432,2=26275,3=66342,5=41123,4=20then 543,2=?
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Answer:
Null Hypothesis, : p = 0.20
Alternate Hypothesis, : p > 0.20
Step-by-step explanation:
We are given that 241 subjects are treated with a drug that is used to treat pain and 54 of them developed nausea.
We have to use a 0.05 significance level to test the claim that more than 20% of users develop nausea.
<em>Let p = population proportion of users who develop nausea</em>
So, <u>Null Hypothesis,</u> : p = 0.20
<u>Alternate Hypothesis</u>, : p > 0.20
Here, <u><em>null hypothesis</em></u> states that 20% of users develop nausea.
And <u><em>alternate hypothesis</em></u> states that more than 20% of users develop nausea.
The test statistics that would be used here is <u>One-sample z proportion</u> test statistics.
T.S. = ~ N(0,1)
where, = proportion of users who develop nausea in a sample of 241 subjects =
n = sample of subjects = 241
So, the above hypothesis would be appropriate to conduct the test.