To help consumers assess the risks they are taking, the Food and Drug Administra

Question

To help consumers assess the risks they are taking, the Food and Drug Administration (FDA) publishes the amount of nicotine found in all commercial brands of cigarettes. A new cigarette has recently been marketed. The FDA tests on this cigarette gave a mean nicotine content of 25.6 milligrams and standard deviation of 2.1 milligrams for a sample of n = 9 cigarettes. The FDA claims that the mean nicotine content exceeds 28.6 milligrams for this brand of cigarette, and their stated reliability is 98%. Do you agree?

Explanation / Answer

mean xbar=25.6

Standard deviation s=2.1

Sample size n=9

Degree of freedom =n-1=9-1=8

We have to test null hypothesis H0: µ<=28.6 versus alternative Ha: µ>28.6

As 98% reliability =0.02 and so right tailed critical t= 2.449

Calculate t=(xbar-28.6)/(s/sqrt(n))

=(25.6-28.6)/(2.1/sqrt(9))

=-4.29

As calculate t<2.449, we do not reject null hypothesis conclude that the mean nicotine content is at most 28.6

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