Hi, I've got this question that I'm unsure about how to approach. I was wondering if anyone could give me a hand or steer me in the right direction.
I've got some sample data.. what it represents isn't really important. The first group is 782, 965, 948, 1181, 1414, 1633, 1852. The second group is 593, 672, 750, 988, 1226, 1462, 1698. The means of each I calculated to be about 1253 and 1055 respectively.
The question is to find a 95% confidence interval for the difference in the means. I'm confused because no standard deviation is given (and usually is in our class), so I'm a bit unclear as how to standardize this to find the CI using the methods I know about. Can anyone help me out?
Also then I am suppose to test this confidence interval to test the hypothesis that the difference in mean is 200 against the alternative that is something else.
2006-12-02 14:29:52 · 3 answers · asked by abe_cooldude 1 in Science & Mathematics ➔ Mathematics
confidence interval (95 %) = mean+- (1.96) * std error of mean
confidence interval (99 %) = mean+- (2.58) * std error of mean
=between --------- and -------
(+-) its plus / minus
2006-12-04 05:04:02 · answer #1 · answered by amit gawande 1 · 0⤊ 0⤋
Um they don't HAVE to always the the standard deviation, often they won't. I assume you have a graphing calculator? Probably a TI-83? Enter the data in L1 and L2 respectively. Then go into Stat, Calc 1-Var Stats for each list. The Sx is what you want. But that's not standard deviation.
You have to use a 2-sample t-mean. (xbar1-xbar2) plus or minus T-star*squareroot(Sx1/n1+Sx2/n2) where xbar is each average, Sx is the value you got earlier, and T-star is 1.984 (calculated from degrees of freedom and sample size)
2006-12-02 22:45:33 · answer #2 · answered by doctorevil64 4 · 0⤊ 0⤋
why, there's a jolly old formula on how to calculate your SD :) something like sqrt (1/(N-1) * sum (x - mean)^2)
as for the difference in means, how about a t-test?
2006-12-02 22:41:25 · answer #3 · answered by Nick C 4 · 0⤊ 0⤋