In a clinical trial of a drug intended to help people quit smoking, 135 subjects
ID: 3322333 • Letter: I
Question
In a clinical trial of a drug intended to help people quit smoking, 135 subjects were treated with the drug for 12 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. 135 subjects were treated with the drug for 12 weeks, and 13 subjects experienced abdominal pain. If someone clams that more than 8% of the drugs users experience that claim is supported with a hypothesis test conducted with a 0.05 significanco level. Using 0.18 as an altenative value of p, the power of the test is 0.96. Interpret this value of the power of the test. abdominal pain The power of 0.06 shows that there is a 1% chance of rejecting the proportion o users who experience abdominal pain is actually Dt en | hypothesis of p #when the true proportion is actualyThat is, if the ereisa % chan eo supporting the claim that the proportion of users who experience abdominal pain is than 0.08 Type integers or decimals. Do not round.)
Explanation / Answer
A test's power is the probability of correctly rejecting the null hypothesis when it is false; a test's power is influenced by the choice of significance level for the test, the size of the effect being measured, and the amount of data available.
So,
The power of 0.96 shows that there is _96_ % chance of rejecting the (ALTERNATIVE; NULL) hypothesis of p = _0.08__ when the true proportion is actually 0.096__. That is, the proportion of users who experience abdominal pain is actually _0.08_, then there is a _96_% chance of supporting the claim that the proportion of users who experience abdominal pain is (LESS; GREATER) than 0.08.