r questions 33 to 37 based on below scenarios of 16 observations were obtained R

Question

r questions 33 to 37 based on below scenarios of 16 observations were obtained Range Mere obtained in an ceperiment. The sample means and anges are shown below 302 33) Assuming standard deviation limits for the mean-chart (round tol decimals standard deviation of the process is 144, what is the 98 (2% risk of Type lerror) lower control a) 492 b) 50.2 c) 483 d) 48.7 c) 51.3 What is the upper limit for the Mean chart if the standard deviation of the process is NOTknown (round to 3 decimals)? a) 50.586 b) 51.686 c) 49.812 d) 49 S12 e) 49 468 35) What is the lower limit for the mean chart if the standard deviation of the process is Nor known (3 decimals a) 50.586 b) 51.686 e) 50.468 d) 51.468 e) 49.494 36) What is the upper limit for the range chart (2 decimals)? a) 3.60 b) 3.29 c) 4 26 d) 3.60 e) 49.47 What is the lower limit for the range chart (2 decimals)? b) e) 81 1.01 e) .80

Explanation / Answer

33) a) 49.2

Estimate of mean X?? = (50.1+49.9+40.5+49.5+50.2)/5 = 50.04

Standard error of mean, s = ??n = 1.44/?16 = 0.36

z value for 0.98 cl = NORMSINV(1-(1-0.98)/2) = 2.33

Lower control limit = X?? - z*??n = 50.04 - 2.33*0.36 = 49.2

34) a) 50.586

If standard deviation is not known, we can calculate control limits using range

For sample size, n =16, Control chart constant A2 = 0.21

R? = (3+4+2+1+3)/5 = 2.6

Upper limit, UCL = X?? + A2*R? = 50.04+0.21*2.6 = 50.586

35) e) 49.494

Lower limit, LCL = X?? - A2*R? = 50.04-0.21*2.6 = 49.494

36) c) 4.26

Control chart constant D4 = 1.64

Upper limit for range chart = D4R? = 1.64*2.6 = 4.26

37) a) 0.94

Control chart constant D3 = 0.36

Upper limit for range chart = D4R? = 0.36*2.6 = 0.94

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