Question

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

Answer 1

Answer:

Null Hypothesis, $H_0$ : p = 0.20  

Alternate Hypothesis, $H_a$ : 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, $H_0$ : p = 0.20  

Alternate Hypothesis, $H_a$ : 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. =  $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$   ~ N(0,1)

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

             n = sample of subjects = 241

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