Question

Suppose 229 subjects are treated with a drug that is used to treat pain and 52 of them develop nausea use a 0.01 significance level to test the claim that more than 20% of users develop nausea identify the Noel and alternative hypothesis for this test

Answer 1

The null hypothesis is
`H_o: p \leq 0.2`

The alternate hypothesis is
`H_a: p > 0.2`

The pvalue of the test is 0.1515 > 0.01, which means that we do not reject the null hypothesis that 20% of less users develop nausea, that is, we have no sufficient evidence that this proportion is larger than 20%.

Explanation:

Test the claim that more than 20% of users develop nausea

This means that the null hypothesis is that 20% or less of the users develop nausea, that is:

`H_o: p \leq 0.2`

And the alternate hypothesis is that more than 20% develop, so:

`H_a: p > 0.2`

The test statistic is:

`z = (X - \mu)/((\sigma)/(√(n)))`

In which X is the sample mean,
`\mu` is the value tested at the null hypothesis,
`\sigma` is the standard deviation and n is the size of the sample.

0.2 is tested at the null hypothesis:

This means that
`\mu = 0.2, \sigma = √(0.2*0.8)`

Suppose 229 subjects are treated with a drug that is used to treat pain and 52 of them develop nausea.

This means that
`n = 229, X = (52)/(229) = 0.2271`

Value of the test statistic:

`z = (X - \mu)/((\sigma)/(√(n)))`

`z = (0.2271 - 0.20)/((√(0.2*0.8))/(√(229)))`

`z = 1.03`

Pvalue of the test an decision:

Probability of finding a proportion above 0.2271, which is 1 subtracted by the pvalue of z = 1.03

Looking at the z-table, z = 1.03 has a pvalue of 0.8485

1 - 0.8485 = 0.1515

The pvalue of the test is 0.1515 > 0.01, which means that we do not reject the null hypothesis that 20% of less users develop nausea, that is, we have no sufficient evidence that this proportion is larger than 20%.