In: Statistics and Probability
Aron | |||||
1 | 2 | 3 | 4 | Average | |
8/6/2017 |
90 | 138 | 118 | 105 | 112.8 |
8/19/2017 | 162 | 101 | 120 | 145 | 132 |
9/16/2017 | 101 | 129 | 132 | 111 | 118.3 |
Average | 117.7 | 122.7 | 123.3 | 120.3 | 121 |
Mjorgan | |||||
1 | 2 | 3 | 4 | Average | |
8/6/2017 | 115 | 88 | 94 | 102 | 99.8 |
8/19/2017 | 89 | 75 | 77 | 90 | 82.8 |
9/16/2017 | 74 | 110 | 117 | 90 | 97.8 |
Average | 92.7 | 91 | 96 | 94 | 93.4 |
Aron believes there is a distinct difference between not only the games played each night, AND within the date on which they bowled, but also between himself and Mjorgan. Construct a statistical test to determine if this might be the case – where are the real variations in their bowling? Interpret your results in a few sentences.
We will be using ANOVA to test the various hypothesis.
a)
For Aron,
Null and Alternative Hypothesis:
H0: µGame 1 = µGame 2 = µGame 3 = µGame 4
H1: Not all Means are equal
Alpha = 0.05
N = 12
n = 3
Degress of Freedom:
dfBetween = a – 1 = 4-1 =3
dfWithin = N-a = 12-4 = 8
dfTotal = N-1 = 12-1 = 11
Critical Values:
Game (dfBetween, dfWithin) : (3,8) = 4.07
Decision Rule:
If F is greater than 4.07, reject the null hypothesis
Test Statistics:
SSBetwen = ?(?ai)2/n - T2/N = 59.33
SSWithin = ?(Y)2 - ?(?ai)2/n = 4798.67
SSTotal = SSBetwen + SSWithin = 4858
MS = SS/df
F = MSeffect / MSerror
Hence,
F = 19.78/599.83 = 0.03297
Result:
Our F = 0.032972, we fail to reject the null hypothesis
Conclusion:
All means are equal. Ie There is no distinction between games played by Aron.
b)
For Aron,
Null and Alternative Hypothesis:
H0: µDay 1 = µDay 2 = µDay 3
H1: Not all Means are equal
Alpha = 0.05
N = 12
n = 4
Degress of Freedom:
dfBetween = a – 1 = 3-1 =2
dfWithin = N-a = 12-3 = 9
dfTotal = N-1 = 12-1 = 11
Critical Values:
Day (dfBetween, dfWithin) : (2,9) = 4.26
Decision Rule:
If F is greater than 4.26, reject the null hypothesis
Test Statistics:
SSBetwen = ?(?ai)2/n - T2/N = 786.5
SSWithin = ?(Y)2 - ?(?ai)2/n = 4071.5
SSTotal = SSBetwen + SSWithin = 4858
MS = SS/df
F = MSeffect / MSerror
Hence,
F = 393.25/452.39 = 0.869
Result:
Our F = 0.869, we fail to reject the null hypothesis
Conclusion:
All means are equal. Ie There is no distinction between games played on different days by Aron.
c)
For Aron & Mjorgan
Null and Alternative Hypothesis:
We will have three hypotheses:
H0: µAron = µMjorgan
H1: Not all Means are equal
H0: µGame 1 = µGame 2 = µGame 3 = µGame 4
H1: Not all Means are equal
H0: An interaction is absent
H1: Interaction is present
Alpha = 0.05
Degress of Freedom:
DfPlayer(A) = a-1 = 2-1 = 1
DfGames (B) = b-1 = 4-1 = 3
df Player * Games (A*B) = (a-1) * (b-1) = 1*3 = 3
df error = N – ab = 24 – 2*4 = 16
df total= N – 1 = 24 – 1 = 23
Decision Rule (3):
We have three hypotheses, so we have three decision rules:
Critical Values:
Player (dfPlayer(A), df error) : (1,16) = 4.49
Games (dfGames (B),df error) : (3,16) = 3.24
Player*Games (df Player * Games (A*B), df error) : (3,16) = 3.24
[Player] If F is greater than 4.49, reject the null hypothesis
[Games] If F is greater than 3.24, reject the null hypothesis
[Interaction] If F is greater than 3.24, reject the null hypothesis
Test Statistics:
SSPlayer = ?(?ai)2/b*n - T2/N = 4565.04
SSGames = ?(?bi)2/a*n - T2/N = 62.12
SSPlayer*Games = ?(?ai * bi)2/n - ?(?ai)2/b*n - ?(?bi)2/a*n + T2/N = 37.46
SSTotal = ?(Y)2 - T2/N = 11851.96
SSError = SSTotal - SSPlayer - SSGames - SSPlayer*Games = 7187.33
MS = SS/df
F = MSeffect / MSerror
Hence,
FPlayer = 4565.04/449.21 = 10.16
FGames = 20.71/449.21= 0.05
FInteraction = 12.49/449.21 = 0.03
Results:
[Player] If F is greater than 4.49, reject the null hypothesis
Our F = 10.16, we reject the null hypothesis.
[Games] If F is greater than 3.24, reject the null hypothesis
Our F = 0.05, we retain the null hypothesis.
[Interaction] If F is greater than 3.24, reject the null hypothesis
Our F = 0.03, we retain the null hypothesis.
Conclusion:
There is distinct difference only between the players.
For a Player, there is no difference in Games or Days