Mathematics, 22.04.2020 04:07, amanramesh2004

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

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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.

answered: Guest

answer: add the !

step-by-step explanation: then we will answer

answered: Guest

answer: the price of one pound cabbage = $ 0.8

the point (10,8) represents that the cost of 10 pounds cabbage is $ 8

step-by-step explanation:

since, here is a proportional relation between the number of pounds of cabbage (x) and the price in dollars (y).

therefore, we can write,

$y\propto x$

⇒ y = k x

where k is the constant of proportionality.

now, point (5,4) is in the line.

thus, it will satisfies the above equation of line.

⇒ 4 = 5 k

⇒ k = 0.8

thus, the equation of the line is,

y = 0.8 x

a) for x = 1,

y = 0.8

therefore, the cost of one pound of cabbage is 0.8 dollars.

b) in the below graph (10,8) represents that the cost of 10 pounds of cabbage is 8 dollars.

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5. The following sample observations on total coating layer thickness (in mm) of eight wire electrod...