In a clinical trial of a drug intended to help people stop smoking, 134 subjects were treated with the drug for 12 weeks, and 10 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 0.19 as an alternative value of p, the power of the test is 0.96. Interpret this value of the power of the test hypothesis of p when the true proportion is actually. That is, if the The power of 0.96 shows that there is a% chance of rejecting the proportion of users who experience abdominal pain is actuallythen there is a% chance of supporting the claim that the proportion of users who experience abdominal pain is (Type integers or decimals. Do not round.) = than 0.08

Holt Mcdougal Larson Pre-algebra: Student Edition 2012
1st Edition
ISBN:9780547587776
Author:HOLT MCDOUGAL
Publisher:HOLT MCDOUGAL
Chapter11: Data Analysis And Probability
Section: Chapter Questions
Problem 8CR
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In a clinical trial of a drug intended to help people stop​ smoking,

134

subjects were treated with the drug for

12

​weeks, and

10

subjects experienced abdominal pain. If someone claims that more than

88​%

of the​ drug's users experience abdominal​ pain, that claim is supported with a hypothesis test conducted with a

0.05

significance level. Using

0.19

as an alternative value of​ p, the power of the test is

0.96

Interpret this value of the power of the test.

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In a clinical trial of a drug intended to help people stop smoking, 134 subjects were treated with the drug for 12 weeks, and 10 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 0.19 as an alternative value of p, the power of the test is 0.96. Interpret this value of the power of the test
hypothesis of p
when the true proportion is actually. That is, if the
The power of 0.96 shows that there is a% chance of rejecting the
proportion of users who experience abdominal pain is actuallythen there is a% chance of supporting the claim that the proportion of users who
experience abdominal pain is
(Type integers or decimals. Do not round.)
=
than 0.08
Transcribed Image Text:In a clinical trial of a drug intended to help people stop smoking, 134 subjects were treated with the drug for 12 weeks, and 10 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 0.19 as an alternative value of p, the power of the test is 0.96. Interpret this value of the power of the test hypothesis of p when the true proportion is actually. That is, if the The power of 0.96 shows that there is a% chance of rejecting the proportion of users who experience abdominal pain is actuallythen there is a% chance of supporting the claim that the proportion of users who experience abdominal pain is (Type integers or decimals. Do not round.) = than 0.08
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