Question

5. The following sample observations on total coating layer thickness (in mm) of eight wire electrodes used for Wire electrical-discharge machining (WEDM): 21 16 29 35 42 24 24 25 Calculate a 99% CI for the variance σ 2 , and the standard deviation σ of the coating layer thickness distribution

Answer 1

Answer:

99% CI for the variance $\sigma^{2}$ , and the standard deviation σ of the coating layer thickness distribution is [23.275 , 477.115] and [4.824 , 21.843] respectively.

Step-by-step explanation:

We are given that the following sample observations on total coating layer thickness (in mm) of eight wire electrodes used for Wire electrical-discharge machining (WEDM) : 21, 16, 29, 35, 42, 24, 24, 25

Firstly, the pivotal quantity for 99% confidence interval for the population variance is given by;

                          P.Q. = $\frac{(n-1)s^{2} }{\sigma^{2} }$  ~ $\chi^{2}__n_-_1$

where,  $s^{2}$ = sample variance = $\frac{\sum (X-\bar X)^{2} }{n-1}$  = 67.43

             n = sample of observations = 8

            $\sigma^{2}$ = population variance

Here for constructing 99% confidence interval we have used chi-square test statistics.

So, 99% confidence interval for the population variance, $\sigma^{2}$ is ;

P(0.9893 < $\chi^{2}_7$ < 20.28) = 0.99  {As the critical value of chi-square at 7

                                          degree of freedom are 0.9893 & 20.28}  

P(0.9893 < $\frac{(n-1)s^{2} }{\sigma^{2} }$ < 20.28) = 0.99

P( $\frac{0.9893 }{(n-1)s^{2} }$ < $\frac{1}{\sigma^{2} }$ < $\frac{20.28 }{(n-1)s^{2} }$ ) = 0.99

P( $\frac{(n-1)s^{2} }{20.28 }$ < $\sigma^{2}$ < $\frac{(n-1)s^{2} }{0.9893 }$ ) = 0.99

99% confidence interval for $\sigma^{2}$ = [ $\frac{(n-1)s^{2} }{20.28 }$ , $\frac{(n-1)s^{2} }{0.9893 }$ ]

                                                   = [ $\frac{7\times 67.43 }{20.28 }$ , $\frac{7\times 67.43 }{0.9893 }$ ]

                                                   = [23.275 , 477.115]

99% confidence interval for $\sigma$  = [ $\sqrt{23.275}$ , $\sqrt{477.115}$ ]

                                                  = [4.824 , 21.843]

Therefore, 99% CI for the variance $\sigma^{2}$ , and the standard deviation σ of the coating layer thickness distribution is [23.275 , 477.115] and [4.824 , 21.843] respectively.