Question

The pre legalization. Of cannabis was collected by testing 1103 patients out of which 45% were males and the rest females. The results Showed that 30% of them tested positive for cannabinoids. On the other hand the post legalization test was done on 1083 patients out of which 47% were males and 53% females. The results showed that 36% of them tested positive for cannabinoids. What statistical test did they use and is their finding accurate? Show the work for which test

Answer 1

Given that,
sample one, x1 =496.35, n1 =1103, p1= x1/n1=0.45
sample two, x2 =509.01, n2 =1083, p2= x2/n2=0.47
finding a p^ value for proportion p^=(x1 + x2 ) / (n1+n2)
p^=0.46
q^ Value For Proportion= 1-p^=0.54
null, Ho: p1 = p2
alternate, H1: p1 != p2
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) = (p1-p2)/√(p^q^(1/n1+1/n2))
zo =(0.45-0.47)/sqrt((0.46*0.54(1/1103+1/1083))
zo =-0.938
| zo | =0.938
critical value
the value of |z α| at los 0.05% is 1.96
we got |zo| =0.938 & | 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 != -0.9381 ) = 0.3482
hence value of p0.05 < 0.3482,here we do not reject Ho
ANSWERS
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null, Ho: p1 = p2
alternate, H1: p1 != p2
test statistic: -0.938
critical value: -1.96 , 1.96
decision: do not reject Ho
p-value: 0.3482
we do not have enough evidence to support the claim that difference of proportion of males