# Thread: Confidence interval. (Difference of two means, Pop. variances unknown)

1. ## Confidence interval. (Difference of two means, Pop. variances unknown)

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.

2. Have you seen the paired t-test?

If you have the data analysis tool in excel you can follow this.

EXCEL Statistics tutorials paired t-test

3. Thanks. Using Excel was a real good way of confirming that my answer was correct. Much appreciated.

4. Just subtract one column from the other, as either weight gain or loss, depending on how you subtract.
Then perform a one-sample t test with 6 dfs.
You cannot use the s-pooled stuff here, since the two sets are highly dependent.