Therefore SSE = 3.2 + 2.0 = 5.2. The df are equal to the sum of the df for and which is 4 + 3 = 7.

Since MSE = SSE/df, MSE = 5.2/7 = 0.743.

The harmonic mean of the n's is:

= 2/(0.2 + 0.25) = 4.4444.

= 0.578.

Finally, a t table can be used to find that the value of t (with 7 df) to be used for the 99% confidence interval is 3.499.

The confidence interval is therefore:

LL = -2.4 - (3.499)(0.578) = -4.42

UL = -2.4 + (3.499)(0.578) = -0.38

-4.42 ≤ µ