Question

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

Category	Observed Frequency	Expected Frequency	Squared Pearson Residual
A	21
B	12
C	16
D	8
E	24

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.)
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 =  
It may be best to use the =CHIDIST() function in Excel to do this calculation.

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

Observed (Oi )	Expected ( Ei)	Oi-Ei	(Oi-Ei)^2	(Oi-Ei)^2/Ei
21	16.2	4.8	23.04	1.4222
12	16.2	-4.2	17.64	1.0889
16	16.2	-0.2	0.04	0.0025
8	16.2	-8.2	67.24	4.1506
24	16.2	7.8	60.84	3.7556

observed frequncies are Oi
21 12 16 8 24
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expected frequencies are Ei = (21 + 12 + 16 + 8 + 24)/5 = 16.2
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and claiming hypothesis is
null, Ho: all 5 categories are equally likely to be selected.
alternative, H1: all 5 categories are not equally likely to be selected.
level of significance, alpha = 0.05
from standard normal table, chi square value at right tailed, ᴪ^2 alpha/2 =9.4877
since our test is right tailed,reject Ho when ᴪ^2 o > 9.4877
we use test statistic ᴪ^2 o = Σ(Oi-Ei)^2/Ei
from the table take sum of (Oi-Ei)^2/Ei, we get ᴪ^2 o = 10.4198
critical value
the value of |ᴪ^2 alpha| at los 0.05 with d.f, n - 1 = 5 - 1 = 4 is 9.4877
we got | ᴪ^2| =10.4198 & | ᴪ^2 alpha | =9.4877
make decision
hence value of | ᴪ^2 o | > | ᴪ^2 alpha| and here we reject Ho
ᴪ^2 p_value =0.0339
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What is the chi-square test-statistic for this data? test statistic: 10.4198
What are the degrees of freedom for this test?    d.f = 4
What is the p-value for this sample? p-value = 0.0339
The p-value is...? less than (or equal to) αα
This test statistic leads to a decision to...? reject the null
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.