STT 825 Fall 01 HOMEWORK #5 SOLUTIONS Due Wednesday, November 21

1. (11.5 points) Given below are claims for expenses for 8 sales representatives.

Expenses listed by rep (Population listing)

Label rep1 rep2 rep3 rep4 rep5 rep6 rep7 rep8

1 30.0 37.4 34.1 22.2 17.6 15.8 29.1 39.6

2 19.0 29.8 47.2 13.7 9.3 22.4 35.7 34.1

3 32.2 28.9 35.1 20.2 5.9 11.6 32.9 26.7

4 29.8 16.2 22.8 12.5 8.4 21.1 35.1 39.9

5 43.7 25.5 23.1 13.6 15.0 21.8 39.5 30.9

6 19.2 24.1 38.5 20.8 13.2 24.6 28.3 37.7

7 21.8 23.3 31.4 25.0 9.2 14.0 28.5 30.4

8 28.8 39.7 40.1 29.5 23.5 6.3 37.5

9 15.9 34.2 14.8 11.7 11.8 10.0 31.5

10 28.1 31.6 31.5 17.8 20.0 27.9 25.6

11 23.8 15.2 45.5 16.8 12.8 23.5

12 20.4 40.6 35.5 22.7 22.3 17.9

13 21.7 24.5 28.7 32.8 19.6 19.3

14 13.7 22.7 24.0 13.0 16.1 16.2

15 24.2 33.6 9.6 14.8 24.2

16 33.2 33.0 18.5 11.6 17.3

17 18.4 30.1 13.9 17.1

18 15.8 31.1 8.8 16.9

19 19.2 12.8 20.5

20 28.5 14.0 22.0

21 13.0 8.9

22 25.9 18.4

23 34.6

24 21.2

25 26.8

We selected an SRS of 3 representatives, and got rep #1, rep #5, and rep#8 in the sample. We will then subsample (SRS) the claims, with m1 = 6, m5 = 10, and m8 = 4, giving 20 claims in the sample.

Sample Labels

Rep#1 3, 5, 10, 12, 13, 14

Rep#5 1, 6, 9, 10, 11, 14, 16, 18, 21, 22

Rep#8 1, 2, 4, 7

(Hint: you will have to compute all the summary statistics for the observations in your samples.)

Data Display

Row samp1 samp5 samp8

1 32.2 17.6 39.6

2 43.7 13.2 34.1

3 28.1 11.8 39.9

4 20.4 20.0 30.4

5 21.7 12.8

6 13.7 16.1

7 11.6

8 16.9

9 8.9

Variable N Mean Median TrMean StDev SE Mean

samp1 6 26.63 24.90 26.63 10.53 4.30

samp5 10 14.73 14.65 14.80 3.57 1.13

samp8 4 36.00 36.85 36.00 4.59 2.29

Variable N Mean Median TrMean StDev SE Mean

Mi 3 15.67 18.00 15.67 7.77 4.48

ti 3 351.8 324.1 351.8 116.2 67.1

a. Compute the unbiased estimator for the total amount of the claims for all sales representatives.

(8/3) (3x 351.82) = 2814.56

b. Compute the standard error of your estimate in (a).

First estimate the variance. The first term = (8)2(1-3/8) (116.18)2 / 3 = 179970.57

The second term = 8/3{ (1-5/18)182 (10.534)2/5 + (1-9/22) 222 (3.57)2/9 + (1-4/7) 72 (4.588)2/4 }

= 8/3 { 6461.82} = 11838.08. First + Second = 191808.10.

The SE = 438.

c. Compute the ratio estimator for the total amount of the claims for all sales representatives.

= K (351.82/15.67) = 132(22.452) = $2973.64

d. Compute the standard error of your estimate in (c).

Only need to compute the first term again (then add to the second term in (b).

For the first term, compute ei = yi - (22.452) Mi for i in the sample, then get their sample standard deviation.

Variable N Mean Median TrMean StDev SE Mean

ei 3 0.1 89.1 0.1 147.52 90.2

(there’s some rounding error here).

First term = 82 (1-3/8) (147.52)2/3 = 290162.01

Estimated variance = 290162.01 + 11828.08 = 302000.18, SE = 549.6

e. Give proportional allocation for 20 observations to the 3 selected psu’s (sales reps).

n1 = (18/47) x 20 = 7.66, use 8

n2 = (22/47) x 20 = 9.36, use 9

n3 = (7/47) x 20 = 2.98, use 3

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2. ( 9 points) Consider the population below with 6 psu's. We will take an unequal probability sample of size 1.

psu ti Mi yi ti/yi

1 10 5 .1 __100______

2 15 5 .1 __150______

3 16 7 .1 __160_____

4 28 16 .2 ___140_____

5 33 17 .2 ___165_____

6 60 30 .3 ___200_____

a. Find t for the population.

t= 162

b. What is the probability that psu #6 is selected for the sample?

p6 = y6 = .3 (since n=1)

c. Fill in the ti/yi column above.

d. Compute E() directly from the table above. For full credit, show your work.

E() = 100(.1) + 150 (.1) + 160 (.1) + 140 (.2) + 165 (.2) + 200 (.3) = 162

e. Compute V() directly from the table above. For full credit, show your work.

= (100 - 162)2 (.1) + (150 - 162)2 (.1) + (160 - 162)2 (.1) + (140 -162)2 (.2) + (165 - 162)2 (.2)

+ (200 - 162)2 (.3) = 931

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3. (13.5 points) Refer to the population in #2 and the same selection probabilities, but this time select a with replacement unequal probability sample of n=3 psu's.

a. What is the probability that psu #6 is selected for the sample?

p6 = 1-(1-y6)3 = 1- (1-.3)3 = .657

b. Compute V().

= (1/3) (931) = 310.33 (the 931 is from your answer in 2e)

c. If we took an SRS of size 3, find V(). Then compare your answer with (b).

You have to compute S2 for the population of 6 t-values: S2 = 336.

Then V() = 62 (1-3/6) 336/3 = 2016

d. Set up the following random numbers for selecting the sample:

psu random numbers

1 1

2 2

3 3

4 4,5

5 6,7

6 8,9,0

i. Suppose we used the sequence of random numbers 8 2 0 6 8 1 0 3 3 9. Starting on the left and not skipping any numbers, which psu's would be in your with replacement sample (be sure to list any repeats).

Random Number 8 2 0

psu Number 6 2 6

ii. Answer (i) with this sequence of random numbers: 4 6 1 3 9 8 0 3 5 4 5 9.

Random Number 4 6 1

psu Number 4 5 1

e. Our sample of size 3 gave psu's 3, 5, 5.

i. Compute .

Sample psu ti yi ti/yi

3 16 .1 160

5 33 .2 165

= (160 + 165 + 165)/3 = 163.33

ii. Compute the standard error of .

Estimated variance = 1/3{ { (160 - 163.33)2 + 2 (165 - 163.33)2 ]/2 }

= 2.778

SE = 1.667

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4. (16 points) Considering the same population as in #1.

a. Give the PPS probabilities.

Mi 5 5 7 16 17 30 Sum = 80

yi 5/80 5/80 7/80 16/80 17/80 30/80

(or .0625 .0625 .0875 .200 .2125 .375)

i. Give the cum-size method for selecting a PPS sample (fill in the column below).

ii. Give the Lahiri method for selecting a PPS sample (fill in the column below).

psu random numbers (Cum-Sum) random numbers (Lahiri)

(Alternate solution)

1 ____1-5______( 1-625)______101-105______

2 ____6-10______( 625-1250)______201-205______

3 ____11-17_____(1251-2125)______301-307______

4 ____18-33____(2126-4125)______401-418______

5 ____34-50_____(4126-6250)______501-511______

6 ____51-80_____(6251-0000)______601-630______

c. i. Use the sequence of random numbers below and the cum-sum random numbers in b(i), and give the psu's (with repeats) in your with replacement PPS sample of size 3.

915675259527958301345140248038652988099730356497909994912720044299310655120529

91, 56, 75, 25 (alternate solution) 9156, 7525, 9527

NA, psu 6 psu 6 psu 4 psu 6, psu 6, psu 6

ii. For the sample in (i) compute and its standard error.

= (200+ 200 + 140)/3 = 180, estimated variance = 1/3{1200) = 400, SE = 20

(alternate solution) = 160, SE = 0

d. Use the sequence of random numbers below and the Lahiri method to select a with replacement PPS sample of size 3. List the psu's (with repeats) in your sample.

624735846873943429923415282104840279904849207772338845390379312832317420796579

624, 735, 846, 873, 943, 429, 923, 415, 282, 104

psu 6 na na na na na na psu 4 na psu 1 Ans: psu’s 1,4,6

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