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<\mu<1.3152`

Explanation:

Confidence interval for population mean is given by :-

`\overline{x}-z_(\alpha/2)(\sigma)/(√(n))< mu< \overline{x}+ z_(\alpha/2)(\sigma)/(√(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)(0.2)/(√(20))<\mu<1.2+ (2.576)(0.2)/(√(20))\\\\=1.2- 0.1152<\mu<1.2+ 0.1152\\\\=1.0848<\mu<1.3152`

Hence, the 99 percent two-sided confidence interval for the mean nicotine content of a cigarette:
`1.0848<\mu<1.3152`