Question

A technician in a quick oil change shop had the following service times in minutes) for 20 randomly selected cars: Sample 2 4 4.5 4.2 4.2 4.3 4.3 4.6 4.5 4.4 4.7 4.3 4.5 4.6 4.4 4.4 4.6 4.7 4.6 4.8 4.5 4.9 a) Determine the mean and range of each sample b) Determine the mean of the sample means (grand mean) and the ranges. c) Assuming a standard deviation of 0.086 and a z value of 3, calculate the upper and lower control limits for the sample mean and range. Indicate if the process is in control and why or why not. d) Assume standard deviation is not known. Use the alternate method and calculate the upper and lower control limits for the sample mean and range. Indicate if the process is in control and why or why not.

Answer 1

# SAMPLE	1	2	3	4	5	AVERAGE	RANGE
1	4.5	4.2	4.2	4.3	4.3	4.3	0.3
2	4.6	4.5	4.4	4.7	4.3	4.5	0.4
3	4.5	4.6	4.4	4.4	4.6	4.5	0.2
4	4.7	4.6	4.8	4.5	4.9	4.7	0.4
AVERAGE						4.5	0.325

C. MEAN / X-BAR = 4.5

STANDARD DEVIATION = 0.086

Z = 3

SAMPLE SIZE(N) = 5

UCL = X-BAR + (Z * STDEV / SQRT(N)) = 4.5 + (3 * 0.086 / SQRT(5)) = 4.6154

LCL = X-BAR - (Z * STDEV / SQRT(N)) = 4.5 - (3 * 0.086 / SQRT(5)) = 4.3846

NO, THE PROCESS IS NOT IN CONTROL AS THE MEAN OBSERVATIONS ARE EXCEEDING THE UCL

D. RANGE = LARGEST OBSERVATION - SMALLEST OBSERVATION

XBAR = 4.5

RBAR = 0.325

A2 VALUE CORRESPONDING TO N = 5 = 0.577

D3 & D4 VALUES CORRESPONDING TO N = 5 = 0, 2.115

CONTROL LIMITS FOR XBAR

UCL X = XBAR + (A2 * RBAR) = 4.5 + (0.577 * 0.325) = 4.6875

LCL X = XBAR - (A2 * RBAR) = 4.5 - (0.577 * 0.325) = 4.3125

NO, BECAUSE THERE ARE POINTS EXCEEDING BOTH THE UCL AND LCL

CONTROL LIMITS FOR RBAR

UCL R = RBAR * D4 = 0.325 * 2.115 = 0.6874

LCL R = RBAR * D3 = 0.325 * 0 = 0

YES, SINCE ALL THE POINTS ARE WITHIN THE CONTROL LIMITS

**THE QUESTION DOES NOT MENTION THE SAMPLE SIZE FOR ANY PART. I HAVE ASSUMED A SIZE OF N = 5 GIVEN FOR 4 SAMPLES, IF THERE ARE ANY PROBLEMS YOU FACE, JUST LEAVE A COMMENT ANC I WILL GET BACK TO YOU AS SOON AS POSSIBLE.