*It is claimed that a new diet will reduce a person's by 4.5 kilograms on the aver age in a period of 2 weeks. the weights or 7 women who followed this diet were recorded before and after a 2-week period*

$\displaystyle

\begin{tabular}{|c c c|}

\hline

Woman &Weight Before & Weight After\\

\hline

1&58.5&60.0\\

2&60.3&54.9\\

3&61.7&58.1\\

4&69.0&62.1\\

5&64.0&58.5\\

6&62.6&59.9\\

7&56.7&54.4\\

\hline

\end{tabular}}

$

Test the manufacturer's claim by computing a 95% confidence interval for the mean difference in the weight. Assume the differences of the weights to be approximately normally distributed.

**Ok, maybe I'm reading something wrong, but for this problem the population variances aren't known, correct? If so, I should use a T-test, right? My problem is that this question doesn't seem to indicate whether $\displaystyle \sigma^{2}_{1}=\sigma^{2}_{2} $ and all of the other problems I've been given in class have always made it point to mention this. Do I proceed in solving this under the assumption that the two population variances are equal or should a I solve assuming that the population variances aren't equal? Thanks**

**Edit:**

I think I'm reading the question wrong. This isn't a difference of two means problem. It's just a regular T-test problem. Sorry, it's been a long day.