detects the cancer at a rate different from the 85% rate claimed by its manufacturer?solution:
I . Let p = the true proportion of men with prostate cancer who test positive.
H 0 : p = 0.85.H (^) A : p ≠ 0.85.
II . We want to use a one-proportion z -test. 175(0.85) = 148.75 > 5 and 175(1 – 0.85) = 26.25
5, so the conditions are present to use this test (the conditions are met whether we use 5 or
10). There is no randomization here. But we can do the test to see if the data provide
evidence that, among these patients, the difference from 0.85 is too big to be reasonably
attributed to random variation.
III .
Using the TI-83/84, the same answer is obtained by using 1-PropZTest in the STAT TESTS
menu.
IV . Because P is reasonably large, we do not have sufficient evidence to reject the null
hypothesis. The evidence is insufficient to challenge the company’s claim that the test is 85%
effective.example: Maria has a quarter that she suspects is out of balance. In fact, she thinks it turns up
heads more often than it would if it were fair. Being a senior, she has lots of time on her hands,
so she decides to flip the coin 300 times and count the number of heads. There are 165 heads in
the 300 flips. Does this provide evidence at the 0.05 level that the coin is biased in favor of
heads? At the 0.01 level?
solution:I . Let p = the true proportion of heads in 300 flips of a fair coin.H 0 : p = 0.50 (or H 0 : p ≤ 0.50).H (^) A : p > 0.50.
II . We will use a one-proportion z -test.
300(0.50) = 150 > 5 and 300(1 – 0.50) = 150 > 5, so the conditions are present for a one-
proportion z -test.