Question

You are conducting a multinomial hypothesis test ( α = 0.05) for the claim that all 5 categories are equally likely to be selected.

Category	Observed Frequency
A	23
B	17
C	13
D	19
E	5

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

Solution:

Claim: all 5 categories are equally likely to be selected.

Level of significance = 0.05

Hypothesis are:

H0: all 5 categories are equally likely to be selected.

Vs

H1: At least one categories is different from others.

Part 1) What is the chi-square test-statistic for this data?

Chi square test statistic for goodness of fit

Where

O_i = Observed Counts

E_i =Expected Counts = N / k = 77 / 5 = 15.400

Thus we need to make following table:

Category	Oi: Observed frequency	Ei: Expected frequency	Oi^2/Ei
A	23	15.400	34.351
B	17	15.400	18.766
C	13	15.400	10.974
D	19	15.400	23.442
E	5	15.400	1.623
	N = 77

Thus

Part 2) What are the degrees of freedom for this test?
d.f. = k - 1

d.f. = 5 - 1

d.f. = 4

Part 3) What is the p-value for this sample?

Use following Excel command:

=CHISQ.DIST.RT(x,df)

=CHISQ.DIST.RT( 12.156, 4)

=0.0162

Thus

p-value = 0.0162

Part 4) The p-value is:

p-value = 0.0162 < α = 0.05

thus correct answer is:
less than (or equal to) α

Part 5) This test statistic leads to a decision to

reject the null

Part 6) 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