Question

please help fast

You are conducting a multinomial hypothesis test (a = 0.05) for the claim that all 5 categories are equally likely to be selected. Complete the table. Observed Expected Category Frequency Frequency А 13 B 5 с 14 D 15 E 8 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.) 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- Is the p-value less than a? yes no 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

1)

applying chi square goodness of fit test:

	relative	observed	Expected		Chi square
Category	frequency(p)	Oi	Ei=total*p		R2i=(Oi-Ei)2/Ei
1	1/5	13	11.00		0.364
2	1/5	5	11.00		3.273
3	1/5	14	11.00		0.818
4	1/5	15	11.00		1.455
5	1/5	8	11.00		0.818
total	1.00	55	55		6.73
test statistic X2=		6.727

degree of freedom =categories-1=

4

p value =

0.1510

from excel: chidist(6.727,4)

fail to reject the null since p value >0.05

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

2)

OH,: The proportion of people within each treatment group who experienced dizziness are the same He: The proportion of people within each treatment group who experienced dizziness are different

Applying chi square test of homogeneity

Expected	Ei=row total*column total/grand total	Tordol	Placebo	Naproxen	Tylenol	Asvil	Total
	Dizziness	83.8609	37.7030	27.6300	7.8278	6.9783	164
	No Dizziness	4062.1391	1826.2970	1338.3700	379.1722	338.0217	7944
	total	4146	1864	1366	387	345	8108
chi square    χ2	=(Oi-Ei)2/Ei	Tordol	Placebo	Naproxen	Tylenol	Asvil	Total
	Dizziness	0.009	0.863	2.536	1.022	0.150	4.5782
	No Dizziness	0.000	0.018	0.052	0.021	0.003	0.0945
	total	0.0090	0.8805	2.5879	1.0426	0.1527	4.6727
test statistic X2=		4.67
p value =	0.3226	from excel: chidist(4.6727,4)

fail to reject the null

There is not sufficient evidence to support the claim that the proportion of people within each treatment group who experienced dizziness are different