Question

A sample of 20 cigarettes is tested to determine nicotine content and the average value observed was 1.2 mg. Compute a 99 percent two-sided confidence interval for the mean nicotine content of a cigarette if it is known that the standard deviation of a cigarette’s nicotine content is σ = .2 mg.

Answer 1

Answer:  $1.0848$

Step-by-step explanation:

Confidence interval for population mean is given by :-

$\overline{x}-z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}< mu< \overline{x}+ z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}$

, where $z_{\alpha/2}$ = two -tailed z-value for ${\alpha$ (significance level)

n= sample size .

$\sigma$ = Population standard deviation.

$\overline{x}$ = Sample mean

By considering the given information , we have

$\sigma=0.2\text{ mg}$

$\overline{x}=1.2\text{ mg}$

n= 20

$\alpha=1-0.99=0.01$

Using z-value table ,

Two-tailed Critical z-value : $z_{\alpha/2}=z_{0.005}=2.576$

The 99 percent two-sided confidence interval for the mean nicotine content of a cigarette will be :-

$1.2- (2.576)\dfrac{0.2}{\sqrt{20}}$

Hence, the 99 percent two-sided confidence interval for the mean nicotine content of a cigarette:  $1.0848$