Question

Suppose 224 subjects are treated with a drug that is used to treat pain and 53 of them develop nausea. Use a 0.01 significance level to test the claim that more than 20% of users develop nausea

Answer #1

Solution :

This is the right tailed test .

The null and alternative hypothesis is

H_{0} : p = 0.20

H_{a} : p > 0.20

n = 224

x = 53

= x / n = 53 / 220 = 0.2366

P_{0} = 0.20

1 - P_{0} = 1 - 0.20 = 0.80

z =
- P_{0} / [P_{0
*} (1 - P_{0} ) / n]

= 0.2366 - 0.20 / [(0.20 * 0.80) / 224]

= 1.37

Test statistic = 1.37.

P(z > 1.37) = 1 - P(z < 1.37) = 1 - 0.9147 = 0.0853

P-value = 0.0853

= 0.01

P-value >

Fail to reject the null hypothesis .

There is not sufficient evidence to test the claim that more than 20% of users develop nausea .

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