Answer to Question #301837 in Statistics and Probability for secret

Question

A random sample of 14 cigarettes of a certain brand has an average nicotine content of 4.4 milligrams and a standard deviation of 1.9 milligrams. What is the margin of error of the 99% confidence interval for the average nicotine content of the cigarettes.

Answer 1

The critical value for $\alpha = 0.01$ and $df = n-1 = 13$ degrees of freedom is $t_c = z_{1-\alpha/2; n-1} =3.012275.$

The corresponding confidence interval is computed as shown below:

$$CI=(\bar{x}-t_c\times\dfrac{s}{\sqrt{n}}, \bar{x}+t_c\times\dfrac{s}{\sqrt{n}})$$

The margin of error is

$$Margin \ of\ error =t_c\times\dfrac{s}{\sqrt{n}}$$

$$=3.012275\times\dfrac{1.9}{\sqrt{14}}\approx1.5296$$

The margin of error of the 99% confidence interval for the average nicotine content of the cigarettes is 1.5296 milligrams.