Question

Suppose

239239

subjects are treated with a drug that is used to treat pain and

5151

of them developed nausea. Use a

0.100.10

significance level to test the claim that more than

2020​%

of users develop nausea.

Identify the null and alternative hypotheses for this test. Choose the correct answer below.

A.

Upper H 0H0​:

pequals=0.200.20

Upper H 1H1​:

pless than<0.200.20

B.

Upper H 0H0​:

pequals=0.200.20

Upper H 1H1​:

pgreater than>0.200.20

C.

Upper H 0H0​:

pequals=0.200.20

Upper H 1H1​:

pnot equals≠0.200.20

D.

Upper H 0H0​:

pgreater than>0.200.20

Upper H 1H1​:

pequals=0.200.20

Identify the test statistic for this hypothesis test.

The test statistic for this hypothesis test is

nothing.

​(Round to two decimal places as​ needed.)

Identify the​ P-value for this hypothesis test.

The​ P-value for this hypothesis test is

nothing.

​(Round to three decimal places as​ needed.)

Answer 1

Solution :

Given that,

= 0.20

1 - = 0.80

n = 239

x = 51

Level of significance = = 0.10

Point estimate = sample proportion = = x / n = 0.213

This a right (One) tailed test.

B)

Ho: p = 0.20

Ha: p 0.20

Test statistics

z = ( - ) / *(1-) / n

= ( 0.213 - 0.20) / (0.20*0.80) / 239

= 0.52

P-value = P(Z>z)

= 1 - P(Z <z )

= 1- P(Z < 0.52)

= 1 - 0.6985

= 0.302

The p-value is p = 0.302, and since p = 0.302 > 0.10, it is concluded that the null hypothesis is fails to reject.