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

Answer:

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

Step-by-step explanation:

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

$\alpha = \frac{1-0.9}{2} = 0.05$

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).