You are conducting a multinomial hypothesis test (αα = 0.05) for
the claim that all 5 categories are equally likely to be
selected.
Category | Observed Frequency |
---|---|
A | 14 |
B | 19 |
C | 11 |
D | 8 |
E | 21 |
What is the chi-square test-statistic for this data?
χ2=χ2=
What are the degrees of freedom for this test?
d.f. =
What is the p-value for this sample? (Report answer accurate to
four decimal places.)
p-value =
Hypothesis:
Ho: All categories are equally likely to be selected
Ha: All 5 categories are Not equally likely to be selected
Total of Observed Frequencies are = 14+ 19 + 11 + 8 + 21 = 73
Expected Frequency of each category = 73/5 = 14.6 category
Computational Table:
Oi | Ei | (Oi-Ei) | (Oi-Ei)2 | (Oi-Ei)2/Ei | |
14 | 14.6 | -0.6 | 0.36 | 0.025 | |
19 | 14.6 | 4.4 | 19.36 | 1.326 | |
11 | 14.6 | -3.6 | 12.96 | 0.888 | |
8 | 14.6 | -6.6 | 43.56 | 2.984 | |
21 | 14.6 | 6.4 | 40.96 | 2.805 | |
Total | 73 | 73 | 8.027 |
Test statistic:
Degrees of Freedom (d.f) = k-1 = 5-1 = 4
Where, K = Number of Categories = 5
P-value: 0.0906 ...........................(Using Chi square Table)
P-value > , i.e 0.0906 > 0.05, That is Fail to Reject Ho at 5% level of significance.
Therefore, All categories are equally likely to be selected
Get Answers For Free
Most questions answered within 1 hours.