Question

Naproxen, a generic anti-inflammatory, claims to have, on average, at least 50 milligrams of active ingredient. An independent lab tests a random sample of 75 tablets and finds the mean content of active ingredient in this sample is 46.6 milligrams with a standard deviation of 9 milligrams. If the lab doesn't believe the manufacturer's claim, what is the approximate p-value for the suitable test?

Answer 1

`t=(46.6-50)/((9)/(√(75)))=-3.272`

The first step is calculate the degrees of freedom, on this case:

`df=n-1=75-1=74`

Since is a one sided test the p value would be:

`p_v =P(t_((74))<-3.272)=0.000812`

Explanation:

Data given and notation

`\bar X=46.6` represent the sample mean

`s=9` represent the sample standard deviation for the sample

`n=75` sample size

`\mu_o =50` represent the value that we want to test

`\alpha` represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

`p_v` represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is at least 50 mg or no, the system of hypothesis would be:

Null hypothesis:
`\mu \geq 50`

Alternative hypothesis:
`\mu < 50`

If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

`t=(\bar X-\mu_o)/((s)/(√(n)))` (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

`t=(46.6-50)/((9)/(√(75)))=-3.272`

P-value

The first step is calculate the degrees of freedom, on this case:

`df=n-1=75-1=74`

Since is a one sided test the p value would be:

`p_v =P(t_((74))<-3.272)=0.000812`