" = 180, Y," bo + bix" has a mean equal to the unknown Ety) and an estimated standard deviation equal to S number of innings pitched by the pitchers in the sample, and the standard deviation of innings pitched. Square the standard devlation to get the variance, and multiply the varlance by (n -1) to find x-8). . [Hint: Use the tool to find R, the mean los ial Use the Distributions tool to help you answer some of the questions that follow. t Distribution Degrees of Freedom 26 .3 .2 1 You go on to tell the team owners that they can be 95% confident that the mean salary for all starting pitchers who pi Session .2 -1 You go on to tell the team owners that they can be 95% confident that the mean salary for all starting pitchers who pitched 180 innings last season is between and $4.736 million $5.62 million Suppose that instead of estima $0.653 million predict the 2008 salary of mlionwher who corresponding to the prediction of the 2008 salary for one specific pltcher who pitched 180 innings in 2007 s salary of all major league starting pitchers who pitched 180 innings in 2007, you vwher who pitched 180 Innings in 2007. The estimated standard deviation f an i demand a salary halfway between the midpoint and the upper limit of the 95% prediction interval for the starting salary of an individual starting pitcher who pitched 180 innings. Your salary demand is Your team's owners, who have hired their own statistician to advise them, respond by saying that the regression results are too imprecise to base your pay on the prediction interval., Their counteroffer is to pay you a base salary plus an add-on. Your base salary is equal to the point estimate of the mean salary for starting pitchers who pitched 180 innings last season. To compute your add-on, they'll letx. equal the value of that provides the most precise confidence and prediction interval estimates. Then theyll calculate the margin of error associated with the 95% confidence interval for E(y"). Your add-on is half the margin of error. (Hint: The value of Sy* is smallest when x R.) The counteroffer made by your team's owners is a salary of

Answer 1

From the output table given, the regression equation is-

Salary = -5.312 + 0.0606 IP

And IP = X i.e Performance during the 2007 season

When X = 204

$\therefore$ Salary = -5.312 + 0.0606 * 204

= 7.0504 ( million dollar)

$\therefore$   Your Statistician predicts the salary of a starting pitcher who pitched 204 innings in 2007 to be 7.0504 (million dollars)

we calculate the sample statistic i.e

t = R * [√(N - 2)/ (1-R²)]

and R = √R² =√0.249 = 0.499

  $\therefore$   t = 0.499 * [ √ (50-2)/ 1- 0.249]

= 0.499 * [ √48/0.751 ]

= 0.499 * 7.99

  $\therefore$  t= 3.987

Using the t- distribution table for two tailed, at t-score = 3.987 and $\alpha$ = 0.10 (because confidence is 90%) and at degree of freedom 26 (given in the graph) , we get critical value of t_critical = 1.706

$\therefore$   the t-critical value used in preceding 90% confidence interval is   1.706

The 90% confidence interval is calculated by-

Sample statistic $\pm$ t_critical * standard error of estimate

Sample statistic i.e t calculated = 3.987

S_e (given) = 4.51685

And t_critical = 1.706

$\therefore$ 3.987 $\pm$ 1.706 * 4.51685

$\therefore$   90% Confidence interval is between   -3.7187 to   11.6927