In a clinical trial of a drug intended to help people stop smoking. 129 subjects were treated with the drug for 11 weeks, and 17 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 The power of 0.00 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 then there is a % chance of supporting the claim that the proportion of users who experience abdominal pain is than 0.08. (Type integers or decimals. Do not round.)

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In a clinical trial of a drug intended to help people stop smoking. 129 subjects were treated with the drug for 11 weeks, and 17 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
The power of 0.00 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 then
there is a % chance of supporting the claim that the proportion of users who experience abdominal pain is
than 0.08.
(Type integers or decimals. Do not round.)
Transcribed Image Text:In a clinical trial of a drug intended to help people stop smoking. 129 subjects were treated with the drug for 11 weeks, and 17 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 The power of 0.00 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 then there is a % chance of supporting the claim that the proportion of users who experience abdominal pain is than 0.08. (Type integers or decimals. Do not round.)
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