You are playing a dice game wi
You are playing a dice game with your friend and he seems to becheating (either that or you are really bad at this game). Youdeduce that the dice is not fair. This is, you expect each of theoutcomes to be equally likely, but they do not seem to be coming upthat way. In order to prove your point, you record the outcomes of120 different die rolls and obtain the following frequencies. Run ahypothesis test to determine if the dice is truly fair using α =0.01.
|
Outcome |
1 |
2 |
3 |
4 |
5 |
6 |
|
Frequency |
10 |
25 |
30 |
20 |
30 |
5 |
Answer:
Chi-square test of goodness of fit
Solution:
Here, we have to use chi square test for goodness of fit.
Null hypothesis: H0: The dice is truly fair.
Alternative hypothesis: Ha: The dice is not trulyfair.
We are given level of significance = α = 0.01
Test statistic formula is given as below:
Chi square = ∑[(O – E)^2/E]
Where, O is observed frequencies and E is expectedfrequencies.
We are given
N = 6
Degrees of freedom = df = N – 1 = 6 – 1 = 5
α = 0.01
Critical value = 15.08627
(by using Chi square table or excel)
Calculation tables for test statistic are given as below:
|
Outcome |
O |
E |
(O – E)^2 |
(O – E)^2/E |
|
1 |
10 |
20 |
100 |
5 |
|
2 |
25 |
20 |
25 |
1.25 |
|
3 |
30 |
20 |
100 |
5 |
|
4 |
20 |
20 |
0 |
0 |
|
5 |
30 |
20 |
100 |
5 |
|
6 |
5 |
20 |
225 |
11.25 |
|
Total |
120 |
120 |
|
27.5 |
Chi square = ∑[(O – E)^2/E] = 27.5
P-value = 0.0000456
(By using Chi square table or excel)
P-value < α = 0.01
So, we reject the null hypothesis
There is sufficient evidence to conclude that the dice is nottruly fair.
There is insufficient evidence to conclude that the dice istruly fair.
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