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.

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Answer:

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

Step-by-step explanation:

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

[tex]\alpha = \frac{1 - 0.99}{2} = 0.005[/tex]

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

That is z with a pvalue of [tex]1 - 0.005 = 0.995[/tex], so Z = 2.575.

Now, find the margin of error M as such

[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]

In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.

[tex]M = 2.575\frac{0.25}{\sqrt{100}} = 0.06[/tex]

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).