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 16 NFL teams for a season

Team	Conf	Yds/Att	Int/Att	Win%
Arizona Cardinals	NFC	6.4	0.044	50.0
Atlanta Falcons	NFC	7.3	0.023	62.6
Carolina Panthers	NFC	7.3	0.033	37.4
Cincinnati Bengals	AFC	6.1	0.027	56.3
Detroit Lions	NFC	7.2	0.024	62.3
Green Bay Packers	NFC	8.9	0.014	93.8
Houstan Texans	AFC	7.3	0.018	62.6
Indianapolis Colts	AFC	5.6	0.025	12.4
Jacksonville Jaguars	AFC	4.8	0.031	31.0
Minnesota Vikings	NFC	5.6	0.034	18.7
New England Patriots	AFC	8.1	0.022	81.3
New Orleans Saints	NFC	8.3	0.021	81.1
Oakland Raiders	AFC	7.6	0.046	49.7
San Francisco 49ers	NFC	6.7	0.011	81.4
Tennessee Titans	AFC	6.9	0.023	56.4
Washington Redskins	NFC	6.5	0.042	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 -<>=Item 9

Answer 1

a)

using excel data analysis tool for regression,steps are: write data>menu>data>data analysis>regression>enter required labels>ok> and following o/p is obtained

Regression Statistics
Multiple R	0.7928
R Square	0.6285
Adjusted R Square	0.6020
Standard Error	14.9013
Observations	16
ANOVA
	df	SS	MS	F	Significance F
Regression	1	5258.9	5258.9	23.68	0.0002
Residual	14	3108.7	222.0
Total	15	8367.6
	Coefficients	Standard Error	t Stat	P-value	Lower 95%	Upper 95%
Intercept	-66.1151	25.0170	-2.6428	0.0193	-119.7712	-12.4591
X	17.4163	3.5787	4.8666	0.0002	9.7407	25.0919

Win%= -66.1 + 17.4*Yds/Att

b)

Regression Statistics
Multiple R	0.6169
R Square	0.3805
Adjusted R Square	0.3363
Standard Error	19.2418
Observations	16
ANOVA
	df	SS	MS	F	Significance F
Regression	1	3184.1	3184.1	8.60	0.0109
Residual	14	5183.5	370.2
Total	15	8367.6
	Coefficients	Standard Error	t Stat	P-value	Lower 95%	Upper 95%
Intercept	93.0661	14.0752	6.6121	0.0000	62.8778	123.2545
X	-1417.0266	483.2027	-2.9326	0.0109	-2453.3934	-380.6599

Win% = 93.1 -1417.0 * Int/Att

c)

SUMMARY OUTPUT
Regression Statistics
Multiple R	0.873337
R Square	0.762717
Adjusted R Square	0.726212
Standard Error	12.35841
Observations	16
ANOVA
	df	SS	MS	F	Significance F
Regression	2	6382.116	3191.058	20.89342	8.69E-05
Residual	13	1985.494	152.7303
Total	15	8367.61
	Coefficients	Standard Error	t Stat	P-value	Lower 95%	Upper 95%	Lower 95.0%	Upper 95.0%
Intercept	-21.1787	26.55292	-0.7976	0.439427	-78.5428	36.18539	-78.5428	36.18539
Yds/Att	14.46541	3.161222	4.575892	0.00052	7.636005	21.29481	7.636005	21.29481
Int/Att	-896.381	330.5452	-2.71183	0.017788	-1610.48	-182.282	-1610.48	-182.282

Win= -21.2 + 14.5*Yds/Att - 896.4 * Int/Att

d)

predicted Win= -21.2 + 14.5*6.2 - 896.4 * 0.036 = 36.24 %

actual percentage = 7/16 = 43.75

predicted percentage < actual