Question

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	20
B	15
C	15
D	12
E	18

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 =


The p-value is...

• less than (or equal to) αα
• greater than αα

This test statistic leads to a decision to...

• reject the null
• accept the null
• fail to reject the null
• accept the alternative

As such, the final conclusion is that...

• There is sufficient evidence to warrant rejection of the claim that all 5 categories are equally likely to be selected.
• There is not sufficient evidence to warrant rejection of the claim that all 5 categories are equally likely to be selected.
• The sample data support the claim that all 5 categories are equally likely to be selected.
• There is not sufficient sample evidence to support the claim that all 5 categories are equally likely to be selected.

Answer 1

H0: P_i = 1/5, where i = A, B, C, D and E

H1: H0 is not true

n = (20 + 15 + 15 + 12 + 18) = 80

Category	Observed Frequency (fi)	Expected Frequency (ei = n * pi)
A	20	80 * (1/5) = 16
B	15	80 * (1/5) = 16
C	15	80 * (1/5) = 16
D	12	80 * (1/5) = 16
E	18	80 * (1/5) = 16

(a)

$\chi_0^2 = \sum_{i = 1}^{5} \frac{(f_i - e_i)^2}{e_i} = \frac{4^2 + 1^2 + 1^2 + 4^2 + 2^2}{16} = 2.375$(Ans)

(b)

d.f = k -1 = 5 - 1 = 4 (Ans)

(c)

p value = 0.66715

(d)

p value is greater than $\alpha = 0.05$

(e)

We fail to reject the null hypothesis

(f)

There is not sufficient evidence to warrant rejection of the claim that all 5 categories are equally likely to be selected

** If the answers do not match please comment