Question

A survey of high school girls classified them by two attributes: whether or not they participated in sports and whether or not they had one or more older brothers.

Older brother	Participated In sports		TOTAL
	Yes	No
Yes	12	8	20
No	13	27	40
TOTAL	25	35	60

1. Use the following data to test the null hypothesis that these two attributes of classification are independent.
2. Suppose we do the below hypothesis test instead:

where,

p1-Propotion of young girls participated in sports of those who has a brother

p2-Propotion of young girls participated in sports of those who does not have a brother.

What would be the test statistic and the p value for this test?

Answer 1

a) to test the independence we Use the Null Hypothesis Ho: These two atributes are independent

The alternate hypothesis is that H1: these two attributes have some relationship

The test statistic we use for the idepence test is the Chi  square test The Calculation is shown in the given below table

The Test Statistic will be where Oi is the observed frequency which is the given values and Ei is the expected frequency we calculate Eij = row total ix column totalj devided by Grand total

The working is shown below

The Chi Square value = 53.78

to calculate p value = df of reedom (r-1) x (c-1)= 2-1x2-1=1 and =0.05 we take

The p value P<.00001 which is very less than 0.05 the significant level so we cannot acceot the null hypothesis of the two attributes are independent so we have to take that there is some relationship between the attributes having elder brother and participated in sports

b) We use the Null Hypothesis as Ho: P1= P2 which is a two tailed test

And we use The test statistic of z

Where z= (p1-P2)/S.E

Where SE =

for this we calculate p as p=

where p1 is the proportion from sample of size n1 and p2 is the proportion from sample of size n2

Since we are assuming that the Z as test statistic , it is a two tailed test was the alternate hypothesis wil have not equal to sign .

So the p value can be calculated of Z score .