Answer:
The null hypothesis is ![H_o: p \leq 0.2](https://tex.z-dn.net/?f=H_o%3A%20p%20%5Cleq%200.2)
The alternate hypothesis is ![H_a: p > 0.2](https://tex.z-dn.net/?f=H_a%3A%20p%20%3E%200.2)
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](https://tex.z-dn.net/?f=H_o%3A%20p%20%5Cleq%200.2)
And the alternate hypothesis is that more than 20% develop, so:
![H_a: p > 0.2](https://tex.z-dn.net/?f=H_a%3A%20p%20%3E%200.2)
The test statistic is:
![z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}](https://tex.z-dn.net/?f=z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D%7D)
In which X is the sample mean,
is the value tested at the null hypothesis,
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}](https://tex.z-dn.net/?f=%5Cmu%20%3D%200.2%2C%20%5Csigma%20%3D%20%5Csqrt%7B0.2%2A0.8%7D)
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](https://tex.z-dn.net/?f=n%20%3D%20229%2C%20X%20%3D%20%5Cfrac%7B52%7D%7B229%7D%20%3D%200.2271)
Value of the test statistic:
![z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}](https://tex.z-dn.net/?f=z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D%7D)
![z = \frac{0.2271 - 0.20}{\frac{\sqrt{0.2*0.8}}{\sqrt{229}}}](https://tex.z-dn.net/?f=z%20%3D%20%5Cfrac%7B0.2271%20-%200.20%7D%7B%5Cfrac%7B%5Csqrt%7B0.2%2A0.8%7D%7D%7B%5Csqrt%7B229%7D%7D%7D)
![z = 1.03](https://tex.z-dn.net/?f=z%20%3D%201.03)
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%.