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.

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okpalawalter8 okpalawalter8 · Answer 1 · 2020-08-15T09:36:35+02:00

Answer:

A

The correct option is B

B

[tex]t = 0.6093[/tex]

C

[tex]p-value = 0.27116[/tex]

D

The correct option is D

Step-by-step explanation:

From the question we are told that

The sample size is [tex]n = 232[/tex]

The number that developed nausea is X = 50

The population proportion is p = 0.20

The null hypothesis is [tex]H_o : p = 0.20[/tex]

The alternative hypothesis is [tex]H_a : p > 0.20[/tex]

Generally the sample proportion is mathematically represented as

[tex]\r p = \frac{50}{232}[/tex]

[tex]\r p = 0.216[/tex]

Generally the test statistics is mathematically represented as

=> [tex]t = \frac{\r p - p }{ \sqrt{ \frac{p(1- p )}{n} } }[/tex]

=> [tex]t = \frac{ 0.216 - 0.20 }{ \sqrt{ \frac{ 0.20 (1- 0.20 )}{ 232} } }[/tex]

=> [tex]t = 0.6093[/tex]

The p-value obtained from the z-table is

[tex]p-value = P(Z > 0.6093) = 0.27116[/tex]

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