Homework SourseRoughly 375000 American adults have lung cancer at any one t
Roughly 375,000 American adults have lung cancer at any one time. The overall 5-year survival rate of adults diagnosed with lung cancer in the United States is 14%. A study is conducted to test the effectiveness of a new procedure for the treatment for lung cancer. Out of 3000 randomly selected adult patients from the U.S. that were diagnosed with lung cancer and then treated with the experimental procedure, 480 were alive after 5 years, for a five-year survival rate of 16%.
a) What is the population? (0.5 points)
b) What is the population parameter that we are interested in for this problem? (0.5 points)
c) What is the sample? (0.5 points)
d) What is the sample statistic? (0.5 points)
e) Check the conditions needed for having a Normal sampling distribution. (0.5 points)
f) Based on Question e), what is the sampling distribution of the sample statistic? (0.5 points)
g) If we repeated the sampling process with the same sample size, what proportion of the samples will have more than 480 patients alive after 5 years? (1 point)
Solution
a)
American adults who have lung cancer
b)
The five-year survival rate of all American adults who have lung cancer
c)
3000 randomly selected adult patients from the U.S. that were diagnosed with lung cancer
d)
p^ = 0.16
e)
As 3000 is less than 5% of 375000, that condition is satisfied.
As 3000*0.14 = 420 and 3000*0.86 = 2580 are both greater than 10, the condition is also satisfied.
f)
It has a mean of p = 0.14, and standard deviation of
s(p) = sqrt(p(1-p)/n) = sqrt(0.14*(1-0.14)/3000) = 0.006335087.
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g)
We first get the z score for the critical value:
x = critical value = 480.5
u = mean = np = 420
s = standard deviation = sqrt(np(1-p)) = 19.00526243
Thus, the corresponding z score is
z = (x-u)/s = 3.18332884
Thus, the left tailed area is
P(z < 3.18332884 ) = 0.000727961 [ANSWER]
