Cristian M. answered 03/07/21
MS Statistics Graduate with 5+ Years of Tutoring Experience
(A quick note before I start here: I almost submitted this response with a z-value for the critical value! Before I could submit it, I changed it to a t-value. Remember to take a second to make the correct decision about using the z- or t-distribution for your critical value before doing problems like this.)
Let sample size n be 53, let sample mean x-bar be 232.082, and let sample standard deviation s be 217.129.
Since the population mean and population standard deviation are unknown, we get our critical value from the t-distribution, and it will be with n - 1 = 53 - 1 = 52 degrees of freedom. Let your critical value be tα/2 = t0.01/2 = t0.005 = 2.674.
(This comes from the fact that we're using 99% confidence, so 1% of area is left under the curve. This is turned into a decimal of 0.01 and cut in half. Then you find what t-value with 53 df has 0.005 area to its right. Also, if you're limited to a printed t-table that gives you rows for 50 and 60 df, defer to the row with 50 df. In which case, use 2.678. I use the more precise 2.674 here.)
Then your confidence interval will be set up this way:
x-bar ± tα/2(s/√n)
232.082 ± (2.674)(217.129/√53)
232.082 ± (2.674)(29.825)
232.082 ± 79.752
Find the lower bound 232.082 - 79.752.
Find the upper bound 232.082 + 79.752.
We get an interval of (152.33, 311.83).
(Running this TInterval procedure in a TI-83 or TI-84 gets you an interval of (152.34, 311.83). This is merely a matter of accuracy and rounding.)
With 99% confidence, we estimate the mean amount of caffeine consumed by women is between 152.33 mg and 311.83 mg.
Another way to interpret this interval is to say that we are 99% confident that the mean amount of caffeine consumed by women is between 152.33 mg and 311.83 mg.