## Question

###### The National Football League (NFL) records a variety of performance data for individuals and team...

c. Develop the estimated regression equation that could be used to predict the percentage of games won given the average number of passing yards per attempt and the number of interceptions thrown per attempt (to 1 decimal). Enter negative value as negative number Win% d. The average number of passing yards per attempt for the Kansas City Chiefs was 6.2 and the number of interceptions thrown per attempt was 0.036. Use the estimated regression equation developed in part (c) to predict the percentage of games won by the Kansas City Chiefs. (Note: For a season the Kansas City Chiefs' record was 7 wins and 9 losses.) Compare your prediction to the actual percentage of games won by the Kansas City Chiefs (to whole number) Predicted percentagee .Yds/Att Int/Att Actual percentage Select your answer

## Answers

Answer:

a).

win% = -60.6+16.5*yds/att

b).

win%= 94.2 +(-1487.8)*Int/Att

c). win% = -14.4+13.3*yds/Att +(-889.9)*int/Att

d).

predicted percentage 36 < actual percentage 44

Excel Addon Megastat used.

Menu used: correlation/Regression ---- Regression Analysis.

Regression Analysis

r²

0.603

n

16

r

0.776

k

1

Std. Error of Estimate

15.398

Dep. Var.

winn%Regression output

confidence interval

variables

coefficients

std. error

t (df=14)

p-value

95% lower

95% upperIntercept

a =

-60.597

ydsatt

b =

16.514

3.582

4.610

.0004

8.832

24.197

ANOVA table

Source

SS

df

MS

F

p-valueRegression

5,039.831

1

5,039.831

21.26

.0004

Residual

3,319.473

14

237.105

Total

8,359.304

15

Regression Analysis

r²

0.393

n

16

r

-0.627

k

1

Std. Error of Estimate

19.036

Dep. Var.

winn%Regression output

confidence interval

variables

coefficients

std. error

t (df=14)

p-value

95% lower

95% upperIntercept

a =

94.173

intatt

b =

-1,487.807

494.088

-3.011

.0093

-2,547.521

-428.093

ANOVA table

Source

SS

df

MS

F

p-valueRegression

3,285.905

1

3,285.905

9.07

.0093

Residual

5,073.399

14

362.386

Total

8,359.304

15

Regression Analysis

R²

0.721

Adjusted R²

0.678

n

16

R

0.849

k

2

Std. Error of Estimate

13.402

Dep. Var.

winn%Regression output

confidence interval

variables

coefficients

std. error

t (df=13)

p-value

95% lower

95% upperIntercept

a =

-14.384

ydsatt

b1 =

13.301

3.406

3.905

.0018

5.942

20.660

intatt

b2 =

-889.874

380.047

-2.341

.0358

-1,710.916

-68.833

ANOVA table

Source

SS

df

MS

F

p-valueRegression

6,024.500

2

3,012.250

16.77

.0003

Residual

2,334.805

13

179.600

Total

8,359.304

15

Predicted values for: winn%

95% Confidence Interval

95% Prediction Interval

ydsatt

intatt

Predicted

lower

upper

lower

upper

Leverage6.2

0.036

36.047

25.730

46.363

5.311

66.782

0.127