In a clinical trial of a drug intended to help people stop smoking, 125 subjects were treated with the drug for 12 weeks, and 14 subjects experienced abdominal pain. If someone claims that more than 8% of the drug's users experience abdominal pain, that claim is supported with a hypothesis test conducted with a 0.05 significance level. Using 0.17 as an alternative value of p, the power of the test is 0.95. Interpret this value of the power of the test. ... The power of 0.95 shows that there is a % chance of rejecting the hypothesis of p= when the true proportion is actually That is, if the proportion of users who experience abdominal pain is actually then there is a % chance of supporting the claim that the proportion of than 0.08. users who experience abdominal pain is (Type integers or decimals. Do not round.)

MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
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Chapter1: Starting With Matlab
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In a clinical trial of a drug intended to help people stop smoking, 125 subjects
were treated with the drug for 12 weeks, and 14 subjects experienced
abdominal pain. If someone claims that more than 8% of the drug's users
experience abdominal pain, that claim is supported with a hypothesis test
conducted with a 0.05 significance level. Using 0.17 as an alternative value
of p, the power of the test is 0.95. Interpret this value of the power of the test.
The power of 0.95 shows that there is a % chance of rejecting the
hypothesis of p = when the true proportion is actually
That is, if the proportion of users who experience abdominal pain is actually
then there is a % chance of supporting the claim that the proportion of
than 0.08.
users who experience abdominal pain is
(Type integers or decimals. Do not round.)
Transcribed Image Text:In a clinical trial of a drug intended to help people stop smoking, 125 subjects were treated with the drug for 12 weeks, and 14 subjects experienced abdominal pain. If someone claims that more than 8% of the drug's users experience abdominal pain, that claim is supported with a hypothesis test conducted with a 0.05 significance level. Using 0.17 as an alternative value of p, the power of the test is 0.95. Interpret this value of the power of the test. The power of 0.95 shows that there is a % chance of rejecting the hypothesis of p = when the true proportion is actually That is, if the proportion of users who experience abdominal pain is actually then there is a % chance of supporting the claim that the proportion of than 0.08. users who experience abdominal pain is (Type integers or decimals. Do not round.)
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