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.

The degrees of freedom are given by:

The p value would be given by:

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

Step-by-step explanation:

Information given

represent the sample mean

represent the sample standard deviation

sample size  

represent the value to verify

represent the significance level

t would represent the statistic

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:  

Alternative hypothesis:  

The statistic would be given by:

 (1)  

Replacing the info we got:

The degrees of freedom are given by:

The p value would be given by:

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

46 dollars is the answer.

step-by-step explanation:

6*6 = 36,         36+10=46

$53.50

step-by-step explanation:

you have to subtract the amount of money chuck earned ($72.50) by the cost of the gas for the mower ($11.25) and the new string for the trimmer he bought ($7.75).

72.50-11.25=61.25

and then subtract $61.25 by the second thing he bought (the new string for the trimmer)

61.25-7.75=53.5

so in total it's $53.50 (53 dollars and 50 cents)