Question

A soft-drink machine is regulated so that the amount of drink dispensed is approximately normally distributed with standard deviation equal to 0.15 deciliter. Find a 95 % confidence level for the mean of all drinks dispensed by this machine if a random sample of 36 drinks has an average content of 2.25 deciliters.

Answer 1

The 95 % confidence level for the mean of all drinks dispensed by this machine is between 2.2010 deciliters and 2.2990 deciliters.

Explanation:

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

`\alpha = (1-0.95)/(2) = 0.025`

Now, we have to find z in the Ztable as such z has a pvalue of
`1-\alpha`.

So it is z with a pvalue of
`1-0.025 = 0.975`, so
`z = 1.96`

Now, find 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 = 1.96*(0.15)/(√(36)) = 0.049`

The lower end of the interval is the mean subtracted by M. So it is 2.25 - 0.049 = 2.2010 deciliters

The upper end of the interval is the mean added to M. So it is 2.25 + 0.049 = 2.2990 deciliters

The 95 % confidence level for the mean of all drinks dispensed by this machine is between 2.2010 deciliters and 2.2990 deciliters.