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: A work measurement analyst in Dorben Company took 10 observations of a high production job. He performance-rated each cycle and then computed the mean normal time for each element. The element with the greatest dispersion had a mean of 0.30 min and a standard deviation of 0.03 min. If it is desirable to have sampled data within 5 percent of the true data, how many observations should this time study analyst take of this operation?

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opudodennis

Answer:

21 Observations

Explanation:

Sample size (n)= 10 observations

Sample mean [tex]X_{mean}[/tex] is 0.3

Standard deviation S= 0.03

Probability of 0.05=10 observations -1 observation=9 observations

Number of observations

Let table value from appendix 3 Table A3-3 of 13th Edition of Niebel's Methods, Standards, & Work Design be t where k=5%=0.05

N=[tex](\frac {ts}{kX_{mean}})^{2}[/tex]

From the table, t=2.262

Number of observations =[tex](\frac {2.262*0.03}{0.5*0.3})^{2}[/tex] =20.46658

Number of observations=21 observations

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