Question

A random sample of 14 cigarettes of a certain brand has an average nicotine content of 3.7 milligrams and a standard deviation of 1.5 milligrams. What is the standard 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 standard error is

$$SE_{\bar{X}} =\dfrac{s}{\sqrt{n}}=\dfrac{1.5}{\sqrt{14}}\approx0.4009$$

The standard error of the 99% confidence interval for the average nicotine content of the cigarettes is 0.4009 milligrams.