Question

Suppose 237 subjects are treated with a drug that is used to treat pain and 51 of them developed nausea. Use a 0.01

significance level to test the claim that more than20% of users develop nausea

The test statistic for this hypothesis test is??? round three decimal

The P-value for this hypothesis test is???

Answer #1

Solution :

This is the two tailed test .

The null and alternative hypothesis is

H_{0} : p = 0.20

H_{a} : p > 0.20

= x / n = 51/237 = 0.2152

P_{0} = 0.20

1 - P_{0} = 1 - 20 = 0.80

**Test statistic = z**

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

=0.2152 - 0.20 / [0.20(1- 0.20) /237 ]

**= 0.585**

P(z >0.585 ) = 1 - P(z <0.585 ) = 0.2794

**P-value = 0.2794**

= 0.01

p = 0.2794 ≥ 0.01, it is concluded that the null hypothesis is not rejected

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subjects are treated with a drug that is used to treat pain
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H0:=0.20
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Choose the correct answer below.

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