Question

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

Answer 1

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