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.CategoryObservedFrequencyExpectedFrequency A10 B16 C8 D20 E14 What is the chi-square test-statistic for this data? (Report answer accurate to three decimal places.)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) ααgreater than ααThis test statistic leads to a decision to...reject the nullaccept the nullfail to reject the nullaccept the alternativeAs 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.

As per our guidelines we can solve first three subpart and rest can be reposted. Solution:- We find, (a) Test statistic =? (b) Degree of freedom (c) p-value=? We use χ2-test Here, we want to test H0: All 5 categories are equally likely to be selected. Vs Ha: All 5 categories are not equally likely to be selected. We prepare the following table: Category Observed Freq (Oi) Expected Freq (fi) (Oi^2/ei) A 10 13.6 7.353 B 16 13.6 18.824 C 8 13.6 4.706 D 20 13.6 29.412 E 14 13.6 14.412 Total 68 68 74.706 N=∑Oi=∑ei=68,∑Oi2ei=74.706 Expected frequencies can calculated as follows, =SumofallfrequenciesTotalNumberofFrequency=685=13.6 The test statistic is χStat2=∑i=15(oi-ei)2ei=∑i=15(oi2)ei-N =74.706-68 =6.706 (b) For degree of freedom: d.f.=k-p-1=5-0-1=4 Where, p-Number of parameters estimated for fitting the probability distribution. k- Total number of observed frequencies. (c) p-value:- P-value=P[χd.f.2>χstat2]=P[χ42>6.706]=0.1523 By using MS-Excel Command "=CHIDIST(6.706,4)" Here, p-value=0.1523>α=0.05 (Level of significance) hence we fail to reject H0. The p-value is greater than α This test statistic leads to a decision to fail to reject the null Option (3) The sample data support the claim that all 5 categories are equally likely to be selected is correct

Answered: You are conducting a multinomial…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

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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	10
B	16
C	8
D	20
E	14

What is the chi-square test-statistic for this data? (Report answer accurate to three decimal places.)
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.

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