Suppose 241 subjects are treated with a drug that is used to treat pain and 54 of them developed nausea. Use a 0.05 significance level to test the claim that more than 20% of users develop nausea. Identify the null and alternative hypotheses for this test. Choose the correct answer below. A. Upper H 0: pequals0.20 Upper H 1: pnot equals0.20 B. Upper H 0: pequals0.20 Upper H 1: pless than0.20 C. Upper H 0: pgreater than0.20 Upper H 1: pequals0.20 D. Upper H 0: pequals0.20 Upper H 1: pgreater than0.20

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Alcaa Alcaa · Answer 1 · 2020-04-16T13:40:10+02:00

Answer:

Null Hypothesis, [tex]H_0[/tex] : p = 0.20

Alternate Hypothesis, [tex]H_a[/tex] : 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.

Let p = population proportion of users who develop nausea

So, Null Hypothesis, [tex]H_0[/tex] : p = 0.20

Alternate Hypothesis, [tex]H_a[/tex] : p > 0.20

Here, null hypothesis states that 20% of users develop nausea.

And alternate hypothesis states that more than 20% of users develop nausea.

The test statistics that would be used here is One-sample z proportion test statistics.

T.S. = [tex]\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] ~ N(0,1)

where, [tex]\hat p[/tex] = proportion of users who develop nausea in a sample of 241 subjects = [tex]\frac{54}{241}[/tex]

n = sample of subjects = 241

So, the above hypothesis would be appropriate to conduct the test.