Consider the following sample data. 200 176 120 168 136 31 The sample variance i

Question

Consider the following sample data. 200 176 120 168 136 31 The sample variance is, 1016 1020 1024 1028 32 The smallest observation in the sample deviates from the mean by _standard deviations 1.20 1.25 1.35 33 Compute the z-scores for each of the five data points above Which of the following is the correct set of z-values? 1.25 0.50 1.25 0.75 1.25 1.25 0.50 1.25 0.25 0.75 1.25 0.50 0.75 1.25 0.50 1.25 0.75 1.25 0.25 0.50 34 The mean of the z-scores in the previous question is, 1.00 0.25 -0.25 0 35 The standard deviation of the z-scores above is, 1.25 1.5

Explanation / Answer

ANSWER: ALL THE OPTIONS GIVEN IN THE CHOICES ARE WRONG AND I HAVE SOLVED EVERY PART AND THESE ARE THE RIGHT ANSWERS.

31) FIRST WE NEED TO FIND THE MEAN OF THE SAMPLE.

MEAN = ( 200 + 176 + 120 + 168 + 136 ) / 5 = 800 / 5 = 160

IN ORDER TO FIND THE VARIANCE , WE WILL NEED TO SUM THE SQUARE OF THE DIFFERENCE BETWEEN THE SAMPLE AND MEAN AND THEN DIVIDE BY THE TOTAL

VARIANCE = ((200 - 160) ^ 2 + (176 - 160) ^ 2 + ( 120 - 160 ) ^ 2 + ( 168 -160) ^2 + ( 136 - 160 ) ^ 2 ) / 5

VARIANCE = ((40) ^ 2 + ( 16) ^ 2 + (-40) ^ 2 + (8) ^ 2 + (-24) ^ 2 ) / 5

VARIANCE = (1600 + 256 + 1600 + 64 + 576) / 5 = 4096 / 5 = 819.2

32) SMALLEST OBSERVATION IS 120 AND THE MEAN IS 160.

DIFFERENCE = ( 120 - 160) = -40

STANDARD DEVIATION = SQUARE OF VARIANCE = SQUARE ROOT OF 819.2 = 28.62

SO THE DEVIATION FROM MEAN = DIFFERENCE / STANDARD DEVIATION = -40 / 28.62 = -1.39

33) Z SCORE FORMULA = (XI - MEAN ) / STD

Z SCORE OF 1ST - 160 ) / 28.62 = 40 / 28.62 = 1.39

Z SCORE OF 2ND 176 - 160) / 28.62 = 16 / 28.62 = 0.55

Z SCORE OF 3RD 120 - 160) / 28.62 = -40 / 28.62 = -1.39

Z SCORE OF 4TH -160) / 28.62 = 8 / 28.62 = 0.28

Z SCORE OF 5TH 136 - 168 ) / 28.62 = -32 / 28.62 = -1.11

1.39 , 0.55, -1.39 , 0.28 , -1.11

34) MEAN OF Z SCORES = ( 1.39 + 0.55 -1.39 + 0.28 -1.11) / 5 = -0.0559

35) STANDARD DEVIATION:

IN ORDER TO FIND THE STANDARD DEVIATION , WE WILL NEED TO SUM THE SQUARE OF THE DIFFERENCE BETWEEN THE SAMPLE AND MEAN AND THEN DIVIDE BY THE TOTAL AND THEN FIND THE UNDER ROOT

VARIANCE =( ( 1.39 - ( -0.559))^ 2) + (0.55 - (-0.559)^2 )+ (-1.39 - ( -0.559)^2) +( 0.28 - (-0.559)^2) + (-1.11- (0.559)^ 2) ) / 5

VARIANCE = 1.10

STANDARD DEVIATION = UNDER ROOT OF VARIANCE = 1.05

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