Answer:
Null Hypothesis,  : p = 0.20
 : p = 0.20   
Alternate Hypothesis,  : p > 0.20
 : 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
 : p = 0.20   
<u>Alternate Hypothesis</u>,  : p > 0.20
 : 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)
   ~ N(0,1) 
where,   = proportion of users who develop nausea in a sample of 241 subjects =
 = 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.