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 55 ``` Compute 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 = (mean xf - mean xp)+/- z * sqrt[sf/nf + sp/np] = (83 - 70) +/- 1.645 sqrt[ 12/45 + 70/55] = 13 +/- 1.645 sqrt[ 1.5394 ] = 13 +/- 1.645 * 1.241 = 13 +/- 2.041 mean score interval is 10.959 to 15.041 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 : (83 - 70) +/- 1.96 * sqrt[ (122)/45 + (152)/55] Also The z-value 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 Go to Math Central