Students may choose between a 3-semester-hour course in physics without labs and a 4-semester-hour course with labs. The final written examination is the same for each section. If 12 students in the section with labs made an average grade of 84 with a standard deviation of 4, annd 18 students in the section without labs made an average grade of 77 with a standard deviation of 6, find a 99% confidence interval for the difference between the average grades for the two courses. Assume the populations to be approximately normally distributed with equal variances.

Attempt:

\(
\nu = \frac{(s_1^2/n_1 + s_2^2/n_2)^2}{[(s_1^2/n_1)^2/(n_1-1)] + [(s_2^2/n_2)^2/(n_2-1)]} = \frac{(16/12 + 36/18)^2}{[(16/12)^2/11] + [(36/18)^2/17]} = 27.99
\)

\(\nu=28\) gives a t-value of 2.763 when leaving 0.005 to the right.

\(
7-2.763 sqrt{\frac{s_1^2}{n_1} +\frac{s_2^2}{n_2}} < \mu_1 - \mu_2 < 7 + 2.763 sqrt{\frac{s_1^2}{n_1} +\frac{s_2^2}{n_2}}
\)

\(
1.96 < \mu_1 - \mu_2 < 12.04
\)

According to my book, the correct answer is

\(
1.5 < \mu_1 - \mu_2 < 12.5
\)

Have I made a mistake somewhere?

 

I'm a little rusty with this, but I believe that with following:
http://www.stat.psu.edu/~resources/Flowcharts/mxc_02/sld001.htm
We would use the bottom equation

So the pooled variance:

\(s^{2}_{p}\) = ((12 - 1) * 4^2) + ((18 - 1) * 6^2)) / ((12 - 1) + 18-1)
\(s^{2}_{p}\) = 28.14

To find the t-value you can just look up .005 at df = 12 + 18 - 2

Our Error is then:

E = 2.763 * √ ((28.14 / 12) + (28.14 / 18))
E = 5.46

7 ± 5.46 = 1.54 and 12.46

When rounded to one decimal spot it seems to match up with your textbook answer.

Of course I could be completely wrong, it's 2AM here