9.24. Cancer Therapy. Researchers in cancer therapy often report only the number
ID: 2923486 • Letter: 9
Question
9.24. Cancer Therapy. Researchers in cancer therapy often report only the number of patients who survive for a specified period of time after treat- ment rather than the patients' actual survival times. Suppose that 40% of the patients who undergo the standard treatment are known to survive 5 years. A new treatment is administered to 200 patients, and 92 of them are still alive after a period of 5 years (a) Formulate the hypotheses for testing the validity of the claim that the new treatment is more effective than the standard therapy (b) Test with = 0.05 and state your conclusion; use the rejection-region method (c) Perform the test by finding the p-value (d) What is the power of the test in (a) against the alternative H1: p= 0.5? (e) What sample size is needed so that effect pi - po 0.1 is found sig- nificant in the = 0.05 level testing with the power of 90%? As before, Po = 0.4Explanation / Answer
Given that,
possibile chances (x)=92
sample size(n)=200
success rate ( p )= x/n = 0.46
success probability,( po )=0.4
failure probability,( qo) = 0.6
null, Ho:p=0.4
alternate, H1: p>0.4
level of significance, = 0.05
from standard normal table,right tailed z /2 =1.64
since our test is right-tailed
reject Ho, if zo > 1.64
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.46-0.4/(sqrt(0.24)/200)
zo =1.7321
| zo | =1.7321
critical value
the value of |z | at los 0.05% is 1.64
we got |zo| =1.732 & | z | =1.64
make decision
hence value of | zo | > | z | and here we reject Ho
p-value: right tail - Ha : ( p > 1.73205 ) = 0.04163
hence value of p0.05 > 0.04163,here we reject Ho
ANSWERS
---------------
a.
null, Ho:p=0.4
alternate, the
new treatment is more effective than the standard therapy H1: p>0.4
btest statistic: 1.7321
critical value: 1.64
b.
decision: reject Ho
c.
p-value: 0.04163
we have evidence to support the
new treatment is more effective than the standard therapy
d.
power of test = 1 - Beta
Beta = Type II error = Accept Ho/ When it is false
when size is 200, so we get
P(X < 0.5) = (0.5-0.46)/0.0352
= 0.04/0.0352= 1.1364
= P ( Z <1.1364) From Standard Normal Table
= 0.8721
The power of a hypothesis test is the probability of making the correct decision if the alternative hypothesis is true. That is, the power of a hypothesis test is the probability of rejecting the null hypothesis H0 when the alternative hypothesis HA is the hypothesis that is true.
Power of test = 1 - 0.8721 = 0.1279
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