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lozanna [386]
3 years ago
13

Fluoxetine, a generic anti-depressant, claims to have, on average, at least 20 milligrams of active ingredient. An independent l

ab 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
Mathematics
1 answer:
san4es73 [151]3 years ago
6 0

Answer:

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

df=n-1=80-1=79  

p_v =P(t_{(79)}  

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)}  

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