SOLUTIONS TO IDS 460 SPRING '95 FINAL EXAMINATION
This examination was cumulative, comprehensive and multiple choice. The questions vary in style and level of difficulty.
The distribution of ages for N = 8700 graduate and professional students at Minnesota State University is as follows.
------------------------------------------------------------------- Age 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 % of students 1 10 5 5 5 5 5 5 6 7 8 9 10 9 8 1 1 Cumulative % 1 11 16 21 26 31 36 41 47 54 62 71 81 90 98 99 100 -------------------------------------------------------------------1. How many of the students are exactly 22 years of age?
(A) 10 (B) 11 (C) 87 (D) 870 (E) 957 students
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10% of 8700
2. What percentage of the population is over 30 years of age?
(A) 38 (B) 46 (C) 53 (D) 54 (E) 62 %
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3. What is the mean age in this population?
(A) 28.95 (B) 29.15 (C) 29.95 (D) 32.95 (E) 33.95 years of age
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Mean = sum of products = 21(.01) + 22(.10) + ... + 37(.01)
= 0.21 + 2.20 + ... + 0.37 = 29.15
4. What is the median time to completion?
(A) 40 (B) 45 (C) 50 (D) 55 (E) 60 minutes
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5. What is the lower quartile?
(A) 40 (B) 45 (C) 50 (D) 55 (E) 60 minutes
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6. What is the upper quartile?
(A) 50 (B) 55 (C) 60 (D) 65 (E) 80 minutes
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(A) 1/16 (B) 1/4 (C) 1/2 (D) 3/4 (E) 15/16
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ANS.: 4 x 3/(4x4) = 3/4
(A) p (B) q (C) pq (D) 1/2 (E) 1/4 -----9. What is the variance of y?
(A) p (B) q (C) pq (D) 1/2 (E) 1/4
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10. What is the maximum value of the variance of y?
(A) p (B) .7071 (C) pq (D) 1/2 (E) 1/4
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11. What is the maximum value of the standard deviation of y?
(A) p (B) .7071 (C) pq (D) 1/2 (E) 1/4
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12. What is the expected value of (1-y) ?
(A) p (B) q (C) pq (D) p-q (E) q-p
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13. What is the expected value of (y-p) ?
(A) 0 (B) 1 (C) p (D) q (E) q-p -----
(A) .25 (B) 1.0 (C) 8.67 (D) 10.0 (E) 100.0
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SOLUTION: Var(mean) = (var/n)[(N-n)/(N-1)]
= 100**2/100 (400-100)/(400-1), or about 75;
s.e.mean = square root of this, or about 8.67 .
(A) 1.96(100/49) (B) 1.96(100/7) (C) 1.96(100/49)(490-49)/(490-1) (D) 1.96(100/49)(.950) (E) 1.96(100/7)(.950) ---------------------16. About what sample size n would be required (instead of 49) so that the margin of error A would be 20?
(A) 8 (B) 50 (C) 83 (D) 101 (E) 111
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(A) 1.96(.3)(.7)/1600 (C) 1.96(.3)(.7)/40 (B) 1.96(.458)/1600 (D) 1.96(.458/1600)(10000-1600)/(10000-1) (E) 1.96(.458/40)(.9166) ------------------------18. About what sample size n is required (instead of 1600) to make the margin of error A = .01 ? [Use p = .5 in the formula to get a conservative estimate of the required sample size.]
(A) 50 (B) 400 (C) 500 (D) 600 (E) 5000
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From the formula in the textbook,
n = Npq/[(N-1)B**2/4 + pq] = 10000(.5)(.5)/[9999*.01**2/4 + (.5)(.5)]
or about 5000.
(A) $53.20 (B) $234.80 (C) $388.60 (D) $623.40 (E) $676.60
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From one of the Problems:
ROWS: MSTATUS COLUMNS: GENDER
0 1 ALL
0 540.7 929.3 825.7
1 487.5 1164.1 1119.7
ALL 527.4 1078.0 994.9
CELL CONTENTS --
JOBINC:MEAN
$540.70-487.50 = $53.20
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Stratum, i 1 2
Stratum sizes 4000 6000
Standard deviation 3 4
Cost per observation $16 $9
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A sample of size n = 100 can be afforded. What is the optimal
allocation to the two strata?
(A) 50, 50 (B) 40, 60 (C) 33, 67 (D) 27, 73 (E) 25, 75
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n1:n2::N1(sigma1)/sqrt(c1):N2(sigma2)/sqrt(c2)
N1(sigma1)/sqrt(c1) = 4000(3)/4 = 3000
N2(sigma2)/sqrt(c2) = 6000(4)/3 = 8000
n1:n2::3000:8000, i.e., n1=100 x 3/11 = 27, n2 = 73.
21. (continuation) If the mean of stratum 1 is 64 and the
mean of stratum 2 is 68, what is the population mean?
(A) 66.0 (B) 66.1 (C) 66.2 (D) 66.3 (E) 66.4
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.4(64) + .6(68) =66.4
22. (continuation) What is the variance of the population?
(A) 9.0 (B) 13.2 (C) 15.0 (D) 16.0 (E) 17.04
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N x pop.var. = BGSS + WGSS
= 4000[(64-66.4)**2 + 3**2] + 6000[(68-66.4)**2 + 4**2]
= 4000[ 5.76 + 9 ] + 6000[ 2.56 + 16 ]
= 4000[ 14.76 ] + 6000[ 18.56 ]
= 4000[ 14.76 ] + 6000[ 18.56 ] = 170400,
so pop.var. = 17.04 .
PART 10. PPS SAMPLING
23. Use Table 4.6 to find the probability that the estimate
of the total is equal to 10.
[Note: This is Table 8.3, p. 316, in the new 5th edition.]
(A) .01 (B) .08 (C) .16 (D) .24 (E) .25
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Prob(estimate = 10) = .08+.01+.16
24. What is the probability that the estimate of the total is
equal to the actual value of the total?
(A) .01 (B) .08 (C) .16 (D) .24 (E) .25
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The true value is 10.
PART 11. PROBLEM 7: ZIPCODE DATA
25. How many of the 1000 zipcodes have a SportsPI at least
two standard deviations above the mean?
(A) 16 (B) 18 (C) 20 (D) 22 (E) 30
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From solution to Problem 7:
MTB > # 5. List the high SportsPI zipcodes: All zipcodes having a
MTB > # SportsPI at least two standard deviations above the mean.
MTB > #
MTB > ERASE C21-C40
MTB > LET K21 = MEAN('SportsPI')
MTB > LET K22 = STDEV('SportsPI')
MTB > LET K23 = K21 + 2*K22
MTB > LET K24 = MAX('SportsPI')
MTB > COPY C1-C20 to C21-C40;
SUBC> USE rows with 'SportsPI' = K23:K24.
MTB > PRINT C21-C40
ROW C21 C22 C23 C24 C25 C26 C27 C28
1 31 1 14 60090 59852 21660 31.2 41593
2 347 2 30 3033 2052 669 31.7 38570
3 363 2 31 7970 5994 1789 30.1 62358
4 369 2 31 8536 5820 3346 34.4 37414
5 375 2 34 10518 762 247 33.3 56568
6 396 2 34 12570 3278 1082 30.9 34050
7 413 2 34 13732 8295 2580 28.9 37172
8 478 2 38 18968 356 125 33.2 31833
9 554 3 9 33322 24945 11118 56.4 27374
10 555 3 9 33445 28337 14535 64.5 24403
11 557 3 9 33548 5848 3266 67.2 39552
12 637 3 20 21082 1009 338 34.7 50901
13 760 3 43 77077 37253 14860 31.4 49553
14 791 3 45 22066 10346 3305 32.7 74308
15 872 4 5 92056 1771 726 34.4 26880
16 892 4 5 94807 1993 831 31.1 33826
17 912 4 6 80134 14662 4494 31.1 52796
18 0 4 50 83013 336 141 32.9 29167
PART 12. GEO-DEMOGRAPHICS
26. The U.S. population is about how many millions?
(A) 2.6 (B) 26 (C) 27 (D) 260 (E) 2600
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27. The number of zipcodes in the U.S. is about ?
(A) 400 (B) 4,000 (C) 40,000 (D) 400,000 (E) 4,000,000
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PART 13. PROBLEM 8: PROBABILITY; TWO-WAY TABLES
28. In Little Town one-fourth of all dog owners have cats and
one-half of all cat owners have dogs. What is the ratio of the
number of dog owners to cat owners?
(A) 3:2 (B) 2:1 (C) 5:2 (D) 3:1 (E) 7:2
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DOG?
| Yes No |
----|-----------------|----
Yes | a b | R1
CAT? | |
No | c d | R2
----|-----------------|----
| C1 C2 |
a = (C1)/4 = (R1)/2; (C1) = 4(R1)/2 = 2(R1): C1:R1::2:1
PART 14. RATIO ESTIMATION
29. If the sugar content of a sample of oranges weighing 50
pounds is 5 pounds, and the total weight of the truckload of
oranges is 1000 pounds, what is the ratio estimate of the sugar
content of the truckload of oranges?
(A) 10 lbs. (B) 15 lbs. (C) 20 lbs. (D) 25 lbs. (E) none of these
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5:50::x:1000, x = 100 lbs.
PART 15. DIFFERENCE ESTIMATION
Consider the following data, similar to Example 6.10 (in the 4th
edition).
The population contains N = 200 inventory items with a stated total
book value of $12,000. Thus the true (population) mean book value
(mean of x) is $60. A simple random sample of n = 6 items yields the
results shown in the table.
____________________________________________________________________
Item, i Audit Value, y Book Value, x Difference, d
i i i
____________________________________________________________________
1 9 10 -1
2 15 12 +3
3 7 8 -1
4 29 26 +3
5 45 44 +1
6 54 50 +4
____________________________________________________________________
Sums: 159 150 +9
____________________________________________________________________
30. Consider the following statements.
I. The sample variance of the differences is 4.70.
TRUE
II. The sample standard deviation of the differences is $2.17
(to the nearest cent).
TRUE
III. The estimated standard error of the sample mean is 0.846
(to the nearest tenth of a cent).
FALSE: Var of mean difference = (4.70/n)(N-n)/N
= (4.70/6) (200-6)/200 = 0.760; sqrt(0.760) = $0.872.
(A) Only I is true.
(B) Only II is true.
(C) Only III is true.
(D) All three statements are true.
(E) None of the above.
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31. (continuation) Construct "two-sigma" confidence limits (to the
nearest cent) for the true difference between the means of y and x.
The limits are ?
(A) (-0.24,3.24) (B) (-1.69, 1.69) (C) (-0.19, 3.19)
---------------
(D) (-0.19, 0.19) (E) (-3.19, 3.19)
mean diff = 1.5; conf.int. is (mean diff - A, mean diff + A),
where A = 2 x $0.872 = $1.74.
32. (continuation) The difference-method estimate of the true
mean of y is ?
(A) 60.50 (B) 61.00 (C) 61.50 (D) 74.50 (E) None of these
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Estimate = true mean of x + mean difference = 60.0 + 1.5 = 61.5
33. (continuation) The estimated variance of the estimator of
the mean of y is ? (to three decimals)
(A) 0.590 (B) 0.764 (C) 0.780 (D) 0.975 (E) None of these
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Same as variance of mean difference =
0.760, computed above
PART 16. TWO-STAGE CLUSTER SAMPLING
34. In Example 9.5, pp. 350-351 of the 5th edition (was page
299 in the 4th edition), what would be the optimal value of m if
the cost of cutting a battery open were 24 times (instead of six
times) the cost of measuring a plate?
(A) 2.09 (B) 4.18 (C) 8.36 (D) 16.72 (E) 33.44
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PART 17. INVERSE SAMPLING
35. A fair coin is tossed until the fifth Head occurs.
What is the probability that this requires exactly 8 tosses?
(A) 1/32 (B) 1/8 (C) 35/256 (D) 35/128 (E) 56/256
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Pr(4 Heads in the first 7 tosses and Heads on the 8th toss) =
C(7,4)(1/2)**8 = 7x6x5/(3x2x1)(1/2)**8 = 35/128
PART 18. RANDOMIZED RESPONSE MODEL
36. A simple random sample of 200 dog owners was selected from
the population of dog owners in the city. Two questions were written
down:
Question 1. Has your dog been vaccinated against rabies?
Question 2. Is the last digit of your social security number odd?
A sheet of random numbers was used. The following procedure was
followed: (i) A two-digit random number was chosen. If the random
number was 80 or above, Question 1 was answered. (ii) If the random
number was 79 or below, Question 2 was answered. There were 90
responses of "yes" and 110 responses of "no". From this, estimate
the percentage of the dog owners whose dogs had been vaccinated (to
the nearest one-tenth of a percent).
(A) 57.1% (B) 87.4% (C) 87.5% (D) 87.6% (E) None of these
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P(Q1) = .2 P(Yes|Q1) = p, to be estimated
P(Q2) = .8 P(Yes|Q2) = 1/2
Set 90/200 = P(Yes) = .2p + .8(1/2), giving p = (.45-.4)/.2 = .25
or 25%.
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20-Nov-95
latest revision 16:31, 23-Apr-96