In: Statistics and Probability
Given that,
possible chances (x)=1
sample size(n)=15
success rate ( p )= x/n = 0.067
success probability,( po )=0.5
failure probability,( qo) = 0.5
null, Ho:p=0.5
alternate, H1: p!=0.5
level of significance, alpha = 0.05
from standard normal table, two tailed z alpha/2 =1.96
since our test is two-tailed
reject Ho, if zo < -1.96 OR if zo > 1.96
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.06667-0.5/(sqrt(0.25)/15)
zo =-3.357
| zo | =3.357
critical value
the value of |z alpha| at los 0.05% is 1.96
we got |zo| =3.357 & | z alpha | =1.96
make decision
hence value of | zo | > | z alpha| and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != -3.35659
) = 0.00079
hence value of p0.05 > 0.0008,here we reject Ho
ANSWERS
---------------
null, Ho:p=0.5
alternate, H1: p!=0.5
test statistic: -3.357
critical value: -1.96 , 1.96
decision: reject Ho
p-value: 0.00079
we have enough evidence to support the claim that whether or not
there was a relationship between how the judges ranked dogs on
their form compared to
how they ranked them on their groom