In: Statistics and Probability
Given that,
possible chances (x)=3
sample size(n)=10
success rate ( p )= x/n = 0.3
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,left tailed z alpha/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=0.3-0.5/(sqrt(0.25)/10)
zo =-1.265
| zo | =1.265
critical value
the value of |z alpha| at los 0.05% is 1.645
we got |zo| =1.265 & | z alpha | =1.645
make decision
hence value of |zo | < | z alpha | and here we do not reject
Ho
p-value: left tail - Ha : ( p < -1.26491 ) = 0.10295
hence value of p0.05 < 0.10295,here we do not reject Ho
ANSWERS
---------------
null, Ho:p=0.5
alternate, H1: p<0.5
test statistic: -1.265
critical value: -1.645
decision: do not reject Ho
p-value: 0.10295
we do not have enough evidence to support the claim that whether
cholesterol was reduced after using a certain brand of
margarine
as part of a low fat, low cholesterol die.