University of Illinois at Chicago
College of Business Administration

MBA 503: Statistics
Instructor: Prof. Stanley L. Sclove
Textbook: Levine, Berenson & Stephan, SFMUME, updated ed. (1998)

Solutions to Final Exam of 16-October-1998

The exam was "Open book" and multiple choice. Two hours were allowed.

Introduction and Data Collection

1. The classification of students by class (freshman, sophomore, junior, senior) is an example of _______ measurement.

(A) nominal (B) ordinal (C) interval (D) ratio
These categories are ordered.

Presenting Data in Tables and Charts

2. In analyzing categorical data, the following graphical device is not appropriate.

(A) bar chart (B) pie chart (C) Pareto diagram (D) stem and leaf display
Stem-and-leaf is for numerical data.

3. A type of vertical bar chart in which the categories are plotted in the descending rank order of the magnitude of their frequencies is called a

(A) bar chart (B) pie chart (C) Pareto diagram (D) stem and leaf display

Histogram with most frequent category on the left, down to the least frequent category on the right.

Summarizing and Describing Numerical Data

4. For the sample of n = 3 observations, X₁ = -2, X₂ = 0 and X₃ = +2, what is the standard deviation?

(A) 1.63 (B) 2 (D) 8/3 (E) 4

variance = (4 + 0 + 4 )/(3-1) = 8/2 = 4; std.dev. = 2

Sampling Distributions

5. For the population of size N=4 with individuals {A,B,C,D} and values u₁ = 3, u₂ = 2, u₃ = 1, and u₄ = 6, the mean is 3 and the variance is 3.5. What is the variance of the mean of n = 2 observations (sampled without replacement)?

(A) 1 (B) 1.08 (C) 7/6 (D) 7/4

var.of mean = (var/n)(FPC), where FPC = (N-n)/(N-1) = (4-2)/(4-1) = 2/3, so var.of mean = (3.5/2)(2/3) = 3.5/3 = 7/6

Likert Scales

6. What is the standard deviation of the following distribution?

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

(A) .500 (B) 1.000 (C) 1.118 (D) 1.250

mean = .1(1) + .1(2) .1(3) .6(4) .1(5) = 3.5; variance = .1(6.25) + .1(2.25) + .1(0.25) + .6(0.25) + .1(2.25) = 1.250, std.dev. = 1.118

7. What is the largest possible standard deviation for a 5-point Likert scale coded as 1, 2, 3, 4, 5?

(A) 1.0 (B) 1.5 (C) 2.0 (D) 4.0
This value corresponds to a completely polarized distribution with .5 at 1 and .5 at 5.

Decision Analysis

8. In the new toy marketing example, perform sensitivity analysis on the value of p, the probability of a favorable market (current value = .4): For what value is EMV(A₁) = EMV(A₂) ?

(A) 39/81 (B) 11/27 (C) 12/27 (D) 15/27
EMV(A1) = (p)(+45) + (1-p)(-36) = 81p - 36 = -3 = EMV(A2). This gives p = 33/81 = 11/27 .

Normal Tables

9. What is the right-central area corresponding to Z = 0.6745 ?

(A) .1500 (B) .1587 (C) .2500 (D) .3413
In the Table, look at the row 0.60 and the column 0.07 .

Normal Distribution

10. If GMAT scores are normally distributed with a mean of 500 and a standard deviation of 100, what is the 95th percentile?

(A) 567 (B) 628 (C) 664 (D) 696
500 + 1.645(100), or about 664 .

11. (continuation) What is the 90th percentile ?

(A) 567 (B) 628 (C) 664 (D) 696
500 + 1.282(100), or about 628 .

Variance of the Return on a Portfolio

12. If Var(X) = 400, Var(Y) = 9 and Cov(X,Y) = -36, what is the numerical value of SD(.5X+.5Y), where SD denotes standard deviation ?

(A) 4.85 (B) 5.85 (C) 9.18 (D) 11.32
.25(Var(X) + Var(Y) + 2Cov(X,Y)) = .25[400 + 9 + (2)(-36)] = .25(337) = 84.25; SD = 84.25^1/2, or about 9.18 .

13. (continuation) What is the numerical value of SD(.6X + .4Y) ?

(A) 4.85 (B) 5.85 (C) 9.18 (D) 11.32
.36Var(X) + .16Var(Y) + 2(.6)(.4)Cov(X,Y)) = .36(400) + .16(9) + 2(.6)(.4)(-36) = 128.16; SD = 128.16^1/2 = 11.32

Control Chart Limits

14. If the target is 368 gm., the standard deviation of individual observations is 15 gm. and the sample size n is 4, and three-sigma limits are used, what is the UCL?

(A) 371.75 (B) 383 (C) 390.5 (D) 413 gm.
SD(mean) = 15/4^1/2 = 15/2 = 7.5 gm.; 368 + (3)(7.5) = 390.5 gm.

Estimating a Population Proportion

15. If p is about .5 and the sample size is 625, then the "margin of error"of a 95% confidence interval is about

(A) 2% (B) 3% (C) 4% (D) 5%
SD(p_s) = [(.5)(.5)/625]^1/2 = .02; e = 1.96(.02) = .04 or 4%

Stratified Sampling

16. Let four strata be denoted by A,B,C,D. If the sampling costs are equal, the standard deviations are 1, 2, 3, and 4, respectively, and the stratum sizes are 100, 200, 300, and 400, respectively, which stratum should get the largest sample?

(A) Stratum A (B) Stratum B (C) Stratum C (D) Stratum D
The sample size is proportional to the std. dev. and to the stratum size.

Statistical Applications in Quality and Productivity Management

17. Which management style emphasizes the continuous improvement of the quality of a product?

(A) Management by Doing
(B) Management by Directing
(C) Management by Control
(D) Management by Process

18. Variation signaled by individual fluctuations or patterns in the data is called

(A) special or assignable causes
(B) common or chance causes
(C) explained variation
(D) the standard deviation

19. Which of the following is not one of Deming's fourteen points ?

(A) belief in mass inspection
(B) creating constancy of purpose for improvement of product or service
(C) adopting and instituting leadership
(D) driving out fear

Pooled Two-Sample t-Test

20. If the sample sizes are 20 and 30, what is the number of degrees of freedom for t ?

(A) 19 (B) 48 (C) 49 (D) 50
n₁ + n₂ - 2

Analysis of Variance

21. What does "interaction" between Shift and Day-of-Week mean ?

(A) Shift is significant, Day-of-Week is not.
(B) Day-of-Week is significant, Shift is not.
(C) The effect of Shift depends upon Day-of-Week.
(D) There is lack of homoscedasticity.

Curvilinear Regression

22. If the regression of MPG on SPEED (in MPH) is

^MPG = 22.4 + 0.786 SPEED - 0.0138 SPEED²

what is the predicted MPG for a speed of MPH?

(A) 19.88 (B) 20.88 (C) 21.88 (D) 23.885 MPG
22.4 + 0.786(55) - 0.0138(55²) = 23.885

23. (continuation) The optimum SPEED is

(A) 15 (B) 17 (C) 28.48 (D) 32.85 MPH
- b₁/2b₂ = -0.786/2(-0.0138) = 28.48 MPH

24. (continuation) The corresponding optimum MPG is

(A) 15 (B) 17 (C) 32.85 (D) 33.59 MPG

22.4 + 0.786(28.48) - 0.0138(28.48²) = 33.59
Alternatively, the maximum is b₀ - b₁²/4b₂ = 22.4 - 0.786² / 4(-0.0138) = 33.59 MPG

Multiple Regression

25. If the mean value of Y is

2 + 3 X₁ + 4 X₂ + 5 X₁ X₂ ,

what is the mean value of Y when X₂ = 1 ?

(A) 3 (B) 6 + 3 X₁ (C) 3 + 8 X₁ (D) 6 + 8 X₁

Time Series Analysis: Exponential Smoothing

26. Actual sales in 1997 were 500. The predicted sales for 1997, using a smoothing constant of .4, were 400. Using exponential smoothing with a smoothing constant of .4, what is the predicted value of Sales for 1998 ?

(A) 400 (B) 420 (C) 440 (D) 460

.4(500) + .6(400) = 200 + 240 = 440 .

Time Series Analysis: Autoregression

27. Consider the predicting equation ^Y_t+1 = 0.6 Y_t + 0.4 Y_t-1 . Predict Y₁₃ at time t=12, given that Y₁₂= 6 and Y₁₁ = 5.

(A) 2.0 (B) 3.6 (C) 5.6 (D) 11.6

^Y₁₃ = 0.6(6) + 0.4(5) = 5.6

Chi-square Goodness-of-Fit Test

28. Consider the data in this table.

-------------------------------------------
value:       1      2      3     4     5
frequency:  10     10     20     5     5
-------------------------------------------

Does the distribution (.1, .2, .4, .2, .1) fit the data ? That is, what is the value of the chi-square test statistic for testing the goodness-of-fit of this distribution?

(A) 2.22 (B) 3.33 (C) 7.5 (D) 50
(f_O-f_E) = -5 for category 1 with an E of 5 and +5 for category 4, with an E of 10. The difference (f_O-f_E) is 0 for the other 3 cells. The chi-square statistic is 25/5 + 25/10 = 5 + 2.5 = 7.5 .

29. (continuation) What is the number of degrees of freedom ?

(A) 2 (B) 3 (C) 4 (D) 5

no. of categories - 1 = 5-1 = 4

Chi-square Test of Independence

Four surgical procedures currently are used to install pacemakers. If the patient does not need to return for followup surgery, the operation is called a "clear" operation. A heart center wants to compare the proportion of clear operations for the four procedures and collects the following data from their own records.

 
                         PROCEDURE
              |     A     B     C     D  |  Total
   -----------|--------------------------|-------  
   Clear      |    27    41    21     7  |    96
   Return     |    11    15     9    11  |    46
   -----------|--------------------------|--------
   Total      |    38    56    30    18  |   142

They will use this information to test for a difference among the proportions of clear operations using a chi-square test with a level of significance of .05 .

30. The number of degrees of freedom will be

(A) 3 (B) 4 (C) 5 (D) 8

(r-1)(c-1) = 3

31. Among all 142 operations, the proportion that were clear is

(A) .32 (B) .68 (C) .71 (D) .73

96/142 = .676

32. Among the 38 operations done with Procedure A, the proportion that were clear is

(A) .32 (B) .68 (C) .71 (D) .73

27/38 = .7105

33. The expected frequency for the Procedure A/Clear cell is

(A) 5.83 (B) 25.69 (C) 38.00 (D) 96.00

(38)(96)/142 = 25.69

Geodemographics

34. What does PRIZM stand for?

(A) Popular Rating Index by Zip Markets
(B) Potential Rating Index by Zip Markets
(C) Potential Recruiting Index by Zip Markets
(D) Popular Recruiting Index by Zip Markets

35. The number of clusters now used by Claritas Corporation in clustering America is

(A) 40 (B) 48 (C) 50 (D) 62 .

17-Oct-1998

University of Illinois at Chicago College of Business Administration

MBA 503: Statistics Instructor: Prof. Stanley L. Sclove Textbook: Levine, Berenson & Stephan, SFMUME, updated ed. (1998)