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?

So the p value obtained was a very low value and using the significance level given [tex]\alpha=0.01[/tex] we have [tex]p_v<\alpha[/tex] 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

[tex]\hat p=0.021[/tex] estimated proportion of interest

[tex]p_o=0.012[/tex] is the value that we want to test

[tex]\alpha=0.01[/tex] represent the significance level

Confidence=99% or 0.99

z would represent the statistic (variable of interest)

[tex]p_v[/tex] 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:[tex]p \leq 0.012[/tex]

Alternative hypothesis:[tex]p > 0.012[/tex]

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

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

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

Calculate the statistic

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

[tex]z=\frac{0.021 -0.012}{\sqrt{\frac{0.012(1-0.012)}{1655}}}=3.363[/tex]

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 [tex]\alpha=0.01[/tex]. The next step would be calculate the p value for this test.

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

[tex]p_v =P(z>3.363)=0.00039[/tex]

So the p value obtained was a very low value and using the significance level given [tex]\alpha=0.01[/tex] we have [tex]p_v<\alpha[/tex] 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%)