Question

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	Squared Pearson Residual
A	9
B	6
C	13
D	9
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.)
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	Expected		Chi square
category		Oi	Ei=total*p		R2i=(Oi-Ei)2/Ei
A		9.0	9.200		0.004
B		6.0	9.200		1.113
C		13.0	9.200		1.570
D		9.0	9.200		0.004
E		9.0	9.200		0.004

chi-square test-statistic X² =2.696

degree of freedom =categories-1=

4

p value =

0.6100

The p-value is greater than α\\

This test statistic leads to a decision to fail to reject the null

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