Pooled t Confidence Interval for Difference in Population Means? For sample 1:
$n = 8,\,
\bar X = 2.20,\,
S = 0.11.$
For sample 2:
$n = 12,\,
\bar X = 2.29,\,
S = 0.07.$
Since confidence interval is 99%, $1 - \alpha = .99,$ so $\alpha$ is .01. Then $\alpha/2 = .005.$
So a Z-score of .995 would be 2.575, according to the table, which would be used in the final calculations.
What I am getting then by using the correct formula is:
$(2.20 - 2.29) \pm \sqrt{(0.11)^2/8 + (0.07)^2/8}(2.575).$
Is this correct?
 A: The question relates to a 'pooled' CI based on Student's t distribution:
'Pooled' because you are told to assume population variances are equal;
'Student's t' (not standard normal) because the (common) population variance
is unknown and sample sizes are much too small to use a normal approximation.
So you are correct that working this problem has nothing to do with the standard
normal distribution.
The statistic
$$T = \frac{\bar X_M - \bar X_E}{S_p\sqrt{\frac{1}{n_M}+\frac{1}{n_E}}}
= \frac{\bar X_M - \bar X_E}{\text{SE}} \sim \mathsf{T}(n_M + n_E - 2),$$
where $S_p^2 = \frac{(n_M-1)S_M^2 + (n_E - 1)S_E^2}{n_M + n_E - 2}.$
Thus a 99% CI for $\mu_M - \mu_E$ is of the form
$$\bar X_m - \bar X_E \pm t^* \text{SE},$$
where $t^*$ cuts probability $.005 = 0.5\%$ from the upper tail of
$\mathsf{T}(n_M + n_E - 2).$ 
So here is what you need to do: 
(1) Find 
$S_p^2 = \frac{7(0.11^2) + 11(0.07^2)}{8+12-2}.$
(2) Find $\text{SE} = S_p\sqrt{1/8 + 1/12}.$
(3) Use row $\text{df} = 18$ of your t-table
to find $t^* = 2.878.$
(4) Use the second displayed formula above to find the confidence interval
and locate the correct answer in your list.
(5) Find all of these formulas (or very similar ones) in your textbook
so you will be able to put all of this together on your own next time.
