The power of 0.96 shows that there is a __% chance of rejecting the (ALTERNATIVE

Question

The power of 0.96 shows that there is a __% chance of rejecting the (ALTERNATIVE; NULL) hypothesis if 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 (GREATER; LESS) than 0.08. In a clinical trial of a drug intended to help people stop smoking, 135 subjects were treated with the drug for 13 weeks, and 20 subjects experienced abdominal pain. f someone claims that more than 8% o the drug's users o erence abdominal pain, that claim s supported with a thesis test onducted with a ons s ca ce 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. The power of 096 shows that there is a%chance of rejecting tho |hypothesis of p-when the true proportion is actuallyThat is, if the proportion o users wno experience abdominal pan is actual y then there isa abdominal pain is | | than 0.08. Type integers or decimals. Do not round.) % chan e of supporting the claim that e proportion of users who experiene

Explanation / Answer

The power of 0.96 shows that there is a 96% chance of rejecting the  NULL hypothesis if p =0.08 when the true proportion is actually  0.18. That is, if the proportion of users who experience abdominal pain is actually 0.18 then there is a 96 % chance of supporting the claim that the proportion of users who experience abdominal pain is GREATER  than 0.08.

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