The operations manager of a large production plant would like to estimate the average amount of time workers take to assemble a new electronic component. After observing a number of workers
assembling similar devices, she guesses that the standard deviation is 6 minutes. How large a sample of workers should she take if she wishes to estimate the mean assembly time to within 10 seconds?
Assume that the confidence level is to be 90%.


Given that the standard deviation $\sigma =6$ minutes $=360$ seconds, margin of error $E =10$ seconds. The confidence coefficient is $1-\alpha=0.9$. Thus $\alpha = 0.1$.

The formula to estimate the sample size required to estimate the mean is

$$ n =\bigg(\frac{z\sigma}{E}\bigg)^2 $$

where $z$ is the $Z_{\alpha/2}$, $\sigma$ is the population standard deviation and $E$ is the margin of error.

Z-critical 0.1
Z-critical 0.1

For $\alpha=0.1$, the critical value of $Z$ is $z=Z_{\alpha/2} = 1.64$.

The minimum sample size required to estimate the mean is

$$ \begin{aligned} n&= \bigg(\frac{z*\sigma}{E}\bigg)^2\\ &= \bigg(\frac{1.64*360}{10}\bigg)^2\\ &=3485.7216\\ &\approx 3486. \end{aligned} $$

Thus, the sample of size $n=3486$ will ensure that the $90$% confidence interval for the mean will have a margin of error $10$.

Further Reading