Question

In: Statistics and Probability

in one study of smokers who tried to quit smoking with nicotine patch therapy, 37 were...

in one study of smokers who tried to quit smoking with nicotine patch therapy, 37 were smoking one year after treatment and 33 were not smoking one year after treatment. Use a .05 significance level to test the claim that among smokers who try to quit with nicotine oatch therapy, the majority are smoking one year after treatment. Do these results suggest that patch therapy is not effective?

Solutions

Expert Solution

Given that,
possibile chances (x)=37
sample size(n)=70
success rate ( p )= x/n = 0.529
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, two tailed z α/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=0.52857-0.5/(sqrt(0.25)/70)
zo =0.478
| zo | =0.478
critical value
the value of |z α| at los 0.05% is 1.96
we got |zo| =0.478 & | z α | =1.96
make decision
hence value of |zo | < | z α | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 0.47809 ) = 0.63259
hence value of p0.05 < 0.6326,here we do not reject Ho
ANSWERS
---------------
null, Ho:p=0.5
alternate, H1: p!=0.5
test statistic: 0.478
critical value: -1.96 , 1.96
decision: do not reject Ho
p-value: 0.63259
we do not have enough evidence to support the claim that among smokers who try to quit with nicotine oatch therapy, the majority are smoking one year after treatment.

orchestra answered 1 year ago
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