Question

Find the P-value for a test of the claim that more than 50% of adults that follow a particular diet will experience increased energy. Of 100 randomly selected subjects who followed the diet, 47 noticed an increase in their energy level. 0.2257 0.7257 0.05486 0.2743

Answer 1

Given that,
possibile chances (x)=47
sample size(n)=100
success rate ( p )= x/n = 0.47
success probability,( po )=0.5
failure probability,( qo) = 0.5
null, Ho:p=0.5
alternate, H1: p>0.5
level of significance, α = 0.05
from standard normal table,right tailed z α/2 =1.645
since our test is right-tailed
reject Ho, if zo > 1.645
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.47-0.5/(sqrt(0.25)/100)
zo =-0.6
| zo | =0.6
critical value
the value of |z α| at los 0.05% is 1.645
we got |zo| =0.6 & | z α | =1.645
make decision
hence value of |zo | < | z α | and here we do not reject Ho
p-value: right tail - Ha : ( p > -0.6 ) = 0.72575
hence value of p0.05 < 0.72575,here we do not reject Ho
ANSWERS
---------------
null, Ho:p=0.5
alternate, H1: p>0.5
test statistic: -0.6
critical value: 1.645
decision: do not reject Ho
p-value: 0.7257
option :B
we have enough evidence to support the claim that more than 50% of adults that follow the particular diet will experience increased energy.