Question

Fluoxetine, a generic anti-depressant, claims to have, on average, at least 20 milligrams of active ingredient. An independent lab tests a random sample of 80 tablets and finds the mean content of active ingredient in this sample is 18.7 milligrams with a standard deviation of 5 milligrams. If the lab doesn't believe the manufacturer's claim, what is the approximate p-value for the suitable test

Answer 1

Answer:

$t=\frac{18.7-20}{\frac{5}{\sqrt{80}}}=-2.326$    

$df=n-1=80-1=79$  

$p_v =P(t_{(79)}<-2.326)=0.0113$  

Step-by-step explanation:

Information given

$\bar X=18.7$ represent the sample mean for the content of active ingredient

$s=5$ represent the sample standard deviation for the sample  

$n=80$ sample size  

$\mu_o =20$ represent the value that we want to test

t would represent the statistic

$p_v$ represent the p value for the test

System of hypothesis

We need to conduct a hypothesis in order to check if the true mean for the active agent is at least 20 mg, the system of hypothesis would be:  

Null hypothesis:$\mu \geq 20$  

Alternative hypothesis:$\mu < 20$  

The statistic would be:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$  (1)  

Now we can calculate the statistic:

$t=\frac{18.7-20}{\frac{5}{\sqrt{80}}}=-2.326$    

P value

The degrees of freedom are calculated like this:

$df=n-1=80-1=79$  

Since is a one left tailed test the p value would be:  

$p_v =P(t_{(79)}<-2.326)=0.0113$