PLEASE SHOW EACH AND EVERY STEP. THAK YOU!!! 971.9 770.6 950.4 970.3 669.8 788.6

ID: 3180432 • Letter: P

Question

PLEASE SHOW EACH AND EVERY STEP. THAK YOU!!!

971.9 770.6 950.4 970.3 669.8 788.6 962.2 914.7 896 627.4 871.9 729.4 885.4 1019 1134.3 523.6 791.9 981.4 701 568 649.8 980.7 1064 703 972.7 623.2 430.7 608.3 A grinding machine at Kruger Industrial Smoothing has been in service for a number of years, and the primary operator of the machine says that the machine has been pretty reliable. You collect a sample oftime between failure observations for the machine. One measure of reliability could be mean time between failures. Another could be the variability of time between failures. For the time between failure data (in operating hours) in homework 6.xlsx, find a 90% confidence interval for the standard deviation of time between failures. Interpret.

Explanation / Answer

The formula for confidence interval for standard deviation is as below:

Ö [ (n-1)*s^2 / X(R)^2] and Ö[ (n-1)*s^2 / X(L)^2]

n = 28

s = sample standard deviation

a = 1 – 0.90 = 0.10

a/2 = 0.10 / 2 = 0.05

(1-a/2) = 1 – 0.05 = 0.95

Degrees of freedom = n – 1 = 27 – 1 = 26

We use Chi-square table to find the critical chi-square value.

X(R)^2 = Critical chi-square value for 27 degrees of freedom at 5% level of significance = 40.113

X(L)^2 = Critical chi-square value for 27 degrees of freedom at 95% level of significance = 16.151

Now find the sample standard deviation:

s = Ö [ sum of ( x – x bar )^2 / (n-1)]

x bar = Sx / n

x bar = sample mean

n = sample size

x bar = Sx / n = 22760.2 / 28 = 812.8643

( x - x bar )

( x - x bar )^2

971.9

159.035714

25292.35833

770.6

-42.264286

1786.269871

950.4

137.535714

18916.07263

970.3

157.435714

24786.00404

669.8

-143.064286

20467.38993

788.6

-24.264286

588.7555751

962.2

149.335714

22301.15548

914.7

101.835714

10370.51265

896

83.135714

6911.546942

627.4

-185.464286

34397.00138

871.9

59.035714

3485.215527

729.4

-83.464286

6966.287037

885.4

72.535714

5261.429805

1019

206.135714

42491.93259

1134.3

321.435714

103320.9182

523.6

-289.264286

83673.82716

791.9

-20.964286

439.5012875

981.4

168.535714

28404.28689

701

-111.864286

12513.61848

568

-244.864286

59958.51856

649.8

-163.064286

26589.96137

980.7

167.835714

28168.82689

1064

251.135714

63069.14685

703

-109.864286

12070.16134

972.7

159.835714

25547.45547

623.2

-189.664286

35972.54138

430.7

-382.164286

146049.5415

608.3

-204.564286

41846.54711

Total =

891646.7843

s = Ö (891646.7843/(28-1)) = Ö33023.9549 = 181.7249

The value of sample standard deviation is 181.7249

Ö [ (n-1)*s^2 / X(R)^2] and Ö [ (n-1)*s^2 / X(L)^2]

Ö [ ((28-1)*181.7249^2)/ 40.113] and Ö[ ((28-1)*181.7249^2)/ 16.151]

149.09 and 234.96

The 90% confidence interval for the standard deviation is 149.09 and 234.96. Here we are 90% confident that the population standard deviation of time between failures will lie in between 149.09 and 234.96.