FINAL EXAM - STATISTICS 3093 - 1996 DECEMBER
SOLUTION NOTES FOR VERSION ONE

QUESTION ONE    [24/80]
             0.0  0.1  1.4  2.2  2.8  3.0  3.2  4.0  4.4  4.7  6.4
QUARTILES              ===            ===            ===
MEAN (AVERAGE)                                                2.93
STANDARD DEVIATION                                            1.95
95% CONFIDENCE INTERVALS
    FOR MEDIAN    X(2) to X(10)                         0.1 TO 4.7
    FOR MEAN                         2.93 +- (2.228*1.95/sqrt(11))
                                     2.93 +-     1.31

QUESTION TWO     [8/80]
a) The expected number of rejections            0.1*16 =  1.6
b) Probability that nobody rejects it           0.9^16 =  0.1853
c) Probability that (exactly) one person rejects it.      0.3294
   This is a binomial distribution computation.

QUESTION THREE   [8/80]
  This is an open question.  Various approaches are correct.
  These are only sample answers:
  A1   1.3  2.4  3.3  2.1. 2.6  1.7  1.3  2.2
  A2   1.3  2.4  3.3  2.1. 2.6  100  1.3  2.2

  B1  Plot of  X   0   1   2   3   4   5
               Y  10  22  29  43  51  59

  B2  Plot of  X   0   1   2   3   4   5
               Y  10  22  29  99  57  59

QUESTION FOUR   [24/80]
    X               1    2    3    4    5     sum of squares of
residuals
    Y              18   18   19   16   34       = 126
    FITTED LINE    15   18   21   24   27     s = sqrt(126/(5-2))
    RESIDUAL        3    0   -2   -8    7       = 6.48

QUESTION FIVE   [16/80]
     X                 1    2     3  TOTAL
    Probability      0.3  0.5   0.2      1
    60/X            60.0 30.0  20.0
    P*60/X          18.0 15.0   4.0     37     37 = E{60/X}
    D = (60/X)-37   23.0 -7.0 -17.0
    D*D            529.0 49.0 289.0
    P*D*D          158.7 24.5  57.8    241    241 = V{60/X}
Alternate approach:
   The population of 60/X is ....  60 60 60  30 30 30 30 30  20 20
   The mean of this is .............................  37
   The variance (divide by 10, not 9) is ........... 247
   a)  X and Y are independent
   E(60/X  30/Y)  =  55.5
   V(60/X  30/Y)  =  V(60/X) +  V(30/Y)    Independence
                  =  V(60/X) +  V(30/X)    X & Y have same distribution
                  =  V(60/X)+0.25V(60/X)   Rules for constants
                  =  241 + 0.25*241        Substitution

   b) 60/X and 30/X are not independent
   E(60/X  30/X)  =  55.5
   V(60/X  30/Y)  =  V(90/X)               Algebra
                  =  2.25 V(60/X)          Rules for constants
                  =  2.25*241              Substitution