University of Illinois at Chicago / College of Business Administration
MBA 503: Statistics / Fall '97 / Sclove

Solutions to Final Exam for Fall Semester, 1997

The exam was "Open book."   Ten parts, 20 points each; 200 points total 
PART 1.  MEAN AND VARIANCE OF A FINITE POPULATION       

PART 2.  POPULATION AND SAMPLES 
                           
PART 3.  DECISION ANALYSIS                               

PART 4.  LIKERT SCALES                                    

PART 5.  NORMAL DISTRIBUTION AND TRUTH IN PACKAGING       

PART 6.  ANALYSIS OF VARIANCE AND COVARIANCE           

PART 7.  REGRESSION                                        

PART 8.  PORTFOLIO ANALYSIS                                

PART 9.  TIME SERIES ANALYSIS                            

PART 10. CONTINGENCY TABLES

PART 1. MEAN AND VARIANCE OF A FINITE POPULATION

Consider a population of size N=4 with individuals {A,B,C,D} and values u1 = 3, u2 = 2, u3 = 1, and u4 = 6.

1.(10 pts.) Compute the population mean.

pop.mean, µ = (3+2+1+6)/4 = 3

2.(10 pts.) Compute the population variance.

pop.var. = {(3-3)2 + (2-3)2 + (1-3)2 + (6-3)2}/4
    = (0 + 1 + 4 + 9)/4 = 14/4 = 3.5

PART 2. POPULATION AND SAMPLES

3.(5 pts.) (continuation) Write down the "sampling distribution of the mean" for samples (without replacement) of size n = 2. 

Sample     Mean  prob.
------     ----  -----
3,2        2.5   1/6
3,1        2.0   1/6
3,6        4.5   1/6
2,1        1.5   1/6
2,6        4.0   1/6
1,6        3.5   1/6

4.(5 pts.) Compute the mean of this sampling distribution.

µxbar = (2.5+2.0+4.5+1.5+4.0+3.5)/6 = 18/6 = 3

[Note that µxbar = µ.]

5.(5 pts.) Compute the variance of this sampling distribution.

((2.5-3)2 + (2.0-3)2 + (4.5-3)2 + (1.5-3)2 + (4.0-3)2 + (3.5-3)2)/6
    = (0.25 + 1 + + 2.25 + 2.25 + 1 + 0.25)/6 = 7/6

6.(5 pts.) Compute the variance of the mean as a function of the population standard deviation, N and n. Verify that you get the same answer as in the preceding question.

(var/n)*FPC =   (var/n)*[(N-n)/(N-1)]   = (3.5/2)[(4-2)/(4-1)] = 3.5/3 = 7/6

PART 3. DECISION ANALYSIS

In the Getz Products case, perform sensitivity analysis on the return   rsu for a small plant in unfavorable market (current value = 580 K$):

7.(9 pts.) Write down EMV(S), the EMV of the net cash flow for a small plant, as a function of this variable.

EMV(S) = .5(100) + .5(r-600) = 50 + .5r - 300 = .5r - 250

8. (9 pts.) What value of this variable makes EMV(S) equal to EMV(L)?

EMV(L) = EMV(S) is the same as +10 = .5r - 250, or .5r = 260, or r = 520 K$.

9. (2 pts.) How much lower is this than the current value?

60 K$ less than its current value of 580 K$ .

PART 4. LIKERT SCALES

10.(6 pts.) Write down a distribution on a 5-point Likert scale that shows a polarized population distribution (like responses on a question on abortion).

TABLE.  Distribution
     ------------------------------
value:  1     2     3     4     5
SAMPLE ANSWER:
prob:  .35   .05   .1    .1    .4
     ------------------------------

11.(7 pts.) Write down a distribution on a 5-point Likert scale that shows low consistency of responses (but is not polarized).

TABLE.  Distribution
     ------------------------------
value:  1     2     3     4     5
SAMPLE ANSWER:
prob:  .2    .2    .2    .2    .2
     ------------------------------

12.(7 pts.) Write down a distribution on a 5-point Likert scale that shows high consistency of responses (but not probability 1 on a single value).

TABLE.  Distribution
     ------------------------------
value:  1     2     3     4     5
SAMPLE ANSWER:
prob:  .1    .1   .6     .1    .1
     ------------------------------

PART 5. NORMAL DISTRIBUTION AND TRUTH IN PACKAGING

Fill levels are normally distributed with a standard deviation of 15 gm. and a mean equal to the target fill level.

13.(5 pts.) According to the lecturer, what does the weight stated on a package mean?

According to FDA rules, that at most 8% of the packages are permitted to weigh less than that.

14.(15 pts.) If the fill weights are normally distributed with a mean equal to the target fill weight and a standard deviation of 15 gm., what should be the target fill weight if the packages are to be labelled "347 gm."?

µ = x - Z*SD = 347 - (-1.405)(15) = 368 gm.

PART 6. ANALYSIS OF VARIANCE AND COVARIANCE

15. (4 pts.) In the example considered in class, what does "interaction" between Shift and Day-of-Week mean ?

That the effect of Shift differs from day to day.

--------------------------------------------------------------------

Suppose the response variable is mean weight of tomatoes. For our purposes, big, heavy tomatoes are considered good. Label one of the next three examples as synergism (positive interaction), another as inhibition (negative interaction), and a third as no interaction.

16.(4 pts.)
  NITROGEN        POTASH
                 low  high
            ---------------    Circle the correct one:
         low       4     6     synergism, inhibition, no interaction
        high       7     9                            ______________
            ---------------
17.(4 pts.)
  NITROGEN        POTASH
                 low  high
            ---------------    Circle the correct one:
         low       4     6     synergism, inhibition, no interaction
        high       7    10     _________
            ---------------
18.(4 pts.)
 NITROGEN        POTASH
                low  high     Circle the correct one:
           ---------------
        low       4     6      synergism, inhibition, no interaction
       high       7     8                 __________
           ---------------

SOLUTION: 8 is less than 4 + (6-4) + (7-4) = 9

____________________________________________________________________

19.(4 pts.) What did the analysis of Job Satisfaction as a function of Salary and Job Category show?

That, even after Salary is taken into account, Job Category still appears to affect Job Satisfaction.

PART 7. REGRESSION

20.(15 pts.) In the regression of MPG on MPH in the Sample Final Exam Questions, what is the predicted MPG for 60 MPH?

MPG = 22.4 + 0.786 MPH - 0.0138 MPHsq
    = 22.4 + 0.786(60) - 0.0138(602)
    = 19.88 MPG

21.(5 pts.) Which speed would be expected to yield higher MPG, 60 MPH or 65 MPH ?

60 MPH would. (The optimum is about 28 mph and the curve is parabolic. Both 60 and 65 are above the optimum, so 60 mph would yield higher MPG than 65 mph.)

PART 8. PORTFOLIO ANALYSIS

In the example from the Sample Final Exam Questions,

22.(10 pts.) Compute Var(.3X+.7Y).

var = (.09)(100) + (.49)(9) + 2(.3)(.7)(-18) = 9 + 4.41 - 7.56 = 5.85.
Note. The std.dev. is sqrt(5.85) = 2.42 .

23. (3 pts.) Compute E(.3X+.7Y) .

.3(10)+.7(6) = 7.2 % return

24.(7 pts.) Compute P(.3X + .7Y > 0).

z = (x-mean)/s.d. = (0-7.2)/2.42 = -2.98; prob of values greater than this is about .9986 .

PART 9. TIME SERIES ANALYSIS

Consider the model

pred.val. of Y(t)   =   0.6 Y(t-1) + 0.4 Y(t-2) .

25.(8 pts.) Predict Y(13) at time t=12, given that Y(12)=12 and Y(11) =11.

pred.val. of Y(13) = 0.6(12) + 0.4(11) = 11.6

26.(12 pts.) Also at time t=12, predict Y(14), given that Y(12)=12 Y(11) =11.

pred.val. of Y(14) = 0.6 Y(13) + 0.4 Y(12)

Using the pred.val. of Y(13), one can estimate this as

0.6(11.6) + 0.4(12) = 11.76

PART 10. CONTINGENCY TABLES

Consider the Smoking and Job Category example from the Sample Final Exam Questions.

TABLE. Severity of smoking behavior, by employment category

                           SMOKING
                 |   None light medium heavy |
   --------------|---------------------------|-----
   Sr. managers  |      4     2      3     2 | 11
   Jr. managers  |      4     3      7     4 | 18
   Sr. employees |     25    10     12     4 | 51
   Jr. employees |     18    24     33    13 | 88
   Secretaries   |     10     6      7     2 | 25
   --------------|---------------------------|-----
           All   |     61    45     62    25 |193
   --------------|---------------------------|-----
   SOURCE:  Greenacre (1984)

For the lower right cell:

27.(5 pts.) What is the Expected frequency, E?

E = (25)(25)/193 = 3.24

28.(5 pts.) What is the value of (O-E) ?

(O-E) = 2 - 3.24 = - 1.24

29.(5 pts.) What is the value of z ?

z = (O-E)/sqrt(E) = -1.24/sqrt(3.24) = -1.24/1.8 = -0.69

30.(5 pts.) What is the one-tailed p-value corresponding to this value of z?

p-value = P(Z less than -0.69) = .245 .

latest revision 17-Oct-97