Question

suppose 212 subjects are treated with a drug that is used to treat pain and 54 of them developed nausea. use a 0.01 significance level to test the claim that more than 20% of users develop nausea.

identify the test statistic

idebtify the p-value

state a conclusion

Answer 1

Solution:

Given:

Sample size = n = 212

x = Number of people treated with a drug that is used to treat pain and developed nausea = 54

Level of significance =

We have to test the claim that: more than 20% of users develop nausea.

Thus hypothesis of the study are:

H0: p = 0.20 Vs H1: p > 0.20   ( Right tailed test)

Part a) Identify the test statistic

where

Thus

Part b) Identify the p-value:

p-value = P(Z > z test statistic value)     

p-value = P(Z > 1.99 )

p-value = 1 - P(Z < 1.99)

Look in z table for z = 1.9 and 0.09 and find area.

Thus from z table we get:

P( Z < 1.99) = 0.9767

Thus

p-value = 1 - P(Z < 1.99)

p-value = 1 -   0.9767

p-value = 0.0233

Part c) state a conclusion

Since p-value = 0.0233 > 0.01 significance level , we fail to reject H0.

Thus at 0.01 significance level , there is not sufficient evidence to support the claim that more than 20% of users develop nausea.