Question

Test the claim that the proportion of men who own cats is significantly different than the proportion of women who own cats at the 0.2 significance level. The null and alternative hypothesis would be: H 0 : p M = p F H 1 : p M ≠ p F H 0 : μ M = μ F H 1 : μ M < μ F H 0 : μ M = μ F H 1 : μ M > μ F H 0 : μ M = μ F H 1 : μ M ≠ μ F H 0 : p M = p F H 1 : p M < p F H 0 : p M = p F H 1 : p M > p F The test is: left-tailed right-tailed two-tailed Based on a sample of 40 men, 45% owned cats Based on a sample of 40 women, 50% owned cats The test statistic is: (to 2 decimals) The p-value is: (to 2 decimals) Based on this we: Fail to reject the null hypothesis Reject the null hypothesis

Answer 1

Given that,
possibile chances (x)=18
sample size(n)=40
success rate ( p )= x/n = 0.45
success probability,( po )=0.5
failure probability,( qo) = 0.5
null, Ho:p=0.5
alternate, H1: p!=0.5
level of significance, α = 0.2
from standard normal table, two tailed z α/2 =1.28
since our test is two-tailed
reject Ho, if zo < -1.28 OR if zo > 1.28
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.45-0.5/(sqrt(0.25)/40)
zo =-0.632
| zo | =0.632
critical value
the value of |z α| at los 0.2% is 1.28
we got |zo| =0.632 & | z α | =1.28
make decision
hence value of |zo | < | z α | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != -0.63246 ) = 0.52709
hence value of p0.2 < 0.5271,here we do not reject Ho
ANSWERS
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null, Ho:p=0.5
alternate, H1: p!=0.5
test statistic: -0.63
critical value: -1.28 , 1.28
decision: do not reject Ho
p-value: 0.527 =0.53
we do not have enough evidence to support the claim that the proportion of men who own cats is significantly different
than the proportion of women who own cats .