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
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
rep1
30.0
19.0
32.2
29.8
43.7
19.2
21.8
28.8
15.9
28.1
23.8
20.4
21.7
13.7
24.2
33.2
18.4
15.8
rep2
37.4
29.8
28.9
16.2
25.5
24.1
23.3
39.7
34.2
31.6
15.2
40.6
24.5
22.7
33.6
33.0
30.1
31.1
19.2
28.5
13.0
25.9
34.6
21.2
26.8
rep3
34.1
47.2
35.1
22.8
23.1
38.5
31.4
40.1
14.8
31.5
45.5
35.5
28.7
24.0
rep4
22.2
13.7
20.2
12.5
13.6
20.8
25.0
29.5
11.7
17.8
16.8
22.7
32.8
13.0
9.6
18.5
13.9
8.8
12.8
14.0
rep5
17.6
9.3
5.9
8.4
15.0
13.2
9.2
23.5
11.8
20.0
12.8
22.3
19.6
16.1
14.8
11.6
17.1
16.9
20.5
22.0
8.9
18.4
rep6
15.8
22.4
11.6
21.1
21.8
24.6
14.0
6.3
10.0
27.9
23.5
17.9
19.3
16.2
24.2
17.3
rep7
29.1
35.7
32.9
35.1
39.5
28.3
28.5
37.5
31.5
25.6
rep8
39.6
34.1
26.7
39.9
30.9
37.7
30.4
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
1
2
3
4
5
6
7
8
9
samp1
32.2
43.7
28.1
20.4
21.7
13.7
samp5
17.6
13.2
11.8
20.0
12.8
16.1
11.6
16.9
8.9
samp8
39.6
34.1
39.9
30.4
Variable
samp1
samp5
samp8
N
6
10
4
Mean
26.63
14.73
36.00
Median
24.90
14.65
36.85
TrMean
26.63
14.80
36.00
StDev
10.53
3.57
4.59
SE Mean
4.30
1.13
2.29
Variable
Mi
ti
N
3
3
Mean
15.67
351.8
Median
18.00
324.1
TrMean
15.67
351.8
StDev
7.77
116.2
SE Mean
4.48
67.1
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
-----------------------------------------------------------------------------------------------------------------2. ( 9 points) Consider the population below with 6 psu's. We will take an unequal probability sample of size 1.
psu
ti
Mi
i
ti/i
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?
6 = 6 = .3 (since n=1)
c. Fill in the ti/i column above.
d. Compute E( tˆ ) directly from the table above. For full credit, show your work.
2
E( tˆ ) = 100(.1) + 150 (.1) + 160 (.1) + 140 (.2) + 165 (.2) + 200 (.3) = 162
e. Compute V( tˆ ) 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
--------------------------------------------------------------------------------------------------------------------------------------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?
6 = 1-(1-6)3 = 1- (1-.3)3 = .657
b. Compute V( tˆ ).
= (1/3) (931) = 310.33 (the 931 is from your answer in 2e)
c. If we took an SRS of size 3, find V( tˆ ). Then compare your answer with (b).
You have to compute S2 for the population of 6 t-values: S2 = 336.
Then V( tˆ ) = 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 tˆ .
Sample psu
ti
i
3
16
.1
5
33
.2
5
33
.2
tˆ = (160 + 165 + 165)/3 = 163.33
ti/i
160
165
165
ii. Compute the standard error of tˆ .
3
Estimated variance = 1/3{ { (160 - 163.33)2 + 2 (165 - 163.33)2 ]/2 }
= 2.778
SE = 1.667
------------------------------------------------------------------------------------------------------------------------------------4. (16 points) Considering the same population as in #1.
a. Give the PPS probabilities.
Mi
5
5
7
16
17
30
Sum = 80
i
5/80 5/80 7/80 16/80 17/80 30/80
(or
.0625 .0625 .0875 .200
.2125 .375)
b.
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
c.
1
random numbers (Cum-Sum)
(Alternate solution)
____1-5_______( 1-625)______
random numbers (Lahiri)
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___________
___101-105___________
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,
NA,
56,
psu 6
75,
psu 6
25
psu 4
(alternate solution)
9156, 7525, 9527
psu 6, psu 6, psu 6
ii. For the sample in (i) compute tˆ and its standard error.
tˆ = (200+ 200 + 140)/3 = 180, estimated variance = 1/3{1200) = 400, SE = 20
(alternate solution) tˆ = 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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