4) Data were collected to explain the number of wins an NFL team has based on the average points per game that they score (PPG) Complete the Table Below Regression Statistics Multiple R R Square Adjusted R Square Standard Error Observations 0.571 11 ANOVA Significance F 0.00430874 MS F df 68.0 Regression Residual 42.7 Total Standard Error Upper 95% t Stat P-value Lower 95% Coefficients -6.758 3.835 0.112 Intercept 0.673 0.178 0.004 PPG Based on the regression results, answer the following questions a) What is the estimated regression equation? Nu mber of wins C.758+0.673 x PPG b) What percentage of the wins is explained by PPG? R2 =1 residual sum of squvare total um of square =| Res.dual sum of 'squart regressicn sum of quarl+res.dal Sum of sauare -0.3857 = 0.6ILR cf wins GI,43'% explained by pP G -42.768142,= %3D c) What is the standard error of the erpor term in the regression equation? SE : Residua! sum of squares n -k uhere nil and ヒ=Z SE =42,7 -2 =42.77 =4,74" d) Is the coefficient on the variable "PPG" statistically significantly different than 0 at the 1% level of significance? How do you know? Ho: B1=0 = 2.1782 nニ! cSerrati^ P-value =o. OG 4 reject the hull hye since ca z0, ol k:1 Ind vari HaiBit o PPG conciusior: theres Sufficent evi to Conciude the variable is sig keg equ : -6,158 KGC73 PPG Win = -G.58+0. G3 PPG win =-6.75810.673 e) What is the predicted number of wins for a team that averages scoring 28 points per game? win=-6,758t18,84 predicted # of wins tor team avy Scores 28 points per win= 12,80G ) What is the effect on wins of increasing the average PPG by 1? game 13 A Interpretation slope: the avg scoring Per game >by I thea predicte d # of wins > by o.c73 Increasing the I then the predicred Effect of nunmber of wins. avg PPG # of wi ns increases oy 6, 673.
4) Data were collected to explain the number of wins an NFL team has based on the average points per game that they score (PPG) Complete the Table Below Regression Statistics Multiple R R Square Adjusted R Square Standard Error Observations 0.571 11 ANOVA Significance F 0.00430874 MS F df 68.0 Regression Residual 42.7 Total Standard Error Upper 95% t Stat P-value Lower 95% Coefficients -6.758 3.835 0.112 Intercept 0.673 0.178 0.004 PPG Based on the regression results, answer the following questions a) What is the estimated regression equation? Nu mber of wins C.758+0.673 x PPG b) What percentage of the wins is explained by PPG? R2 =1 residual sum of squvare total um of square =| Res.dual sum of 'squart regressicn sum of quarl+res.dal Sum of sauare -0.3857 = 0.6ILR cf wins GI,43'% explained by pP G -42.768142,= %3D c) What is the standard error of the erpor term in the regression equation? SE : Residua! sum of squares n -k uhere nil and ヒ=Z SE =42,7 -2 =42.77 =4,74" d) Is the coefficient on the variable "PPG" statistically significantly different than 0 at the 1% level of significance? How do you know? Ho: B1=0 = 2.1782 nニ! cSerrati^ P-value =o. OG 4 reject the hull hye since ca z0, ol k:1 Ind vari HaiBit o PPG conciusior: theres Sufficent evi to Conciude the variable is sig keg equ : -6,158 KGC73 PPG Win = -G.58+0. G3 PPG win =-6.75810.673 e) What is the predicted number of wins for a team that averages scoring 28 points per game? win=-6,758t18,84 predicted # of wins tor team avy Scores 28 points per win= 12,80G ) What is the effect on wins of increasing the average PPG by 1? game 13 A Interpretation slope: the avg scoring Per game >by I thea predicte d # of wins > by o.c73 Increasing the I then the predicred Effect of nunmber of wins. avg PPG # of wi ns increases oy 6, 673.
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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Question
Data were collected to explain the number of wins an NFL team has based on the average points per game that they score (PPG)
*COMPLETE THE TABLE BELOW*
Transcribed Image Text:Efect of number ef winsiincreasing the
4) Data were collected to explain the number of wins an NFL team has based on the average points per
game that they score (PPG)
Complete the Table Below
Regression
Statistics
Multiple R
R Square
Adjusted R Square
Standard Error
0.571
Observations
11
ANOVA
Significance F
0.00430874
MS
F
df
68.0
Regression
Residual
42.7
Total
Upper
95%
Standard
Error
t Stat
P-value
Lower 95%
Coefficients
-6.758
3.835
0.112
Intercept
0.673
0.178
0.004
PPG
Based on the regression results, answer the following questions
a) What is the estimated regression equation?
Nu mber of wins
6.758+0.673 x PPG
b) What percentage of the wins is explained by PPG?
R2 =1 residual sum of squvare total sum of square=1
of wins
Res.dual sum of 'square regression sum of yuarl+res.og! Sum of square
-42.768142,7 : -0.3857 = 0.61B
GI,43', explained
by pPG
c) What is the standard error of the erfor term in the regression equation?
SE : Residua! sum of Squares
n -k where nill and
ヒ=Z
SE =42,711 -2 =4277 :4.744 =2,178Z
Is the coefficient on the variable “PPG" statistically significantly different than 0 at the 1% level
of significance? How do you know?
HoiBi=0
Ha:
n- obserration
reject the hull hye s ince ca z0, o l
conclusion: theres Suffient evi to Conciude the variable is sia
P-value =C, OG4
kE1 Ind vari
keg equ
: -6,158 KGG73PPG
e) What is the predicted number of wins for a team that averages scoring 28 points per game?
Win = -6.58+0. G 3 PPG win =-6.758+0.673
win=-6,758+18,84
predicted # of wins tor team avg Scores 28 points per
win=12,80G
) What is the effect on wins of increasing the average PPG by 1?
game
S13 A
Interpretation slope: the avg scoring Per game > by I thea predicte d
# of wins > by o.c73
avg
I then the Predicred
PPG
# of wi ns increases oy a 673.
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