Lecture 6 - Ch 8b - Confidence Interval for the Mean with *Unknown* Std Dev - Examples and Minitab

Example Problems
We did the following example problems (I'm not sure I captured all the details of the problems, but I think I got the important parts):

Ex1: Sample size, n=17
Find a 95% confidence interval.
A1: 0.95 is 1-alpha, so your alpha is 0.05. Therefore, you want to look up t_0.025,16, which is 2.120

Ex2: Sample size, n=14
Find a 90% confidence interval.
A2: 0.90 is 1-alpha, so your alpha is 0.1. Therefore you look up t_0.05,13, which is 1.771

Ex3: Sample size, n=25. xbar=50, s=8
Find a 95% confidence interval for the mean.
A3: From the t-table, you find t_0.025,24 is 2.064
Therefore, the 95% confidence interval for the mean is 50 +/- 2.064 (8/sqrt(25))

Ex4: (This is problem 8.69 in the book) Sample size, n=50, xbar=5.5, s=0.1
Find a 99% confidence interval for the mean.
A4: 0.99 is 1-alpha, so alpha is 0.01. We look up t_0.005,49 which is 2.68. (If your table doesn't list degrees of freedom for the one you're looking for, just use the closest one. In our case, I think we used 50 instead of 49. Close enough!)
Therefore, the 99% confidence interval for the mean is 5.5 +/- 2.68(0.1/sqrt(50))
= 5.5 +/- 0.0379
= (5.4621,5.5379)

Using Minitab to find Confidence Intervals

In Minitab, pull up your data into a column. We used the teabags.mtw sample data from the textbook problem 8.69. Select Stat - Basic Statistics - 1 sample t from the menu bar. Select your data column in the Samples in columns section. Click Options and set your confidence interval.

The output from Minitab looks like this:
Variable N Mean StDev SE Mean 99% CI
Teabags 50 5.50140 0.10583 0.01497 (5.46129, 5.54151)

Unfortunately, it doesn't give us the critical value of t (which we calculated to be 2.68). But you can see that the confidence interval is pretty much the same as the answer we got above manually. I think the difference is due to the fact that Minitab used more precise values for the mean and std dev.