Question

A nighttime cold medicine’s label indicates the presence of 600 mg of acetaminophen in each fluid ounce of the drug. The FDA randomly selects 65 1-ounce samples and finds the mean content is 595 mg with a standard deviation of 20 mg. Is there evidence that the label is incorrect? Use a= .05.

Answer 1

`t=(595-600)/((20)/(√(65)))=-2.016`

The degrees of freedom are given by:

`df=n-1=65-1=64`

The p value would be given by:

`p_v =2*P(t_((64))<-2.016)=0.048`

And for this case since the p value is lower than the significance level we have enough evidence to reject the null hypothesis and we can conclude that the true mena is different from 600 mg

Explanation:

Information given

`\bar X=595` represent the sample mean

`s=20` represent the sample standard deviation

`n=65` sample size

`\mu_o =600` represent the value to verify

`\alpha=0.05` represent the significance level

t would represent the statistic

`p_v` represent the p value

System of hypothesis

We want to test if the true mean is different from 600 mg, the system of hypothesis would be:

Null hypothesis:
`\mu = 600`

Alternative hypothesis:
`\mu \\eq 600`

The statistic would be given by:

`t=(\bar X-\mu_o)/((s)/(√(n)))` (1)

Replacing the info we got:

`t=(595-600)/((20)/(√(65)))=-2.016`

The degrees of freedom are given by:

`df=n-1=65-1=64`

The p value would be given by:

`p_v =2*P(t_((64))<-2.016)=0.048`

And for this case since the p value is lower than the significance level we have enough evidence to reject the null hypothesis and we can conclude that the true mena is different from 600 mg