How many integrated circuits must be randomly selected and tested for time to failure in order to estimate the mean time to failure? We want 95% confidence that the sample mean is within 2 hr of the population mean, and the population standard deviation is known to be 18.6 hours.


Given that the margin of error $E =2$. The confidence coefficient is $1-\alpha=0.95$. Thus $\alpha = 0.05$.
The population standard deviation is $\sigma = 18.6$.

Z-critical value
Z-critical value

The critical value of $Z$ is $z=Z_{\alpha/2} = 1.96$.

The minimum sample size required to estimate the mean is

$$ \begin{aligned} n &= \bigg(\frac{z* \sigma}{E}\bigg)^2\\ & = \bigg(\frac{1.96*18.6}{2}\bigg)^2\\ & =332.26\\ &\approx 333. \end{aligned} $$

Thus, the sample of size $n=333$ will ensure that the $95$% confidence interval for the mean will have a margin of error $2$.

Further Reading