2. According to a recent survey, of the 84 women who responded to the anonymous
ID: 2932151 • Letter: 2
Question
2. According to a recent survey, of the 84 women who responded to the anonymous questionnaire, 36 said that they had used cannabis intermittently during their pregnancy to treat symptoms of vomiting, nausea, and appetite loss. Based on a study conducted between 1984 and 1987, 45% of mothers were found to use marijuana during pregnancy. Test at the 1% significance level if the current percentage of mothers who have used marijuana during pregnancy is different from 45%. Note that the computed p- value was 0.7032Explanation / Answer
Given that,
possibile chances (x)=36
sample size(n)=84
success rate ( p )= x/n = 0.43
success probability,( po )=0.45
failure probability,( qo) = 0.55
null, Ho:p=0.45
alternate, H1: p!=0.45
level of significance, = 0.01
from standard normal table, two tailed z /2 =2.58
since our test is two-tailed
reject Ho, if zo < -2.58 OR if zo > 2.58
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.42857-0.45/(sqrt(0.2475)/84)
zo =-0.39
| zo | =0.39
critical value
the value of |z | at los 0.01% is 2.58
we got |zo| =0.395 & | z | =2.58
make decision
hence value of |zo | < | z | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != -0.39477 ) = 0.69301
hence value of p0.01 < 0.693,here we do not reject Ho
ANSWERS
---------------
null, Ho:p=0.45
alternate, H1: p!=0.45
test statistic: -0.39
critical value: -2.58 , 2.58
decision: do not reject Ho
p-value: 0.70
percentage of
mothers who have used marijuana during pregnancy is similar to 45%
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