Question

1. A student was asked to compute the mean and standard deviation for the following sample of n= 5 scores: 81, 87, 89, 86, and 87. To simplify the arithmetic, the student first subtracted 80 points from each score to obtain a new sample consisting of 1,7,9,6, and 7. The mean and standard deviation for the new sample are then calculated to be M=6 and s = 3. What are the values of the mean and standard deviation for the original sample? 2. The range is completely determined by the two extreme scores in a distribution, The standard deviation, on the other hand, uses every score. a. Compute the range and the standard deviation for the following sample of n= 5 scores. Note that there are three scores clustered around the mean in the center of the distribution and two extreme values. Scores: 0, 6, 7, 8, 14 b. Now we break up the cluster in the center of the distribution by moving two of the central scores out to the extremes. Once again compute the range and standard deviation. New scores: 0, 0, 7, 14, 14 c. According to the range, how do the two distributions compare in variability? How do they compare according to the standard deviation? 3. For the following population of N = 6 scores: 3, 1, 4, 3, 3, 4 a. Compute SS b. Compute the variance (6) c. Compute the standard deviation (0)

Answer 1

3) 901 given population squares 3,1,4,3,3,4 and N=6 a SS Sum of squares (SS) = EX-(8x) EXE 3+4+1 + 3+3 +4 2x 18 EX= (35+4)*60*+63)970359714) = 9+16+1+9+9+16 Ex = GO SS - EX- (Exjr CO- (18) S = 60 324 = 60-54 6 SS= C b. variance co^= Mr Ecxi-on = EX 18 6

? cxi-8)= (3-351761-3)*704-35+ (3-35+ (3-3)+(9.35 Of(-25+ (1st cost cost cu 4+1+1 Elxi-85= 6 variancelos Å Ecxi-ar txo wariance con=1 c. stanclard deviation ca) ñ ecxin8jY 2x6 a. SS G la.