Question

In a random sample of​ males, it was found that 27 write with their left hands and 223 do not. In a random sample of​ females, it was found that 72 write with their left hands and 464 do not. Use a 0.01 significance level to test the claim that the rate of​ left-handedness among males is less than that among females.

Complete parts​ (a) through​ (b) below.

a) Identify the test statistic. Z= ___

b) Identify the P-value = ____

Answer 1

To test claim that rate of left handedness among males is less than among females we use test of equality of two population proportion. We have, total sample of male =(n1) ={27+223=250}; number of male who write with their left hands (x) ={27} ; Proportion means rate of left handedness among maleis (x/n1) ={27/250}=0.108=(p1) ; total sample of female ={72+464}=536; number of female who write with their left hands (y) =72 ; proportion means rate of left handedness among females ={y/n2}={72/536}=0.1343=(p2) ; we test null hypothesis vs alternative hypothesis as :H0:P1=P2 vreses H1= P1<P2 we have z statistic= Z= (p1-p2) /(√P×Q×[{1/n1}+{1/n2}] where, P is proportion of combined sample P= n1p1+n2p2/(n1+n2) , Q=1-P; by puting all values we get P= 0.1259 and Q=0.8741 then puting all values in formula of z statistic we get (a) value of z statistic is -1. 1087 (b) we find p value using z table ,we have calculated z is -1. 1087 now we look in z table I left column find tenth place of - 1.1087 which is (-1.1) and look in top row find hundred place or 0.01 we get value (0.1335) hence, P value is 0.1335