University of Illinois at Chicago / College of Business Administration

MBA 503: Statistics
Fall, 1997 / Sclove
Textbook: Levine, Berenson & Stephan

Solutions to Midterm Examination

This exam was based on Chapters 1 through 7 of the textbook and the related notes, lectures and homework.

Five parts, 20 points per part.

PART 1. {20 pts.} DISTRIBUTIONS FOR A LIKERT SCALE

1.{3 pts.} Write down a distribution on a five-point Likert scale that shows a lot of consistency of responses. [The distribution was given on the board.]

SOLUTION:
TABLE.  Distribution
     ------------------------------
value:  1     2     3     4     5

prob:  .1    .1    .6    .1     .1
     ------------------------------

2.{3 pts.} Compute the mean of this distribution.

By symmetry, the mean is 3.0.
Alternatively, compute .1(1)+.1(2)+.6(3)+.1(4)+.1(5) = 3.0

3.{3 pts.} Compute the variance of this distribution.

var = .1(1-3)² + .1(2-3)² + .6(3-3)² + .1(4-3)² + .1(5-3)²
= .4 + .1 + 0 + .1 + .4
= 1.0

4.{1 pt.} Take the square root of the variance to obtain the standard deviation.

SD = sqrt(1.0), or 1 .

5.{3 pts.} Write down a distribution on a five-point Likert scale that shows a lack of consistency of responses. [The distribution was given.]

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

prob:  .35   .1    .1     .1   .35
     ------------------------------

6.{3 pts.} Compute the mean of this distribution.

mean = .35(1) + .1(2) + .1(3) + .1(4) + .35(5) = 3.0

7.{3 pts.} Compute the variance of this distribution.

var = .35(1-3)² + .1(2-3)² + .1(3-3)² + .1(4-3)² + .35(5-3)²
= .35(4) + .1(1) + .1(0) + .1(1) + .35(4) = 3.0

8.{1 pt.} Take the square root of the variance to obtain the standard deviation.

SD = sqrt(3), or about 1.73.

PART 2. {20 pts.} RANDOM SAMPLING FOR MEASURED CHARACTERISTICS; THE NORMAL DISTRIBUTION

In quality control, samples are selected from a production line and various quality characteristics are measured in order to check that the process is "in control." Suppose that a bottling process is intended to fill bottles with, on average, 21 fl.oz. of beverage. Variation around this mean follows the normal distribution with a standard deviation of 0.5 fl.oz.

9.{5 pts.} What is the standard error of the mean if n = 25 ?

std.err. = 0.5/sqrt(n). If n = 25, this is 0.1 fl.oz.

10. {15 pts.} If a technician samples 25 bottles (when the process is "in control") and measures the amount of beverage in each, what is the probability that the sample average (for the 25 bottles) will exceed 21.05 fluid ounces?

z = (21.05-21.0)/0.1 = 0.5; P(0 < Z < 0.5) = .1915; P(Z > 0.5) = .3085

PART 3. {20 pts.} STRATIFIED SAMPLING CASE, CONTINUED

Suppose the N's and SD's are as before: N1 = 1550, N2 = 620, N3 = 930 households, SD1 = 5, SD2 = 15, SD3 = 10 hours, but the costs are equal. The optimal sample allocation is then called "Neyman sampling," after the famous statistician Jerzy Neyman, one of the founding fathers of modern statistics.

11. {15 pts.} Compute the optimal sample sizes if the total n is 100.

Stratum    N   SD    N*SD       w     n
------- ---- ----    ----   -----    --

    1   1550    5    7750    .294    29
    2    620   15    9300    .353    35
    3    930   10    9300    .353    35
                    -----   -----    --
                    26350            99

12. { 5 pts.} In this optimal allocation, n2 = n3. Why?

N2*SD2 = N3*SD3; i.e., N3 is 50% larger than N2, but SD2 is 50% larger than SD3.

PART 4. {20 pts.} INTERVAL ESTIMATION OF A PROPORTION

13. { 5 pts.} Briefly, what is the "margin-of-error"?

It is the figure to be added to and subtracted from the point estimate to obtain the upper and lower confidence limits.

14. {15 pts.} In political polls regarding elections with two candidates the percentages are fairly close to 50%. In a typical such poll the sample size is taken to be 1600. Why in such a poll is the margin of error often stated to be about three percent?

e = Z*(est'd std.err. of sample proportion) or about 2*sqrt{.5(.5)/1600} = 2(.5)/40 = .025 for a 95% CI or about 2.576*sqrt{.5(.5)/1600} = 2.576(.5)/40 = .0322 for a 99% CI

Note that p(1-p) is at most 1/4, this upper limit being attained when p = 1/2. Using 1/2 gives a close approximation unless p is very small or very large.

PART 5. {20 pts.} ACCEPTANCE SAMPLING

Let p be the unknown true proportion of defectives in the lot. Consider testing the null hypothesis p = .03 against the alternative p = .07, based on a sample of n = 200 from a lot of N = 2000. Construct the .05 level test, as follows. Let X be the number of defectives in the sample.

15. { 8 pts.} Under the null hypothesis, what is the standard deviation of X ? HINT: Don't forget the finite population correction.

SD(X) = sqrt{np(1-p)}*fpc
= sqrt{200(.03)(.97)}*sqrt{(2000-200)/(2000-1)}
= 2.41(0.949)
= 2.29

16. { 2 pts.} Under the null hypothesis, what is the mean of X ? HINT: This is just np, where n = 200 and p = .03.

mean = np = 200(.03) = 6.0

17. {10 pts.} One accepts the null hypothesis if X is less than or equal to c. This number c is called the "acceptance number." What is the value of c here? HINT: X has a distribution that is well approximated by a normal distribution with the mean and standard deviation you computed above.

c = 6.0 + 1.645*2.29 = 6.0 + 3.8 = 9.8

The acceptance number could be taken to be 9 or 10. The Type I and Type II error rates of the two could be computed and a choice could be made.

Note that the normal approximation is appropriate because np = 6, which is greater than 5 (and also n(1-p) is greater than 5).

18-Sept-1997
mtsolns.503f97