A group of clinical physicians is performing tests on patients to determine the
ID: 3175503 • Letter: A
Question
A group of clinical physicians is performing tests on patients to determine the effectiveness of a new antihypertensive drug. Patients with high blood pressure were randomly chosen and then randomly assigned to either the control group (which received a well-established antihypertensive) or the treatment group (which received the new drug). The doctors noted the proportion of patients whose blood pressure was reduced to a normal level within 1 year.
GROUP
PROPORTION
THAT IMPROVED
NUMBER OF PATIENTS
Treatment
0.45
120
Control
0.32
150
a. State appropriate hypotheses to determine whether the proportion of patients that showed blood pressure improvement using the new drug (treatment group) is significantly different than the proportion that improved using the older drug (control group).
Ho :
H1 :
b. Calculate the test statistic for the above test.
Can the null hypothesis be rejected?If so, at what significance level?Complete the table.
Value of
Critical values
Reject H0?
0.10
0.05
0.01
Make an appropriate statement to interpret your results.
GROUP
PROPORTION
THAT IMPROVED
NUMBER OF PATIENTS
Treatment
0.45
120
Control
0.32
150
Explanation / Answer
Given that,
sample one, n1 =120, p1= x1/n1=0.45
sample two, n2 =150, p2= x2/n2=0.32
null, Ho: p1 = p2
alternate, H1: p1 != p2
we use test statistic (z) = (p1-p2)/(p^q^(1/n1+1/n2))
zo =(0.45-0.32)/sqrt((0.378*0.622(1/120+1/150))
zo =2.189
| zo | =2.189
ANSWERS
---------------
null, Ho: p1 = p2
alternate, H1: p1 != p2
test statistic: 2.189
level of significance, = 0.1
from standard normal table, two tailed z /2 =
since our test is two-tailed
reject Ho, if zo < -1.645 OR if zo > 1.645
critical value
the value of |z | at los 0.1% is 1.645
we got |zo| =2.189 & | z | =1.645
make decision
hence value of | zo | > | z | and here we reject Ho
level of significance, = 0.05
from standard normal table, two tailed z /2 =
since our test is two-tailed
reject Ho, if zo < -1.96 OR if zo > 1.96
critical value
the value of |z | at los 0.05% is 1.96
we got |zo| =2.189 & | z | =1.96
make decision
hence value of | zo | > | z | and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 2.1893 ) = 0.0286
hence value of p0.05 > 0.0286,here we reject Ho
level of significance, = 0.01
from standard normal table, two tailed z /2 =
since our test is two-tailed
reject Ho, if zo < -2.576 OR if zo > 2.576
critical value
the value of |z | at los 0.01% is 2.576
we got |zo| =2.189 & | z | =2.576
make decision
hence value of |zo | < | z | and here we do not reject Ho
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