HW 8, Problem 1, Part 2: Finding Confidence Interval

The problem I'm inquiring about is:

"Construct a .99% confidence interval estimate of the difference μs−μf, where μs is the mean age of student cars and μf is the mean age of faculty cars."

I'm running into problems calculating the confidence interval for double data sets. Based off the In Class problems, I know you can calculate the confidence interval for one set of data two different ways:

1) [-t* , t*] x se + mu

2) x1 = mu + sd / sqrt(n) x T
x2 = mu + sd / sqrt(n) x T

Does anyone know how to correctly apply either of these formulas if more than one data set is being used? Any insight that could be offered on this problem would be greatly appreciated.

Check out Example 5.17 on pg.233 of the book. It's a very similar problem.

Your first formula looks correct, you just need to remember what mu is in this case (μs−μf) and that SE when analyzing the difference of two means using a t-test is not sd/sqrt(n).

Use the t-table to obtain the correct t* value with your df being 1 less than your smallest sample size.

Thanks so much for your response Paul!

I tried the way you suggested and I still am not getting the right answers. I've double and triple-checked my numbers, see if you can catch my mistake:

SE = sqrt[( (3.4^2) / 10 + (3.5^2) / 18 ) = 1.3552

df = 9, it's asking for a 99% confidence, therefore, according to the table, t* = 2.82

mu = μs−μf, so, 7.9 - 5.7 = 2.2

The formula is, mu +/- [-t* , t*] x SE,

so, 1.3552 x 2.82 = 3.8217

3.8217 + 2.2 = 6.02165

and, -3.8217 - 2.2 = -1.6217

So the answers I keep getting are (-1.6217, 6.02165) but they still aren't correct. Do you see where I am making my mistake?

I think that's correct except for t*. For a confidence interval you should use the two-tailed value from the t-table, so 3.25 rather than 2.82.

Thanks Paul, worked perfectly! I owe you one!