Question

Suppose 244 subjects are treated with a drug that is used to treat pain and 51 of them developed nausea. Use a 0.01 significance level to test the claim that more than 20​% of users develop nausea.

Answer 1

The null hypothesis,
`H_(0)` should not be rejected

Explanation:

Out of 244 subjects, 51 subjects develop nausea:

Sample proportion,
`\hat{p} = (51)/(244) \\`

`\hat{p} = 0.209`

We want to test whether Population proportion, p > 20% or not

i.e whether
`p > (20)/(100)` or not

i.e whether p > 0.2 or not

Null hypothesis, H₀ : p ≤ 0.2

Alternative hypothesis,
`H_(a) : p > 0.2`

The test statistic is given by the formula:

`z = \frac{\hat{p} -p_(0) }{\sqrt{(p_(0)(1-p_(0) ) )/(n) } }`

n = 244

`p_(0) = 0.2`

`1 - p_(0) = 1-0.2 = 0.8`

`\hat{p}-p_(0) = 0.209 - 0.2 = 0.009`

`z = \frac{0.009 }{\sqrt{(0.2(0.8 ) )/(244) } }`

z = 0.3515

p(z>0.3515) = 1 - p(z≤0.3515) = 1 - 0.637

p(z>0.3515) = 0.363

p value = 0.363

α = 0.01

Since 0.363 > 0.01

The null hypothesis,
`H_(0)` should not be rejected

There is no evidence that supports the claim that the null hypothesis should be rejected at 0.01 significance level

Answer 2

`z=\frac{0.209 -0.2}{\sqrt{(0.2(1-0.2))/(244)}}=0.351`

`p_v =P(z>0.351)=0.363`

So the p value obtained was a very high value and using the significance level given
`\alpha=0.01` we have
`p_v>\alpha` so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 1% of significance the proportion of subjects with nauseas is not higher than 0.2 or 20% at 1% of significance.

Explanation:

Data given and notation

n=244 represent the random sample taken

X=51 represent the subjects with nausea

`\hat p=(51)/(244)=0.209` estimated proportion of subjects with nausea

`p_o=0.2` is the value that we want to test

`\alpha=0.01` represent the significance level

Confidence=99% or 0.99

z would represent the statistic (variable of interest)

`p_v` represent the p value (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the true proportion is higher than 0.2.:

Null hypothesis:
`p \leq 0.2`

Alternative hypothesis:
`p > 0.2`

When we conduct a proportion test we need to use the z statistic, and the is given by:

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

The One-Sample Proportion Test is used to assess whether a population proportion
`\hat p` is significantly different from a hypothesized value
`p_o`.

Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

`z=\frac{0.209 -0.2}{\sqrt{(0.2(1-0.2))/(244)}}=0.351`

Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level provided
`\alpha=0.01`. The next step would be calculate the p value for this test.

Since is a right tailed test the p value would be:

`p_v =P(z>0.351)=0.363`

So the p value obtained was a very high value and using the significance level given
`\alpha=0.01` we have
`p_v>\alpha` so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 1% of significance the proportion of subjects with nauseas is not higher than 0.2 or 20% at 1% of significance.