A production process fills containers by weight. Weights of containers are approximately normally distributed. Historically, the standard deviation of weights is 5.5 ounces. (This standard deviation is therefore known.) A quality control expert selects 36 containers at random. In this sample, the mean weight is 48.6 ounces. How large a sample would be required in order for the 99% confidence interval for to have a length of 2 ounces?


Given that the standard deviation $\sigma =5.5$ seconds, margin of error $E =1$. The confidence coefficient is $1-\alpha=0.99$. Thus $\alpha = 0.01$.

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.


For $\alpha=0.01$, the critical value of $Z$ is $z=Z_{\alpha/2} = 2.58$.

The minimum sample size required to estimate the mean is

$$ \begin{aligned} n&= \bigg(\frac{z*\sigma}{E}\bigg)^2\\ &= \bigg(\frac{2.58*5.5}{1}\bigg)^2\\ &=201.3561\\ &\approx 202. \end{aligned} $$

Minimum sample size $n =202$ will ensure that the $99$% confidence interval for the mean will have a margin of error $1$.

Further Reading