Subject: mean scores Hi An investigator wants to find out of there are any difference in "skills" between full and part time students. Records show the following: Student Mean Score Std Dev Number     Full time 83 12 45 Part time 70 15 55Compute a 95% confidence interval for the difference in mean scores.
my solution as the number of students is >= 30, the sample size is deemed large does this look right? seems strange to have the 70/55 for sp/np? thanks Hi Murray,I think, you have a small misprint. if you will look over the formula more carefully, you will find that you have to square standard deviations. So, your formula will be : Also The zvalue in the confidence interval formula is z(alpha/2). Thus for a 95% confidence interval this is z(alpha/2) = z(0.025) = 1.96, not 1.65. Andrei
