Question

A coffee vending machine fills 100 cups of coffee before it has to be refilled. On a Monday, the mean number of ounces filled in a cup of coffee was 7.5. The machine is known to have a standard deviation in filling volume of 0.25 ounces. Find a 99% confidence interval for the mean number of ounces dispensed by this machine.

Answer 1

The 99% confidence interval for the mean number of ounces dispensed by this machine is (7.44, 7.56).

Explanation:

We have that to find our
`\alpha` level, that is the subtraction of 1 by the confidence interval divided by 2. So:

`\alpha = (1 - 0.99)/(2) = 0.005`

Now, we have to find z in the Z-table as such z has a p-value of
`1 - \alpha`.

That is z with a pvalue of
`1 - 0.005 = 0.995`, so Z = 2.575.

Now, find the margin of error M as such

`M = z(\sigma)/(√(n))`

In which
`\sigma` is the standard deviation of the population and n is the size of the sample.

`M = 2.575(0.25)/(√(100)) = 0.06`

The lower end of the interval is the sample mean subtracted by M. So it is 7.5 - 0.06 = 7.44 ounces.

The upper end of the interval is the sample mean added to M. So it is 7.5 + 0.06 = 7.56 ounces.

The 99% confidence interval for the mean number of ounces dispensed by this machine is (7.44, 7.56).