Question

In the summer of 2003, The New England Journal of Medicine published results of some Scandinavian research. Men diagnosed with prostate cancer were randomly assigned to either undergo surgery or not. Among the 347 men who had surgery (group 1), 16 eventually died of prostate cancer compared with 31 of the 348 men (group 2) who did not have surgery. The researchers want to determine if surgery increases the chance of survival. Let p1: proportion of men that survived after surgery and p2: proportion of men that survived without surgery.

2. Construct a two-tailed 90% confidence interval for the difference between men’s survival without and with surgery and interpret it in terms of the problem context.

a. Justify assumptions for your choice of CI.

b. Compute the 90% confidence interval and interpret it:

3. Apply a Hypothesis testing.

a. What are the null and alternative hypotheses? H0: vs. Ha:

b. Justify the assumptions before you compute your test statistic and then compute the test statistic.

c. What is the p-value for this test?

d. What would you conclude from this test at 5% level of significance?

Answer 1

(2)

(a)

n1 =347

1 = 331/347 = 0.954

n2 = 348

2 = 317/348 = 911

= 0.10

Assumptions for our choice of Confidence Interval for difference in 2 independent samples proportions Z test:

(i) Each sample is randomly selected.

(ii) Each sample is independent

(iii)

n1p1 = 347 X 0.954 = 331 5

n1q1 = 347 X 0.046 = 16 5

n2p2 = 348 X 0.911 = 317 5

n2q2= 348 X 0.089 = 31 5

So,

all conditions are satisfied.

(b)

(i)

= 0.10

From Table, critical values of Z = 1.645

the 90% confidence interval is computed as follows:

So,

the 90% confidence interval is ( 0.012, 0.074)

(ii)

The 90% confidence interval (0.012, 0.074) is a range of values we are 90% contain will contain the true unknown population proportion difference between men’s survival with and without surgery

(3)

(a)

H0: Null Hypothesis: p1 p2 ( surgery does not increase the chance of survival. )

HA: Alternative Hypothesis: p1 > p2 ( surgery increase the chance of survival. ) (Claim)

(b)

n1 =347

1 = 331/347 = 0.954

n2 = 348

2 = 317/348 = 911

= 0.05

Assumptions for Hypothesis Testing for difference in 2 independent samples proportions Z test:

(i) Each sample is randomly selected.

(ii) Each sample is independent

(iii)

n1p1 = 347 X 0.954 = 331 5

n1q1 = 347 X 0.046 = 16 5

n2p2 = 348 X 0.911 = 317 5

n2q2= 348 X 0.089 = 31 5

So,

all conditions are satisfied.

(c)

Pooled Proportion is given by:

Test Statistic is given by:

By Technology,

the p-value for this test =0.012

(d)

Since p - value = 0.012 is less than = 0.05, the difference is significant. Reject null hypothesis.

Conclusion:
The data suport the clim that surgery increase the chance of survival.