Question 7.1: A time study analyst wants to estimate the time required to ......
A time study analyst wants to estimate the time required to perform a certain job. A preliminary study yielded a mean of 6.4 minutes and a standard deviation of 2.1 minutes. The desired confidence is 95 percent. How many observations will he need (including those already taken) if the desired maximum error is
a. ± 10 percent of the sample mean?
b. One-half minute?
a. s = 2.1 minutes z = 5 1.96
\overline{x} = 6.4 minutes a = 10%
n~=\left({\frac{z s}{a \overline{x}}}\right)^{2}~=~\left({\frac{1.96(2.1)}{.10(6.4)}}\right)^{2}~=~41.36;{\mathrm{round~up~to~42~observations}}b. e = .5 minute n\;=\;\left({\frac{z s}{e}}\right)^{2}\;=\;\left({\frac{1.96(2.1)}{.5}}\right)^{2}\;=\;67.77;\mathrm{round~up~to~68~observations}
Note: When the value of n is noninteger, round up.