The Professor's Response
Hi! I am doing number 48 and for some reason I don't get the answer posted on the hw, i get 1.07, 5.13. However, when I do it on my graphing calculator like you did in the video, I get the right answer. I was wondering if you had a video of you doing the problem w/o the graphing calculator so I could see where I am doing it wrong.
See the professor's answer below.
Hi Ruby,
The mistake you are making is that you are not using the t critical value like you are supposed to use. You are using the z critical value. In step 2 of the problem, you are going to the t table and looking up 0.005 in one tail (for 99% confidence), which is correct, but then you go straight to the bottom of the table to get the critical z value 2.576.
That is your error. Instead of going to the bottom of the t table, you should be stopping in the 28 degrees of freedom row because your sample size is only 29. Remember for 30 or less, we will use the t critical value with n - 1 degrees of freedom, not the z critical value which is found in the last row of the table.
If you use the correct critical value, your margin of error will look like this:
2.763 * 4.25/SquareRoot(29) = 2.18057
Then you can add and subtract this margin of error to and from the mean to get the right interval.
Hope that makes sense,
Professor McGuckian