In: Statistics and Probability
Throwback Question: Suppose we know that the risk of liver cancer among alcoholics without cirrhosis of the liver is 24.5%. A researcher conjectures that the risk of cancer among alcoholics with cirrhosis of the liver is higher. Suppose we sample 90 alcoholics with cirrhosis of the liver and determine 31 have liver cancer. Use this information to answer the following:
(a)
Question:
For each person in the sample, what variable is recorded?
Correct option:
percentage of individuals with liver cancer
(b)
Question:
Completely describe the sampling distribution for the sample proportion of alcoholics with cirrhosis who have liver cancer when samples of size 90 are selected (assuming there is no difference in occurrence between those with and without cirrhosis).
(i) mean = = 0.245
(ii) standard deviation =
(c)
Question:
Checks: np = and n(1 - p) =
np = 90 X 0.245 = 22.05 > 10
n (1 - p) = 90 X (1 - 0.245) = 67.95 > 10
So,
both conditions are satisfied. So, normal distribution approximation to binomial distribution is valid.
(d)
Question:
Test the appropriate claim using the data collected at a significance level of 0.10.
Correct option:
H0: p 0.245
Ha: p > 0.245
= 0.10
= 31/90 = 0.3444
Test Statistic is given by:
By Technology p - value = 0.0141
So,
Probability = 0.0141
Since p - value = 0.0141 is less than = 0.10, the difference is significant. Reject null hypothesis.
Correct option:
Reject H0.
Correct option:
At the 0.10 level, there Select an answer: is significant evidence to support the claim that the incidence rate of liver cancer is higher in alcoholics diagnosed with cirrhosis.