Question

Suppose 236 subjects are treated with a drug that is used to treat pain and 53 of them developed nausea. Use a 0.10 significance level to test the claim that more than 20​% of users develop nausea.

1. Identify the null and alternative hypotheses for this test.

2. Find the test statistic for this hypothesis (round to three decimal)

3. Find the P-value

Answer 1

Solution :

Given that,

= 0.20

1 - = 0.80

n = 236

x = 53

Level of significance = = 0.10

Point estimate = sample proportion = = x / n = 0.225

This a right (One) tailed test.

1) The null and alternative hypothesis is,

Ho: p = 0.20

Ha: p 0.20

2)Test statistics

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

= ( 0.225 - 0.20) / (0.20*0.80) /236

= 0.960

3)P-value = P(Z>z)

= 1 - P(Z <z )

= 1- P(Z < 0.960)

= 1 - 0.8315

= 0.1685

The p-value is p = 0.1685, and since p = 0.1685 >0.10, it is concluded that the null hypothesis is not rejected.