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

Answer:

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

$p_v =P(z>0.351)=0.363$  

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.

Step-by-step explanation:

Data given and notation  

n=244 represent the random sample taken

X=51 represent the subjects with nausea

$\hat p=\frac{51}{244}=0.209$ 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:  

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (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:  

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

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.

Answer 2

Answer:

The null hypothesis, $H_{0}$ should not be rejected

Step-by-step explanation:

Out of 244 subjects, 51 subjects develop nausea:

Sample proportion, $\hat{p} = \frac{51}{244}$

$\hat{p} = 0.209$

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

i.e whether $p > \frac{20}{100}$ 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:

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

n = 244

$p_{0} = 0.2$

$1 - p_{0} = 1-0.2 = 0.8$

$\hat{p}-p_{0} = 0.209 - 0.2 = 0.009$

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

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