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 publishes the amount of nicotine found in all commercial brands of cigarettes. A new cigarette recently been marketed. The FDA tests on this cigarette gave a mean nicotine content of 27.5 milligrams and standard deviation of 2.8 milligrams for a sample of n = 9 cigarettes. Assuming the involved population is approximately normal, construct a 98% confidence interval for the mean nicotine content of this brand of cigarette.

Explanation / Answer

Note that              

Margin of Error E = t(alpha/2) * s / sqrt(n)              
              
where              
alpha/2 = (1 - confidence level)/2 =    0.01          
X = sample mean =    27.5          
t(alpha/2) = critical t for the confidence interval =    2.896459448          
s = sample standard deviation =    2.8          
n = sample size =    9          
df = n - 1 =    8          
Thus,              

Margin of Error E =    2.703362151 = 2.703          

Hence, the confidence interval is X +/- E,

OPTION C: 27.5 +/- 2.703 [ANSWER, C]

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