Managers at an automobile manufacturing plant would like to estimate the mean completion time of an assembly line operation. The managers plan to choose a random sample of completion times and estimate mean via the sample. Assuming that the standard deviation of the population of completion times is 10.3 minutes, what is the minimum sample size needed for the managers to be 95% confident that their estimate is within 2.2 minutes of Mean?


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

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

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

Given that the margin of error $E =2.2$. The confidence coefficient is $1-\alpha=0.95$. Thus $\alpha = 0.05$.

The population standard deviation is $\sigma = 10.3$.

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

$$ \begin{aligned} n &= \bigg(\frac{z* \sigma}{E}\bigg)^2\\ & = \bigg(\frac{1.96*10.3}{2.2}\bigg)^2\\ & =84.2056\\ &\approx 85. \end{aligned} $$

Thus, the sample of size $n=85$ will ensure that the $95$% confidence interval for the mean will have a margin of error $2.2$.

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