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## Explanation & Answer

View attached explanation and answer. Let me know if you have any questions.Here is the jmp output available. Finishing on the manual computations

JMP OUTPUT:

A-E

H-I

Confidence interval for part H and the prediction interval respectively as shown in row 7 for x= 15.In

Summary as shown Below

View attached explanation and answer. Let me know if you have any questions.

Chapter 7

11. The following data represents the number of hits (x) and the number of strikeouts (y) for players on a

college baseball team. Use the provided data below to complete parts a through f.

π

π

41

20

29

23

39

27

11

10

24

12

1

3

a. Use the above dataset to compute π½Μ 0, π½Μ 1, and the least squares line.

(π₯ β π₯Μ
)

(y-π¦Μ
)

(x-π₯Μ
)2

(y-π¦Μ
)2

(x-π₯Μ
)(y-π¦Μ
)

16.83333333

4.833333333

14.83333333

-13.16666667

-0.166666667

-23.16666667

4.166666667

7.166666667

11.16666667

-5.833333333

-3.833333333

-12.83333333

283.3611111

23.36111111

220.0277778

173.3611111

0.027777778

536.6944444

1236.833333

17.36111111

51.36111111

124.6944444

34.02777778

14.69444444

164.6944444

406.8333333

70.13888889

34.63888889

165.6388889

76.80555556

0.638888889

297.3055556

645.1666667

β

π½Μ 1 = 645.167/1236.833 = 0.5216

Xbar =24.16666667 ybar =15.83333333

π½Μ 0 = 15.8333 β 0.5216*24.1667 = 3.2273

b. Calculate π and π
2 . Interpret π
2 in the context of this baseball application.

(π¦Μ)

(y-π¦Μ)

(y-π¦Μ)2

24.61407

18.35453

23.57081

8.965234

15.7464

3.748956

-4.61407

4.645466

3.429187

1.034766

-3.7464

-0.74896

21.28963

21.58035

11.75933

1.070741

14.03548

0.560935

70.29646

Total=

R2 = 1- (70.296456/406.8333) = 0.8272

r= sqrt(R2) = sqrt(0.8272) = 0.9095

The r squared value indicates that 82.72% of the variation in number...