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brittneynguyen
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A random sample of the SATM scores of 100 students, the mean was 504.5 and standard deviation 100. A random sample of the SATM scores of 30 students who too the test for the seond time this year was also obtained. Mean of these 30 scores was 539.1 while the standard deviation was 90. A 95% confidence interval for Mu2Mu 1 is approximately?
 one year ago
 one year ago
brittneynguyen Group Title
A random sample of the SATM scores of 100 students, the mean was 504.5 and standard deviation 100. A random sample of the SATM scores of 30 students who too the test for the seond time this year was also obtained. Mean of these 30 scores was 539.1 while the standard deviation was 90. A 95% confidence interval for Mu2Mu 1 is approximately?
 one year ago
 one year ago

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dctheo Group TitleBest ResponseYou've already chosen the best response.0
first i cheated and went to this link to have the computer calculate the result for me. http://www.graphpad.com/quickcalcs/ttest1.cfm?Format=SD the result was t = 1.6991 with 128 degrees of freedom and a standard error of the mean of 20.364 then i went to this link to see what the formula would be to give me that result. http://lap.umd.edu/psyc200/handouts/psyc200_0812.pdf the answer was on page 3 of 14 where the formula for pooled variance and standard error of the mean were located. once i got those i was able to duplicate the reults. i calculated a pooled variance of 9569.53125 which led to a standard error of 20.3636. that was close enough for me to determine i had the right formulas. once that was done, i used the standard error of the mean to calculate the tscore. that formula was 34.6 / 20.3636 which let to a tvalue of 1.6991 which matched what the calculator told me it should be. to find the confidence interval at 95%, then i needed to find what the limits were at 95% confidence level with 128 degrees of freedom. i used my calculator to determine the limits were at plus or minus 1.978670775 * 20.364 which became plus or minus 40.29365165. if i rounded this to 40.293 i was able to duplicate the computer's answer which said that the interval was from 5.693 to 74.893. it's a lot of work, the most complicated part being the calculation of the pooled variance and then the standard error from that. for that you need the formulas in the second reference i gave you. cheating was so much easier. check out both references so you can see how it was done.
 one year ago
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