An advertising company wishes to estimate the population mean of the distribution of hours of television watched per household per day. Suppose that the population standard deviation of hours watched per household per day is known to be 2.8 hours. The company decides that it wants the 99% confidence interval for the population mean to be no longer than 0.5 (hour). What is the minimum sample size that will result in a small enough confidence interval?


Given that the standard deviation $\sigma =2.8$ seconds, margin of error $E =0.5$. 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

$$ \begin{aligned} n&= \bigg(\frac{z*\sigma}{E}\bigg)^2\\ &= \bigg(\frac{2.58*2.8}{0.5}\bigg)^2\\ &=208.7447\\ &\approx 209. \end{aligned} $$

Minimum sample size $n =209$ will ensure that the $99$% confidence interval for the mean will have a margin of error $0.5$.

Further Reading