Question

In: Statistics and Probability

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

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.

 

Solutions

Expert Solution

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


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