You are conducting a multinomial hypothesis test (α = 0.05) for the claim that all 5 categories are equally likely to be selected. Complete the table.
| Category |
Observed Frequency |
Expected Frequency |
| A |
25 |
|
| B |
11 |
|
| C |
21 |
|
| D |
6 |
|
| E |
9 |
|
Report all answers accurate to three decimal places. But retain unrounded numbers for future calculations.
What is the chi-square test-statistic for this data? (Report answer accurate to three decimal places, and remember to use the unrounded Pearson residuals in your calculations.)
X2= ___________________
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 =___________________
The p-value is...
a) less than (or equal to) α
b) greater than α
This test statistic leads to a decision to...
a) reject the null
b) accept the null
c) fail to reject the null
d) accept the alternative
As such, the final conclusion is that...
a) There is sufficient evidence to warrant rejection of the claim that all 5 categories are equally likely to be selected.
b) There is not sufficient evidence to warrant rejection of the claim that all 5 categories are equally likely to be selected.
c) The sample data support the claim that all 5 categories are equally likely to be selected.
d) There is not sufficient sample evidence to support the claim that all 5 categories are equally likely to be selected.
Stuart T.
The chi-square value is 18.556 and as you have 5 categories you have 4 degrees of freedom. Your p value is .001 which means your results are significant. X2 (4, N=72)=18.556, p=.001. You would reject the null hypothesis. There is sufficient evidence to warrant rejection of the claim that all 5 categories are equally likely to be selected. According to your data category A and C seem much more likely to be selected. I hope this helps. Stuart08/24/19