Question

Use the given degree of confidence and sample data to find a confidence interval for the population standard deviation sigma. Assume that the population has a normal distribution. Round the confidence interval limits to the same number of decimal places as the sample standard deviation. The mean replacement time for a random sample of 20 washing machines is 11.6 years and the standard deviation is 2.8 years. Construct a​ 99% confidence interval for the standard​ deviation, sigma​, of the replacement times of all washing machines of this type.

Answer 1

Answer:
`1.97<\sigma<4.67`

Explanation:

The confidence interval for population standard deviation is given by :-

`s\sqrt{((n-1))/(\chi^2_(n-1,\alpha/2))}<\sigma<s\sqrt{((n-1))/(\chi^2_(n-1,1-\alpha/2))}`

Given : Sample size :
`n=20`

Standard deviation :
`s=2.8\text{ years}`

Significance level :
`\alpha: 1-0.99=0.01`

Critical value using chi-square distribution table :

`\chi^2_(n-1,\alpha/2)=\chi^2_(19,0.005)=38.58`

`\chi^2_(n-1,1-\alpha/2)=\chi^2_(19,0.995)=6.84`

Then , 99% confidence interval for the standard​ deviation of the replacement times of all washing machines of this type will be :

`2.8\sqrt{((19))/(38.58)}<\sigma<2.8\sqrt{((19))/(6.84)}\\\\\\=1.9649600278<\sigma<4.66666666667\\\\\approx1.97<\sigma<4.67`