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

Answer:

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%.

Step-by-step 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:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

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 $\mu = 0.2, \sigma = \sqrt{0.2*0.8}$

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

This means that $n = 229, X = \frac{52}{229} = 0.2271$

Value of the test statistic:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

$z = \frac{0.2271 - 0.20}{\frac{\sqrt{0.2*0.8}}{\sqrt{229}}}$

$z = 1.03$

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%.