*What do you mean by that?This is the only thing given. Suppose you are interest
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Question
*What do you mean by that?This is the only thing given.
Suppose you are interested in testing whether the home field advantage effect is at work at the University of Utah, by looking at the number of points scored by Utah's opponents during the 2016 season. The idea is that if there is a home field advantage effect, the opposing team will average less points scored during Utah home games versus away games. To address this question, perform a hypothesis test at the alpha = 0.05 level. Data can be found at http://www.espn.com/college-football/team/schedule//id 254/year/2016 (@ means an away game, vs. means a home game; include the final game vs. Indiana as an away game)., justify all steps, and state your (practical) conclusion for full credit.Explanation / Answer
Solution
Interpretation of the question
From the given data, we need to test whether Utah University scored better than the opposition in ‘home games’ as compared to ‘away game’. This is same as saying, ‘whether opposition scored worse than Utah University in ‘home games’ as compared to ‘away game’.
We are given the points scored by Utah University and opposition in 12 games, of which 6 were ‘home game’[indicated by VS to the left of opponent name] and 6 were ‘away game’[indicated by @ to the left of opponent name].
Methodology
We will calculate ‘opposition score – Utah score’ for the two sets and then test if the means are the same or different.
Now, to work out solution,
For convenience in explaining and presentation, the given data set is re-tabulated as below:
‘away game’(@)
‘home game’ (VS)
Date
Utah score (U)
Opposition score (O)
O – U =
D1, say
Date
Utah score (U)
Opposition score (O)
O - U=
D2, say
Sept 17
34
17
- 17
Sept 1
24
0
- 24
Oct 1
23
28
5
Sept 10
20
19
- 1
Oct 15
19
14
- 5
Sept 23
31
27
- 4
Oct 22
52
45
- 7
Oct 8
36
23
- 13
Nov 10
49
26
- 23
Oct 29
24
31
7
Nov 26
22
27
5
Nov 19
28
30
2
Total
199
157
- 42
-
163
130
- 33
To test the claim that Utah has a home field advantage effect
We assume D1 ~ N(µ1, 12) and D2 ~ N(µ2, 22) where 12 and 22 are unknown.
We have a sample of 6 observations on each of D1 and D2
H0: µ1 = µ2 Vs HA: µ1 µ2
Test Statistic: t = (D1bar – D2bar)/s{(2/n)}
where n = common sample size for D1 and D2 and D!bar and D2bar are sample means of D1 and D2 respectively and s2 = (s12 + s22), s12 and s22 being the respective sample variances.
Under H0, t ~ t2n-2.
H0 is rejected/accepted at % level of significance if |tcal| >< tcrit
where tcal = calculated value of test statistic, t and
tcrit = upper (/2) percent point of t-distribution with (2n - 2) degrees of freedom. .
Calculations
D1bar = - 7 D2bar = - 5.5 s1 = 11.8068 s2 = 11.5217 s = 11.6651
And |t| = 0.2227
tcrit = t10, 0.025 = 2.228
Since |tcal| < tcrit, H0 is accepted at the given 5% level of significance.
Conclusion: There is no evidence to suggest that there is home field advantage effect DONE
‘away game’(@)
‘home game’ (VS)
Date
Utah score (U)
Opposition score (O)
O – U =
D1, say
Date
Utah score (U)
Opposition score (O)
O - U=
D2, say
Sept 17
34
17
- 17
Sept 1
24
0
- 24
Oct 1
23
28
5
Sept 10
20
19
- 1
Oct 15
19
14
- 5
Sept 23
31
27
- 4
Oct 22
52
45
- 7
Oct 8
36
23
- 13
Nov 10
49
26
- 23
Oct 29
24
31
7
Nov 26
22
27
5
Nov 19
28
30
2
Total
199
157
- 42
-
163
130
- 33
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