Question

The dean of a university wants to use the mean of a random sample to estimate the average amount of time students take to get from one class to the next. She wants to assert with probability 0.95 that her error will be at most 0.25 minutes. If she knows from the past studies that population standard deviation is 15 minutes, how large a sample she will need?

Answer 1

The standard error$(E=|\bar{x}-\mu|)$ is given as,

$E=|\bar{x}-\mu|=Z_{\alpha/2}*(\sigma/\sqrt{n})$

We make the sample size $n$ the subject of the formula in order to determine its value.

Now,

$E=Z_{\alpha/2}(\sigma/\sqrt{n})\implies E/Z_{\alpha/2}=\sigma/\sqrt{n}$

Taking the reciprocal,

$Z_{\alpha/2}/E=\sqrt{n}/\sigma\implies (Z_{\alpha/2}/E)*\sigma=\sqrt{n}$

We then square both sides of the equation,

$(Z_{\alpha/2}/E)*\sigma)^2=(\sqrt{n})^2\implies (Z_{\alpha/2}/E)*\sigma)^2=n$

The data above provides the following values,

$\alpha=1-0.95=0.05,\space \sigma=15,\space E=0.25$

The value for $Z_{\alpha/2}=Z_{0.05/2}=Z_{0.025}$ is obtained from the standard normal tables as,

$Z_{0.025}=1.96$

The value of $n$ is,

$n=((1.96/0.25)*15)^2= 13829.76\approx 13830$

Hence, the sample size she will need is $n=13830$