# A coal company wants to determine a 95% confidence interval estimate for the daily tonnage of coal that they mine. Assuming that the standard deviation of daily output is 200 tons, how many days should they sample so that the margin of error is 39.2 tons?

Apr 30, 2017

100 days

#### Explanation:

The margin of error can be stated as

$M E = S E \cdot c v$ where $c v = {z}_{.95}$ because this corresponds to a two tail test so we need the z value at 95%.

We can now use a look up table to find the corresponding z value to use which should be $1.96$. Now $S E = \setminus \frac{\sigma}{\sqrt{n}}$ leading to
$39.2 = \frac{200}{\sqrt{n}} \cdot 1.96$. Now we just need to solve for $n$

$\sqrt{n} = \frac{200}{39.2} \cdot 1.96 = 10$

$n = {10}^{2}$

so we probably need to sample for about 100 days if we want be within that margin of error