A methods and measurements analyst needs to develop a time standard for a certain task. In a preliminary study, he observed one of his workers perform this task five times, with the following results: $$\begin{array} { l r r r r r }
\text { Observation } & 1 & 2 & 3 & 4 & 5 \
\hline \text { Time (seconds) } & 84 & 76 & 80 & 84 & 76
\end{array}$$ How many observations should be made if the analyst wants to be 99.74 percent confident that the maximum error in the observed time is two seconds? Assume that the standard deviation of each task time is four seconds.
A) 5
B) 6
C) 25
D) 36
E) 49

Question

A methods and measurements analyst needs to develop a time standard for a certain task. In a preliminary study, he observed one of his workers perform this task five times, with the following results: $$\begin{array} { l r r r r r } 
\text { Observation } & 1 & 2 & 3 & 4 & 5 \
\hline \text { Time (seconds) } & 84 & 76 & 80 & 84 & 76
\end{array}$$ How many observations should be made if the analyst wants to be 99.74 percent confident that the maximum error in the observed time is two seconds? Assume that the standard deviation of each task time is four seconds.
A) 5
B) 6
C) 25
D) 36
E) 49

Quizplus · Accepted Answer

The sample size needed to achieve the desired level of confidence and maximum error can be calculated using the formula:n = (Z^2 * s^2) / E^2where:n is the sample sizeZ is the z-score corresponding to the desired level of confidences is the standard deviation of the task timeE is the maximum allowable errorPlugging in the given values, we get:n = (2.576^2 * 4^2) / 2^2 = 36Therefore, the analyst should make 36 observations to be 99.74 percent confident that the maximum error in the observed time is two seconds.