Question

Test the claim that the proportion of men who own cats is smaller than 60% at the .05 significance level.

The null and alternative hypothesis would be:

H0:μ=0.6H0:μ=0.6
H1:μ<0.6H1:μ<0.6

H0:p=0.6H0:p=0.6
H1:p>0.6H1:p>0.6

H0:μ=0.6H0:μ=0.6
H1:μ≠0.6H1:μ≠0.6

H0:p=0.6H0:p=0.6
H1:p<0.6H1:p<0.6

H0:p=0.6H0:p=0.6
H1:p≠0.6H1:p≠0.6

H0:μ=0.6H0:μ=0.6
H1:μ>0.6

Correct The test is:

(A) two-tailed

(B) right-tailed

(C) left-tailed

Correct Based on a sample of 100 people, 53% owned cats:

The test statistic is: _____ (to 2 decimals)

The critical value is: _____ (to 2 decimals)

Based on this we:

(A) Reject the null hypothesis

(B) Fail to reject the null hypothesis

Answer 1

Given that,
possibile chances (x)=53
sample size(n)=100
success rate ( p )= x/n = 0.53
success probability,( po )=0.6
failure probability,( qo) = 0.4
null, Ho:p=0.6
alternate, H1: p<0.6
level of significance, α = 0.05
from standard normal table,left tailed z α/2 =1.645
since our test is left-tailed
reject Ho, if zo < -1.645
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.53-0.6/(sqrt(0.24)/100)
zo =-1.429
| zo | =1.429
critical value
the value of |z α| at los 0.05% is 1.645
we got |zo| =1.429 & | z α | =1.645
make decision
hence value of |zo | < | z α | and here we do not reject Ho
p-value: left tail - Ha : ( p < -1.42887 ) = 0.07652
hence value of p0.05 < 0.07652,here we do not reject Ho
ANSWERS
---------------
null, Ho:p=0.6
alternate, H1: p<0.6
option:C
left tailed test
test statistic: -1.429
critical value: -1.645
decision: do not reject Ho
option:B
p-value: 0.07652
we do not have enough evidence to support the claim that the proportion of men who own cats is smaller than 60%