Question

1. A flashback to a problem from HW5 The #MeToo movement has brought new attention to recent challenges faced by women around the world. One of the original challenges faced by women was the inability to vote. New Zealand was the first country to allow women this right (1893). In the years since, various countries have given this right as well. Blois a table that summarizes each year since 1893 and records how many new countries gave women the right to vote in each year. Do the data in this table appear to roughly follow a Poisson distribution? Number of New Countries Giving Women the Right to Vote in a Given Year 0 How Many Years That Saw That Level of Change 82 25 4 4 This time, we'll answer the above question using a chi-squared approach. Since we must have expected counts at least 5, we'll reorganize the chart so there is hope Number of New Countries Giving Women the Right to Vote in a Given Year 0 How Many Years That Saw That Level of Change 82 25 2 or more a. State null and alternative hypotheses for the analysis to follow. Use problem. It might help to read all the parts belovw 0.362 throughout this b. Find the expected counts for the 0, 1 and 2+ categories assuming the null hypothesis

Please specifically do the d,e,f,g part THX!

Answer 1

(a) null hypothesis H0: observed frequency=expected frequency( follow poission distribution with $\lambda=$0.362)

altenate hypothesis H1: observed frequency$\ne$expected frequency

( does not follow poission distribution with $\lambda=$0.362)

(b)P(X=0)=0.6963 ( using ms-excel =poisson(0,0.362,0)

total frequency=N=82+25+6=113

E(X=0)=P(X=0)*N=0.6963*113=78.7

and similarly for others

P(X>=2)=1-P(X=0)-P(X=1)=1-0.6963-0.2501=0.0516

x	P(X=x)	Expected
0	0.6963	78.7
1	0.2521	28.5
2.00	0.05166	5.8

(c) chi-square=sum((O-E)²/E) =0.6 with k-1=3-1=2 df

x	P(X=x)	Expected(E)	Observed(O)	(O-E)	(O-E)2/E
0	0.696282368	78.7	82	3.3	0.1401
1	0.252054217	28.5	25	-3.5	0.4257
2 or more	0.051663415	5.8	6	0.2	0.0045
	sum=	113.0	113.0	0.0	0.6

(d) your text book problem

(e) p-value=0.7519

(g) since p-value is more than typical value=0.05 of level of sginificnace(alpha), so we fail to reject null hypothesis H0 and conclude that data follow poission distribution with $\lambda=$0.362