Question

Clarinex is a drug used to treat asthma. In clinical tests of this drug, 1655 patients were treated with 5- mg doses of Clarinex, and 2.1% of them experienced fatigue (based on data from the Schering Corporation). Use a 0.01 significance level to test the claim that the percentage of Clarinex users experiencing fatigue is greater than the 1.2% rate for those not using Clarinex. Does it appear that fatigue is an adverse reaction of Clarinex?

Answer 1

Answer:

$z=\frac{0.021 -0.012}{\sqrt{\frac{0.012(1-0.012)}{1655}}}=3.363$  

$p_v =P(z>3.363)=0.00039$  

So the p value obtained was a very low value and using the significance level given $\alpha=0.01$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the proportion of interest is significantly higher than 0.012 (1.2%)

Step-by-step explanation:

Data given and notation

n=1655 represent the random sample taken

$\hat p=0.021$ estimated proportion of interest

$p_o=0.012$ 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 true proportions is higher than 0.012.:  

Null hypothesis:$p \leq 0.012$  

Alternative hypothesis:$p > 0.012$  

When we conduct a proportion test we need to use the z statisitic, 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:  

$z=\frac{0.021 -0.012}{\sqrt{\frac{0.012(1-0.012)}{1655}}}=3.363$  

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:  

$p_v =P(z>3.363)=0.00039$  

So the p value obtained was a very low value and using the significance level given $\alpha=0.01$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the proportion of interest is significantly higher than 0.012 (1.2%)