Problem set 6:
Suppose a die is thrown 180 times by a Casino manager to see if
there was a difference in the
outcomes of the six faces of a die. The outcomes was recorded as
follows; 1 appears 25 times, 2-40
times, 3-37 times, 4-22 times, 5-24 times, 6-32 times. Test at .01
level of significance. Follow the
steps in testing hypothesis.
Ho: There was no difference in the outcomes of the six faces of a die
Ha: There was difference in the outcomes of the six faces of a die
Level of Significance = .01, two-tailed
DF: 6-1= 5 degrees of freedom.
So,
Chi-square test:
Critical value = 15.09
Do not reject the null hypothesis if the computed value is -
15.09<x2<15.09
Where,
fo = observed frequency in any category
fe = expected frequency in any category
Faces of die fo fe Fo-fe (Fo-fe)2(Fo-fe)2/fe
1 25 30 -5 25 0.833…
2 40 30 10 100 3.333…
3 37 30 7 49 1.633….
4 22 30 -8 64 2.133…
5 24 30 -6 36 1.20
6 32 30 2 4 0.133…
X2=9.267
The computed X2 value 9.267 is less than 15.09. The two-tailed P value
equals 0.0989. Thus, we do not reject the null hypothesis. This
difference is considered to be not quite statistically significant