University of Illinois at Chicago
Instructor           Prof. Stanley L. Sclove

SOLUTIONS TO EXERCISES IN STATISTICS REVIEW


PART 1.   COMPUTATIONS FOR A SAMPLE

For a sample of n = 3 observations, the sum is 6 and the sum of squares is 50.
1.1. The sum of squared deviations (from the sample mean) is   ? SOLUTION:Use the computational formula for the SSD:
SSD = SSQ - SUM2/n 
    = 50 - 62/3 
    = 50 - 12 = 38

PART 2.   CALCULATIONS FROM A SAMPLE

2.1.     A sample of three observations has a sum of 6 and a sum of squares equal to 50.   One observation is -3. What are the   other two observations?
 
[1]: X1 + X2 + X3 = 6
[2]: X12 + X22 + X32= 50
[3]: X3 = -3
Use [3] in [1] and [2]:
[1']: X1 + X2 = 9; X2 = 9 - X1
[2']: X12+ X22 = 41,
[2'']: X12 + (9-X1)2 = 41
       X12 + 81 - 18 X1 + X12 = 41
      2*X12 - 18*X1 -40 = 0
         (X1 - 4)(X1 - 5) = 0
X1 = 4 or 5
Say X1 = 4; then X2 = 9-X1 = 5.
(Many people obtain the solution by trial-and-error.)


2.2. A sample of 10 observations has a mean of 100. The sum of 9 of the observations is 900. What is the value of the other observation?

SOLUTION:  Sum of all ten = 10 x mean = 10 x 100 = 1000. Sum of nine = 900
Difference = value of other observation = 100
 

2.3.   A sample of   n = 14 has a standard deviation of 3.1.
What is the sum of squared deviations?

SOLUTION:

Sample variance, s2 = 3.1 = 9.61 =  SSD/(n-1) =  SSD/13; SSD =  (9.61)(13) = 124.93



2.4.    Consider the following table. 

TABLE.  Distribution of Number of Magazine Subscriptions in Households 

-----------------------------------------------
Number of subscriptions    0    1    2    3 
Number of households      10   40   30    f
-----------------------------------------------
Find the value of   f such that the mean number of subscriptions per household is 2.0.     f = ?

SOLN:  2.0 = mean = sum/number
           = [10(0)+40(1)+30(2)+3f ] / (10+40+30+f ); f = 60

PART 3.  PROBABILITY: EQUALLY LIKELY CASES

A custodian is asked to rank four brands (A, B, C, D) of common household cleanser according to his preference, number 1 being the cleanser he prefers most, and so on. Suppose the custodian really has no preference among the four brands and hence all orders are equally likely to occur.

3.1.   What is the probability that C is first and D is third in the ranking? C D  There are two ways to fill in the blanks, A then ___ ___ ___ ___ B or B then A. The total number of possibilities 1st 2nd 3rd 4th is 4!, or 24. The prob. is 2/24 = 1/12 = .0833.

3.2.  What is the probability that A is ranked either second or third? These are 2 out of 4 equally probable rankings for A; hence the prob is 2/4, or 1/2.

PART 4. PROBABILITY: COMPOUND EVENTS

4.1.  A state highway department has contracted for the delivery of sand, gravel, and cement at a construction site. Due to other work commitments and labor force problems, contracting firms cannot always deliver items on the agreed delivery date. Based on past evidence, the probabilities that sand, gravel, and cement will be delivered on the promised delivery dates by the contracting firms are .3, .6 and .8, respectively. Assume that the delivery or nondelivery of one material is independent of another. Find the probability that all three materials will be delivered on time.

.3 x .6 x .8 = .144

PART 5. EXPECTED VALUE OF A RANDOM VARIABLE

5.1. Consider the following probability distribution of a random variable x.
-----------------------------
v         -3    5     10
P(x = v)  .2    p   .8-p
-----------------------------


What is the value of  p  so that the expected value of x is 5.0?

5.0 = E(x) = .2(-3) + 5p + 10(.8-p), p = .48
 

PART 6.  DISCRETE DISTRIBUTIONS

6.1.   Often the values 1, 2, 3, 4 and 5 are assigned to categories such as "Strongly Disagree," "Disagree Somewhat," "Neutral," "Agree Somewhat," "Strongly Agree." This question has to do with such a "scaling." Of all random variables taking the values 1, 2, 3, 4, 5, the one with P(x=1) = 1/2 and P(x=5) = 1/2 has maximum standard deviation. What is the value of this standard deviation?

SOLUTION: Variance of a random variable = probability-weighted average of the squared deviations from the mean
    = .5(1-3) + .5(5-3)2 =4;   std.dev. = 2.

PART 7.   BINOMIAL DISTRIBUTIONS

7.1.   Suppose a binomial distribution has a mean of 6 and a variance of 3. Then what are the values of the parameters n and p of the distribution?  mean = np = 6; 3 = variance = npq = 6; q = 1/2, p = 1/2,   n = 12
 

PART 8. SAMPLING FROM A FINITE POPULATION


8.1.   A sample of n = 400 is to be drawn (without replacement) from a population of  N = 2000 with a standard deviation of $4000. What is the standard deviation of the  sample mean?

Var(y) = [pop.var./n][(N-n)/(N-1)] = [40002/400][(2000-400)/(2000-1)] = [40002/400](1600/1999) _ SD(y) = (4000/20)*sqrt(1600/1999) = 200*sqrt(.8004) = 200*.895 = $178.93
 

PART 9. NORMAL DISTRIBUTIONS: PERCENTILES

9.1.   What is the 95th percentile of the standard normal distribution? 95th pctile = value exceeded by only 5% of the distribution: z = 1.645 14. What is the 75th percentile of the standard normal of the standard normal distribution?

z = 0.6745

PART 10. 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 fluid ounces of beverage. Variation around this mean follows the normal distribution with a standard deviation of 0.5 fluid ounces.

10.1.  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 21.2 fluid ounces?

s.e. mean = 0.5/sqrt(25) = 0.1 oz.; z = (21.2-21)/0.1 = +2.00, P(x>2.00) = .0228

PART 11. NORMAL APPROXIMATION TO THE BINOMIAL

Statistics released by the National Highway Traffic Safety Administration and the National Safety Council show that on an average weekend night, 1 out of every 10 drivers on the road is drunk. If 400 drivers are randomly checked next Saturday night, what is the probability that the number of drunk drivers will be

11.1.   More than 49?

mean = np = 400(.1) = 40; variance = npq = 400(.1)(.9) = 36; SD = sqrt(36) = 6. z = (49.5-40)/6 = 9.5/6 = 1.583, P(z > 1.583) = .057

11.2.   Exactly 40? z = (40.5-40)/6 = .5/6 = 1/12 = .0833 P(0 < z < .0833) = .033 P(-.0833 < z < .0833) = 2(.033) = .066

11.3.   Exactly 49 ? z = (49.5-40)/6 = 9.5/6 = 1.583, P(z > 1.583) = .057
z = (48.5-40)/6 = 8.5/6 = 1.4167, P(z > 1.4167) = .078
P(exactly 49) is approx. .078-.057 = .021 .


 

PART 12. INTERVAL ESTIMATION OF A BINOMIAL PROBABILITY

In a random sample of n = 625 persons, three hundred (300) favor Paul Parrot for President.

12.1.   What is the estimate of the standard deviation of the sampling distribution of the sample proportion ("p-hat")?

Sample proportion = 300/625 = .48 Var = pq/n, est'd by .48(.52)/625 = .000399; est of SD is sqrt of this, or .01998, very close to .02.

PART 13.  SAMPLE SIZE DETERMINATION FOR A CONFIDENCE INTERVAL

Suppose that GMAT scores have a known standard deviation of 100.  A sample of scores of UIC MBA students is to be taken to estimate the mean in that population. Determine how large a sample is required to form a 95% confidence interval with a margin of error (half-width) of 25 points.

13.1.    The required sample size is about ?

25 = B = 1.96 SD(y) = (1.96)(100)/sqrt(n); sqrt(n) = (1.96)(100)/25 = 7.84; n = 61.47 (use 61 or 62)


PART 14.  OBSERVED SIGNIFICANCE LEVEL: p-VALUE

14.1.   In a test of the null hypothesis that the mean is 18 versus alternatives that the mean is greater than 18, a sample of n = 100 observations gave a mean of 19.5 and standard deviation of 6.00. What is the p-value (i.e., the achieved level of significance)?

z = (19.5 - 18)/0.6 = +2.5, p-value = .006


PART 15. CONFIDENCE INTERVAL FOR A DIFFERENCE BETWEEN TWO MEANS

Lifetimes of two types of batteries were compared. Summary statistics are given in the table. Note that the means are given in hours and the standard deviations are given in minutes.
 
TABLE. Statistics from samples of battery lifetimes
                     n        mean         std.dev.
--------------------------------------------------
Battery A           64        7 hr          30 min
Battery B          100        6 hr          15 min
----------------------------------------------------
In what follows, do not pool the variances since, judging from the ratio of 2 between the two sample standard deviations, it appears that the population variances differ. Also, use the normal distribution (rather than t) due to the large sample sizes.

15.1.   What is the estimate of the standard deviation of the difference between sample means?

Var(diff of uncorrelated variables) = sum of variances = Var(mean1) + Var(mean2) = 302/64 + 152/100 = 16.3125 SD(mean1-mean2) = sqrt(16.3125) = 4.039 min.

15.2.   What is the 95% confidence interval for the difference between means?  95% C.I. = (diff-B,diff+B), where diff = mean1-mean2 = 7 hr. - 6 hr.= 1 hr. = 60 min., and B = 1.96(4.039 min.) = 7.9 min., so 95% C.I. = (52.1, 67.9) min.


PART 16. CHI-SQUARE GOODNESS OF FIT TEST

16.1.  The numbers of accidents per day were recorded in a plant for 100 days. The data are tabulated below. Compute the value of chi-square for testing the hypothesis that the following data came from the distribution (.4, .3, .2, .1) over the values, 0, 1, 2, 3 or more.
Number of accidents012more than 2
Number of days with this many accidents4525 2010
Hypothesized prob of this no. of accidents.4.3.2.1

The value of the chi-square test statistic is ?
 

O          45        25  20   10
E          40        30  20   10
O-E         5        -5   0    0
(O-E)2   25        25   0    0
(O-E)2/E  0.625 0.833   0    0
Value of chi-sq test statistic = .625 + .833 = 1.46

[The number of degrees of freedom is 3 and the p-value is .6915, which is much greater than conventional cut-off values such as .10 or .05 and as such indicates a satisfactory fit.]


PART 17. CONTINGENCY TABLES

                    OWN A CAR?
                   | Yes No |
                    _________
                ___|_______ |___
         Full time | 16   1 | 17
EMPLOYMENT
         Part time | 68  15 | 83
              None | 50  19 | 69
               ____|________|_____
                   | 134 35 | 169


17.1.     Of all 169 people, what percentage are employed full time and own a car?  16/169 = 9.47%  25. Of those who are employed full time, what percentage own a car?

16/17 = 94%

17.2.    Of those who own a car, what percentage are employed full time?

16/134 = 11.9%

17.3.    What is the number of degrees of freedom associated with the chi-square test statistic for this table?  (r-1)(c-1) = 2  28.  Compute the value of the chi-square test statistic. (Answer: 4.586)
 

MTB >  chisquare c1 c2
Expected counts are printed below observed counts
C1     C2      Total
-----------------------
1        16     1     17
         13.48  3.52
2        68     15     83
         65.81  17.19
3        50     19     69
         54.71  14.29
-----------------------
Total   134     35   169
 
ChiSq = 0.471 + 1.805 + 0.073 + 0.279 + 0.405 + 1.552 = 4.586
df = 2
1 cells with expected counts less than 5.0
17.4.  (continuation)  Find the corresponding p-value.
MTB > cdf 4.586;  SUBC> chisq 2. 4.5860 0.8990
MTB > let k1 = 1-.8990
MTB > print k1  K1 0.101000


PART 18.  REGRESSION

Suppose that for speeds between 5 and 95 mph the miles per gallon (G) and speed (S) are related according to the regression equation
Y = 7.216 - 1.073 X ,
where Y = ln G and X = ln S.

18.1.   If the speed (S) is 65, then the predicted gasoline mileage (G) is ?

pred.val. of ln G is 7.216 - 1.073 ln S = 7.216 - 1.073(4.174) = 2.737;  pred.val. of G is exp(2.737) = 15.44 mpg


Created     1995: Feb 12        latest update      2004: Dec 24