Question

In a sample of 28 cups of coffee at the local coffee shop, the temperatures were normally distributed with a mean of 162.5 degrees with a sample standard deviation of 16.7 degrees. What would be the 95% confidence interval for the temperature of your cup of coffee?

Answer 1

The 95% confidence interval is (135.0285 degrees, 189.9715 degrees).

Explanation:

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

`\alpha = (1-0.9)/(2) = 0.05`

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
`1-0.05 = 0.95`, so
`z = 1.645`

Now, find M as such

`M = z*s`

In which s is the standard deviation of the sample. So

`M = 1.645*16.7 = 27.4715`

The lower end of the interval is the mean subtracted by M. So it is 162.5 - 27.4715 = 135.0285 degrees

The upper end of the interval is the mean added to M. So it is 162.5 + 27.4715 = 189.9715 degrees

The 95% confidence interval is (135.0285 degrees, 189.9715 degrees).