Question

The National Football League (NFL) records a variety of performance data for individuals and teams. To investigate the importance of passing on the percentage of games won by a team, the following data show the conference (Conf), average number of passing yards per attempt (Yds/Att), the number of interceptions thrown per attempt (Int/Att), and the percentage of games won (Win%) for a random sample of  NFL teams for a season

Team	Conf	Yds/Att	Int/Att	Win%
Arizona Cardinals	NFC	6.7	0.044	49.9
Atlanta Falcons	NFC	7.3	0.024	62.5
Carolina Panthers	NFC	7.3	0.032	37.3
Cincinnati Bengals	AFC	6.0	0.028	56.1
Detroit Lions	NFC	7.1	0.023	62.3
Green Bay Packers	NFC	8.8	0.015	93.9
Houstan Texans	AFC	7.5	0.019	62.2
Indianapolis Colts	AFC	5.6	0.027	12.5
Jacksonville Jaguars	AFC	4.4	0.034	31.5
Minnesota Vikings	NFC	5.7	0.035	18.9
New England Patriots	AFC	8.5	0.022	81.1
New Orleans Saints	NFC	8.2	0.020	81.0
Oakland Raiders	AFC	7.4	0.045	50.1
San Francisco 49ers	NFC	6.3	0.012	81.5
Tennessee Titans	AFC	6.5	0.025	56.5
Washington Redskins	NFC	6.6	0.043	31.4

a. 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 (to 1 decimal). Enter negative value as negative number.

WIN%=______+_______*Yds/Att

b. Develop the estimated regression equation that could be used to predict the percentage of games won given the number of interceptions thrown per attempt (to 1 decimal). Enter negative value as negative number.

      WIN%=______+_______*INT/Att

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%=______+_______*Yds/Att+______*INT/Att

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 percentage		Actual percentage
	- Select your answer < > =

Answer 1

a)

Using Excel

data -> data analysis -> regression

SUMMARY OUTPUT
Regression Statistics
Multiple R	0.73309
R Square	0.537421
Adjusted R Square	0.50438
Standard Error	16.57176
Observations	16
ANOVA
	df	SS	MS	F	Significance F
Regression	1	4466.784	4466.784	16.26513	0.001234
Residual	14	3844.726	274.6233
Total	15	8311.509
	Coefficients	Standard Error	t Stat	P-value	Lower 95%	Upper 95%
Intercept	-49.2552	26.00747	-1.89388	0.07909	-105.036	6.525323
Yds/Att	15.07536	3.737998	4.033005	0.001234	7.058157	23.09257

Win% ^ = -49.3 + 15.1 Yds/Att

b)

SUMMARY OUTPUT
Regression Statistics
Multiple R	0.665319
R Square	0.442649
Adjusted R Square	0.402839
Standard Error	18.19031
Observations	16
ANOVA
	df	SS	MS	F	Significance F
Regression	1	3679.084	3679.084	11.11884	0.004913
Residual	14	4632.425	330.8875
Total	15	8311.509
	Coefficients	Standard Error	t Stat	P-value	Lower 95%	Upper 95%
Intercept	97.74139	13.80055	7.082427	5.49E-06	68.14216	127.3406
Int/Att	-1551.7	465.3487	-3.33449	0.004913	-2549.78	-553.628

In%^ = 97.7 - 1551.7 Int/Att

c)

SUMMARY OUTPUT
Regression Statistics
Multiple R	0.851424
R Square	0.724923
Adjusted R Square	0.682604
Standard Error	13.26158
Observations	16
ANOVA
	df	SS	MS	F	Significance F
Regression	2	6025.207	3012.604	17.12978	0.000227
Residual	13	2286.302	175.8694
Total	15	8311.509
	Coefficients	Standard Error	t Stat	P-value	Lower 95%	Upper 95%
Intercept	4.258101	27.50143	0.154832	0.879332	-55.1551	63.67132
Yds/Att	11.689	3.200347	3.652416	0.002924	4.775069	18.60293
Int/Att	-1080.47	362.9652	-2.97679	0.010706	-1864.61	-296.331

Win%^ = 4.3 + 11.7 Yds/Att -1080.5 Int/Att

d)

Win%^ = 4.3 + 11.7 Yds/Att -1080.5 Int/Att

= 37.83%

Actual = 7/(7+9) = 0.4375 = 43.75%