A certain drug is used to treat asthma. In a clinical trial of
the drug,
24 of260 treated subjects experienced headaches (based on data from themanufacturer). The accompanying calculator display shows results from a test of the claim that less than11% of treated subjects experienced headaches. Use the normal distribution as an approximation to the binomial distribution and assume a0.01 significance level to complete parts (a) through (e) below. |
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a. Is the test two-tailed, left-tailed, or right-tailed?
Left-tailed test
Right tailed test
Two-tailed test
b. What is the test statistic?
z=nothing
(Round to two decimal places as needed.)
c. What is the P-value?
P-value=nothing
(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.
Upper H 0 : p not equals 0.11H0: p≠0.11
B.
Upper H 0 : p greater than 0.11H0: p>0.11
C.
Upper H 0 : p equals 0.11H0: p=0.11
D.
Upper H 0 : p less than 0.11H0: p<0.11
Decide whether to reject the null hypothesis. Choose the correct answer below.
A.
Reject
the null hypothesis because the P-value is
less than or equal to
the significance level,
α.
B.
Fail to reject
the null hypothesis because the P-value is
less than or equal to
the significance level,
α.
C.
Reject
the null hypothesis because the P-value is
greater than
the significance level,
α.
D.
Fail to reject
the null hypothesis because the P-value is
greater than
the significance level,
α.
e. What is the final conclusion?
A.There
is not
sufficient evidence to warrant rejection of the claim that less than
11%
of treated subjects experienced headaches.
B.There
is
sufficient evidence to warrant rejection of the claim that less than
11%
of treated subjects experienced headaches.
C.There
is
sufficient evidence to support the claim that less than
11%
of treated subjects experienced headaches.
D.There
is not
sufficient evidence to support the claim that less than
11%
of treated subjects experienced headaches.
Solution :
This is the left tailed test .
The null and alternative hypothesis is
H0 : p = 0.11
Ha : p < 0.11
= x / n = 24 / 260 = 0.0923
P0 = 0.11
1 - P0 = 0.89
Test statistic = z
= - P0 / [P0 * (1 - P0 ) / n]
= 0.0923 - 0.11/ [(011 * 0.89) /260 ]
= -0.91
P-value = 0.1809
= 0.01
P-value >
Fail to reject the null hypothesis .
D.
Fail to reject the null hypothesis because the P-value is greater than the significance level
A.There is not sufficient evidence to warrant rejection of the claim that less than 11% of treated subjects
experienced headaches.
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