Do you believe the difference between the groups is practically significant? %Pr

Question

Do you believe the difference between the groups is practically significant? %Practical versus Statistical Significance In clinical trials for treatment of a skin disorder, 642 of 2105 patients receiving the current standard treatment were cured of the disorder and 697 of 2115 patients receiving a new proposed treatment were cured of the disorder (a) Does the new procedure cure a higher proportion of patients at the = 0.05 level of significance? (b) Do you think that the difference in success rates is practically significant? What factors might influence your decision?

Explanation / Answer

26.

a.
Given that,
sample one, x1 =642, n1 =2105, p1= x1/n1=0.305
sample two, x2 =697, n2 =2115, p2= x2/n2=0.33
null, Ho: p1 = p2
alternate, H1: p1 < p2
level of significance, = 0.05
from standard normal table,left tailed z /2 =1.645
since our test is left-tailed
reject Ho, if zo < -1.645
we use test statistic (z) = (p1-p2)/(p^q^(1/n1+1/n2))
zo =(0.305-0.33)/sqrt((0.317*0.683(1/2105+1/2115))
zo =-1.714
| zo | =1.714
critical value
the value of |z | at los 0.05% is 1.645
we got |zo| =1.714 & | z | =1.645
make decision
hence value of | zo | > | z | and here we reject Ho
p-value: left tail - Ha : ( p < -1.7142 ) = 0.04325
hence value of p0.05 > 0.04325,here we reject Ho
ANSWERS
---------------
null, Ho: p1 = p2
alternate, H1: p1 < p2
test statistic: -1.714
critical value: -1.645
decision: reject Ho
p-value: 0.04325
we have enough evidence to support that the new procedure cure is higher proportion of patients
b.
Given that,
sample one, x1 =642, n1 =2105, p1= x1/n1=0.305
sample two, x2 =697, n2 =2115, p2= x2/n2=0.33
null, Ho: p1 = p2
alternate, H1: p1 != p2
level of significance, = 0.05
from standard normal table, two tailed z /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 =(0.305-0.33)/sqrt((0.317*0.683(1/2105+1/2115))
zo =-1.714
| zo | =1.714
critical value
the value of |z | at los 0.05% is 1.96
we got |zo| =1.714 & | z | =1.96
make decision
hence value of |zo | < | z | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != -1.7142 ) = 0.0865
hence value of p0.05 < 0.0865,here we do not reject Ho
ANSWERS
---------------
null, Ho: p1 = p2
alternate, H1: p1 != p2
test statistic: -1.714
critical value: -1.96 , 1.96
decision: do not reject Ho
p-value: 0.0865
we do not have enough evidence to support that the new procedure cure is higher proportion of patients

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