The body mass index () is a factor in assessing the health of a person. The data given are the for random samples of 18 women and 20 men selected in a town. Assume that the for such women are normally distributed with variance 25, while the for such men are normally distributed with the variance of 16. Construct a 96% confidence interval for the difference between the mean of the women and men in the town. Interpret your results.
The above is the Question given, from my understanding the variance given is the sample variance but which test should I use because I don't know whether the population variance is equal or unequal.
The question has provided a set of data but didn't mention it is population data or sample data and request to pick the data randomly from the dataset.
Thanks for explaining why you're having trouble with this problem.
It is seems likely that the data given are for samples of 18 women and 20 men. If so, find the sample means $\bar X$ and $\bar Y$ for women and men, respectively. (Because population variances are given, there is no point in estimating them from sample data.)
The standard error for the difference $\bar X -\bar Y$ of sample means of women and men is $$\mathrm{SE}= \sqrt{\frac{\sigma_x^2}{n_x}+\frac{\sigma^2_y}{n_y}}= \sqrt{\frac{25}{18}+\frac{16}{20}}.$$
So a 95% confidence interval for $\mu_x - \mu_y$ is of the form $$\bar X - \bar Y \pm 1.96\mathsf{SE}.$$
Can you find formulas similar to these in your text or class notes? For a 90% CI you would use a different probability factor (instead of $1.96$) from standard normal tables.