Given that,
possibile chances (x)=21
sample size(n)=893
success rate ( p )= x/n = 0.024
success probability,( po )=0.019
failure probability,( qo) = 0.981
null, Ho:p=0.019
alternate, H1: p>0.019
level of significance, α = 0.1
from standard normal table,right tailed z α/2 =1.282
since our test is right-tailed
reject Ho, if zo > 1.282
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.02352-0.019/(sqrt(0.018639)/893)
zo =0.989
| zo | =0.989
critical value
the value of |z α| at los 0.1% is 1.282
we got |zo| =0.989 & | z α | =1.282
make decision
hence value of |zo | < | z α | and here we do not reject
Ho
p-value: right tail - Ha : ( p > 0.98853 ) = 0.16145
hence value of p0.1 < 0.16145,here we do not reject Ho
ANSWERS
---------------
null, Ho:p=0.019
alternate, H1: p>0.019
np>10,n(1-p)>10
test statistic: 0.989
critical value: 1.282
decision: do not reject Ho
p-value: 0.16145
we do not have enough evidence to support the claim that more than
1.9% of this drug user experience flulike symptoms as a side
effect
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