Question

7. A certain drug is used to treat asthma. In a clinical trial of the​ drug, 20 of 286 treated subjects experienced headaches​ (based on data from the​ manufacturer). The accompanying calculator display shows results from a test of the claim that less than 12​% of treated subjects experienced headaches. Use the normal distribution as an approximation to the binomial distribution and assume a 0.05 significance level to complete parts​ (a) through​ (e) below.

a. Is the test​ two-tailed, left-tailed, or​ right-tailed?

Right tailed test

​Left-tailed test

​Two-tailed test

b. What is the test​ statistic?

Z=

​(Round to two decimal places as​ needed.)

c. What is the​ P-value?

​P-value=______

​(Round to four decimal places as​ needed.)

d. What is the null​ hypothesis, and what do you conclude about​ it?

Identify the null hypothesis.

A:Ho<0.09

B:Ho>0.09

C:Ho:p≠0.09

D:Ho:p=0.09

Decide whether to reject the null hypothesis. Choose the correct answer below.

A.Reject the null hypothesis because the​ P-value is greater than the significance​ level, alpha.

B.Reject the null hypothesis because the​ P-value is less than or equal to the significance​ level, alpha.

C.Fail to reject the null hypothesis because the​ P-value is greater than the significance​ level, alpha.

D.Fail to reject the null hypothesis because the​ P-value is less than or equal to the significance​ level, alpha.

e. What is the final​ conclusion?

A.There is not sufficient evidence to support the claim that less than 9​% of treated subjects experienced headaches.

B.There is not sufficient evidence to warrant rejection of the claim that less than 9​% of treated subjects experienced headaches.

C.There is sufficient evidence to warrant rejection of the claim that less than 9​% of treated subjects experienced headaches.

D.There is sufficient evidence to support the claim that less than 9​% of treated subjects experienced headaches.

Answer 1

7.

i.

Given that,
possible chances (x)=20
sample size(n)=286
success rate ( p )= x/n = 0.07
success probability,( po )=0.09
failure probability,( qo) = 0.91
null, Ho:p=0.09
alternate, H1: p<0.09
level of significance, α = 0.05
from standard normal table,left tailed z α/2 =1.645
since our test is left-tailed
reject Ho, if zo < -1.645
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.06993-0.09/(sqrt(0.0819)/286)
zo =-1.186
| zo | =1.186
critical value
the value of |z α| at los 0.05% is 1.645
we got |zo| =1.186 & | z α | =1.645
make decision
hence value of |zo | < | z α | and here we do not reject Ho
p-value: left tail - Ha : ( p < -1.18601 ) = 0.11781
hence value of p0.05 < 0.11781,here we do not reject Ho
ANSWERS
---------------
a.
left tailed test
b.
test statistic: -1.186
critical value: -1.645
decision: do not reject Ho
c.
p-value: 0.11781
option:C
p value is greater than the significance​ level, alpha.
d.
null, Ho:p=0.09
alternate, H1: p<0.09
option:A
we do not have enough evidence to support the claim that less than 9​% of treated subjects experienced headaches.
e.
option:A
There is not sufficient evidence to support the claim that
less than 9​% of treated subjects experienced headaches.

ii.
Given that,
possible chances (x)=20
sample size(n)=286
success rate ( p )= x/n = 0.07
success probability,( po )=0.12
failure probability,( qo) = 0.88
null, Ho:p=0.12
alternate, H1: p<0.12
level of significance, α = 0.05
from standard normal table,left tailed z α/2 =1.645
since our test is left-tailed
reject Ho, if zo < -1.645
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.06993-0.12/(sqrt(0.1056)/286)
zo =-2.606
| zo | =2.606
critical value
the value of |z α| at los 0.05% is 1.645
we got |zo| =2.606 & | z α | =1.645
make decision
hence value of | zo | > | z α| and here we reject Ho
p-value: left tail - Ha : ( p < -2.60572 ) = 0.00458
hence value of p0.05 > 0.00458,here we reject Ho
ANSWERS
---------------
null, Ho:p=0.12
alternate, H1: p<0.12
test statistic: -2.606
critical value: -1.645
decision: reject Ho
p-value: 0.00458
we have enough evidence to support the claim that less than 12​% of treated subjects experienced headaches.