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

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

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 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.