In a clinical trial of a drug intended to help people stop smoking, 127 subjects were treated with the drug for 13 weeks, and 15 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 0.16 as an alternative value of p, the power of the test is 0.95. Interpret this value of the power of the test.

Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
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ISBN:9780079039897
Author:Carter
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Chapter4: Equations Of Linear Functions
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In a clinical trial of a drug intended to help people stop smoking, 127 subjects were treated with the drug for 13 weeks, and 15 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 0.16 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
v hypothesis of p=
when the true proportion is actually
That is, if the
proportion of users who experience abdominal pain is actually
v than 0.08.
(Type integers or decimals. Do not round.)
then there is a
% chance of supporting the claim that the proportion of users who experience abdominal pain is
Transcribed Image Text:In a clinical trial of a drug intended to help people stop smoking, 127 subjects were treated with the drug for 13 weeks, and 15 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 0.16 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 v hypothesis of p= when the true proportion is actually That is, if the proportion of users who experience abdominal pain is actually v than 0.08. (Type integers or decimals. Do not round.) then there is a % chance of supporting the claim that the proportion of users who experience abdominal pain is
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