Question

Suppose 221 subjects are treated with a drug that is used to treat pain and 51 of them developed nausea. Use a 0.10 significance level to test the claim that more than 20​% of users develop nausea.

Answer 1

`z=\frac{0.231 -0.2}{\sqrt{(0.2(1-0.2))/(221)}}=1.152`

The p avlue for this case is given by:

`p_v =P(z>1.152)=0.125`

The p value for this case is higher than the significance level so then we have enough evidence to FAIL to reject the null hypothesis and we can't conclude that the true proportion is higher than 0.2 or 20%

Explanation:

Information provided

n=221 represent the random sample taken

X=51 represent the people with nausea

`\hat p=(51)/(221)=0.231` estimated proportion of people with nausea

`p_o=0.21` is the value to test

`\alpha=0.1` represent the significance level

z would represent the statistic

`p_v` represent the p value

Hypothesis to verify

We want to check if the true population is higher than 0.20, the system of hypothesis are.:

Null hypothesis:
`p \leq 0.2`

Alternative hypothesis:
`p > 0.2`

The statistic is given:

`z=\frac{\hat p -p_o}{\sqrt{(p_o (1-p_o))/(n)}}` (1)

Replacing the info given we got:

`z=\frac{0.231 -0.2}{\sqrt{(0.2(1-0.2))/(221)}}=1.152`

The p avlue for this case is given by:

`p_v =P(z>1.152)=0.125`

The p value for this case is higher than the significance level so then we have enough evidence to FAIL to reject the null hypothesis and we can't conclude that the true proportion is higher than 0.2 or 20%