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?

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Answer:

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

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{1-0.9}{2} = 0.05[/tex]

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

So it is z with a pvalue of [tex]1-0.05 = 0.95[/tex], so [tex]z = 1.645[/tex]

Now, find M as such

[tex]M = z*s[/tex]

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

[tex]M = 1.645*16.7 = 27.4715[/tex]

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).