If you do not have access to a package like SPSS which is outstanding in its human factors, (clarity, tutorials, stat coach etc.), you might google web calculators. They would then need no special equipment except their web browsers which are available on most campuses.
A couple of guesses. It looks as if the TI is using about 8.1 df. (or the harmonic mean of [10,7] ) which gives fractional df. Since one SD is twice the other, the Levene F for non homogeneous variances is about 4. So the "separate variance" interval estimate would be more appropriate and would result in small differences. I didn't have your raw data so I wrote the following syntax new file. * this program substitutes values of t for df form 6 to 15. input program. loop df = 6 to 8.4 by .1. compute t = idf.t(.975, df). end case. end loop. end file. end input program. compute lcl = (349 - 383) - t * sqrt( ((19.62**2)/10) + ((39.52**2)/7)). compute ucl = (349 - 383) + t * sqrt( ((19.62**2)/10) + ((39.52**2)/7)). formats df (f4.1) t (f5.3) lcl ucl (f7.2). list variables = df t lcl ucl. execute. which resulted in DF T LCL UCL 6.0 2.447 -73.58 5.58 Yours 6.1 2.437 -73.42 5.42 6.2 2.428 -73.27 5.27 6.3 2.419 -73.13 5.13 6.4 2.410 -72.99 4.99 6.5 2.402 -72.85 4.85 6.6 2.394 -72.72 4.72 6.7 2.386 -72.60 4.60 6.8 2.379 -72.48 4.48 6.9 2.372 -72.36 4.36 7.0 2.365 -72.25 4.25 7.1 2.358 -72.14 4.14 7.2 2.351 -72.03 4.03 7.3 2.345 -71.93 3.93 7.4 2.339 -71.83 3.83 7.5 2.333 -71.74 3.74 7.6 2.327 -71.64 3.64 7.7 2.322 -71.55 3.55 7.8 2.316 -71.47 3.47 7.9 2.311 -71.38 3.38 8.0 2.306 -71.30 3.30 8.1 2.301 -71.22 3.22 TI's 8.2 2.296 -71.14 3.14 8.3 2.292 -71.07 3.07 8.4 2.287 -70.99 2.99 Hope this helps. Art [EMAIL PROTECTED] Social Research Consultants University Park, MD USA 301-864-5570 Quadratic wrote: > I am teaching introductory statistics even though statistics is not > really my specialty. I encourage my students to use a TI-83 > calculator(even though most of them cannot afford one). However there > is one statistical procedure that always comes up with a different > result when I use the TESTS menu on the TI-83 versus using a > t-distribution table and the formula in the textbook, namely a > non-pooled t-confidence interval for the difference between two means. > (Every other statistical procedure, e.g. a pooled or non-pooled > t-test, or a pooled confidence interval, comes out the same when I do > it either way.) > > For example: x1bar = 349, x2bar = 383 (sample means), s1 = 19.6, s2= > 39.5 (sample standard deviations), n1 = 10, n2 = 7 (sample sizes), > find a 95% confidence interval for mu1 - mu2 (difference in population > means). When I use the formula in the textbook with critical t = > 2.447 (df = 6, using one less than the smaller sample size) from the > table, I get 349 - 383 +/- 2.447*sqrt[19.6^2/10 + 39.5^2/7] which > comes out to the interval (-73.56, 5.56). However, when I put the > same information into the TI-83 (option "0" on the TESTS menu) and > choose "no" for pooled, it gives me the interval (-71.21, 3.21). If I > do a t-test with the same data, it comes out the same either way. > What am I doing wrong? > > Thank you. > > Howard Wachtel > Delaware Valley College > [EMAIL PROTECTED] . . ================================================================= Instructions for joining and leaving this list, remarks about the problem of INAPPROPRIATE MESSAGES, and archives are available at: . http://jse.stat.ncsu.edu/ . =================================================================
