In a clinical trial of a drug intended to help people stop smoking, 131 subjects were treated with the drug for 13 weeks, and 13 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 0.18 as an alternative value of p, the power of the test is 0.96. Interpret this value of the power of the test.
In a clinical trial of a drug intended to help people stop smoking, 131 subjects were treated with the drug for 13 weeks, and 13 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 0.18 as an alternative value of p, the power of the test is 0.96. Interpret this value of the power of the test.
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
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In a clinical trial of a drug intended to help people stop smoking, 131 subjects were treated with the drug for 13 weeks, and 13 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 0.18 as an alternative value of p, the power of the test is 0.96.
Interpret this value of the power of the test.

Transcribed Image Text:15. In a clinical trial of a drug intended to help people stop smoking, 131 subjects were treated with the drug for 13 weeks, and 13 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 0.18 as an alternative value of p.
the power of the test is 0.96. Interpret this value of the power of the test.
% chance of rejecting the (1)
That is, if the proportion of users who experience abdominal pain is actually
than 0.08.
The power of 0.96 shows that there is a
proportion of users who experience abdominal pain is (2)
(Type integers or decimals. Do not round.)
(1) O alternative (2) O less
O null
O greater
hypothesis of p =
then there is a
when the true proportion is actually
% chance of supporting the claim that the
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