Question

The Following table is based on a time study for a metal working operation

		Obs (minutes per cycle)
Element	Performance Rating	1	2	3	4	5	5
A	108%	1.2	1.25	1.30	1.28	1.26	1.24
B	112%	0.83	0.81	0.78	0.81	0.83	0.79
C	105%	0.58	0.61	0.59	0.62	0.58	0.61

MUST SHOW ALL WORK TO RECEIVE FULL OR PARTIAL CREDIT.

a) Based on the number of observation, determine the standard time for the operation, assuming an allowance of 12 percent of job time.

b) How many observations would be needed to estimate the mean time for element B within 2 percent of its true value with a 95.5 percent confidence

Answer 1

(a)

For element A, Observed time = (1.2 + 1.25 + 1.3 + 1.28 + 1.26 + 1.24) / 6 = 1.255

For element B, Observed time = (0.83 + 0.81 + 0.78 + 0.81 + 0.83 + 0.79) / 6 = 0.808

For element C, Observed time = (0.58 + 0.61 + 0.59 + 0.62 + 0.58 + 0.61) / 6 = 0.598

Normal time for A, = 1.255*1.08 = 1.355

Normal time for B, = 0.808 * 1.12 = 0.905

Normal time for C, = 0.598* 1.05 = 0.628

Standard time for A = 1.355*(1+0.12) =  1.52

Standard time for B = 0.905*(1+0.12) = 1.01

Standard time for C = 0.628*(1+0.12) = 0.70

(b)

= (0.83+0.81+0.78+0.81+0.83+0.79)/6 = 0.81

s = sqrt[ (0.83-0.81)2 + (0.81-0.81)2 + (0.78 - 0.81)2 + (0.81 - 0.81)2 + (0.83 - 0.81)2 + (0.79 - 0.81)2 / (6-1) ]  = 0.0204

n = [ 2* 0.0204/ 0.02*0.81]2 = 6.34 = 7 observations

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