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suter [353]
3 years ago
12

Suppose 232subjects are treated with a drug that is used to treat pain and 50of them developed nausea. Use a 0.01significance le

vel to test the claim that more than 20​%of users develop nausea. Identify the null and alternative hypotheses for this test.
A. Upper H0?: p equals 0.20
Upper H1?: p not equals 0.20
B. Upper H0?: p equals 0.20
Upper H1?: p greater than 0.20
C. Upper H0?: p greater than 0.20
Upper H1?: p equals 0.20
D. Upper H0?: p equals 0.20
Upper H1?: p less than 0.20
Identify the test statistic for this hypothesis test. Identify the​ P-value for this hypothesis test.
Identify the conclusion for this hypothesis test.
A. Reject Upper H 0. There is sufficient evidence to warrant support of the claim that more than 20?% of users develop nausea.
B. Fail to reject Upper H 0. There is sufficient evidence to warrant support of the claim that more than 20?% of users develop nausea.
C. Reject Upper H 0. There is not sufficient evidence to warrant support of the claim that more than 20?% of users develop nausea.
D. Fail to reject Upper H 0. There is not sufficient evidence to warrant support of the claim that more than 20?% of users develop nausea.
Mathematics
1 answer:
Margarita [4]3 years ago
5 0

Answer:

A

   The  correct option is B

B

   t =  0.6093

C

 p-value  =  0.27116

D

The  correct option is  D

Step-by-step explanation:

From the question we are told that

    The  sample size is  n  =  232

    The  number that developed  nausea  is X =  50

    The population proportion is  p  =  0.20  

 

The  null hypothesis is   H_o : p  =  0.20

The  alternative hypothesis is  H_a :  p > 0.20

Generally the sample proportion is mathematically represented as

     \r p  =  \frac{50}{232}

     \r p  =  0.216

Generally the test statistics is mathematically represented as

 =>           t =  \frac{\r p  -  p }{ \sqrt{ \frac{p(1- p )}{n} } }

=>           t =  \frac{ 0.216 - 0.20 }{ \sqrt{ \frac{ 0.20 (1- 0.20 )}{ 232} } }

=>        t =  0.6093

The  p-value obtained from the z-table is

       p-value  =  P(Z >  0.6093) =  0.27116

  Given that the  p-value >  \alpha  then we fail to reject the null hypothesis

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