Question

The nicotine contents of five cigarettes of a serration brand, measured in milligrams, are 21, 19, 23, 19, 23. Establish a 99% confidence interval estimate of the average nicotine content of this brand of cigarette.

Answer 1

Answer on Question 37685, Math, Statistics Let us find mean:

$(21+19+23+19+23)/5=21$

The standard deviation is

$\sigma=\sqrt{\frac{21^{2}+19^{2}+23^{2}+19^{2}+23^{2}-(21+19+23+19+23)^{2}/5}{N-1}}=2$

Our degree of freedom is

$df=5-1=4$

Our $\alpha$ is

$\alpha=(1-0.99)/2=0.005$

For df=4 and $\alpha=0.005$ we find coefficient from t-distribution table, it is equal to

$t=4.604$

Now, the 90% interval is

$21\pm t\cdot\frac{\sigma}{\sqrt{N}}=21\pm 4.604\cdot\frac{2}{\sqrt{21}}\approx 21\pm 2$