Stat401B Lab#2 Solution Key Fall 2014 Problem 1:
Full Data set (Population)
Rating
Quantiles
100.0%
99.5%
97.5%
90.0%
75.0%
50.0%
25.0%
10.0%
2.5%
0.5%
0.0%
maximum
quartile
median
quartile
minimum
7850
6495.4
4063
2467
1445
833
581.5
370
226.25
74.2
43
Summary Statistics
Mean
Std Dev
Std Err Mean
Upper 95% Mean
Lower 95% Mean
N
Stem and Leaf
1185.7386
1003.002
55.297292
1294.5207
1076.9565
329
Problem 2: Answers will vary due to random sampling
Sample size of 12 selected using random digit table
rating
Quantiles
100.0%
99.5%
97.5%
90.0%
75.0%
50.0%
25.0%
10.0%
2.5%
0.5%
0.0%
maximum
5158
5158
5158
4728.4
1940
1434
600.25
410
398
398
398
quartile
median
quartile
minimum
Summary Statistics
Mean
Std Dev
Std Err Mean
Upper 95% Mean
Lower 95% Mean
N
1704.1667
1435.227
414.31434
2616.0664
792.26694
12
Stem and Leaf
Problem 3
Distribution Sample 1 Distribution Sample 2 Distribution Sample 3 Rating Rating Rating Quantiles Quantiles Quantiles 100.0% 99.5% 97.5% 90.0% 75.0% 50.0% 25.0% 10.0% 2.5% 0.5% 0.0% maximum quartile median quartile minimum 2586 2586 2586 2185.8 1139 819 775.25 395.5 352 352 352 100.0% 99.5% 97.5% 90.0% 75.0% 50.0% 25.0% 10.0% 2.5% 0.5% 0.0% maximum quartile median quartile minimum 2530 2530 2530 1902 1315.75 773.5 648.5 462.5 314 314 314 100.0% 99.5% 97.5% 90.0% 75.0% 50.0% 25.0% 10.0% 2.5% 0.5% 0.0% maximum quartile median quartile minimum 4142 4142 4142 2689 2024.5 888 601.5 463.6 391 391 391 Summary Statistics Summary Statistics Summary Statistics Mean Std Dev Std Err Mean Upper 95% Mean Lower 95% Mean N 987.41667 562.78891 162.46316 1344.9957 629.83765 12 Mean Std Dev Std Err Mean Upper 95% Mean Lower 95% Mean N 1015.25 541.33219 110.49897 1243.8345 786.66546 24 Mean Std Dev Std Err Mean Upper 95% Mean Lower 95% Mean N 1359.25 972.14824 162.02471 1688.1776 1030.3224 36 IMPORTANT COMMENT: In the above normal quantile plots, for the purpose of interpretation of the shape of the distribution, they must be viewed from the direction of the left margin Problem 3 (continued)
Statistic
n
Sample 1
12
Sample 2
24
Sample 3
36
𝑦 987.41667
1015.25
1359.25
𝑠 ! 316731.3
293040.5
945072.2
s
R
Q(0.25)
Q(0.5)
Q(0.75)
IQR
562.78891
2234
775.25
819
1139
363.75
541.33219
2216
648.5
773.5
1315.75
667.25
972.14824
3751
601.5
888
2024.5
1423
Problem 4
(a) Compute the sampling error in the estimate 𝑦 of 𝜇 for each of the 3 samples.
Sample 1: | 𝑦 - µ | = |987.41667 - 1185.7386| = 198.3219
Sample 2: | 𝑦 - µ | = |1015.25 - 1185.7386| = 170.4886
Sample 3: |𝑦 - µ | = |1359.25 - 1185.7386| = 173.5114
(b) Compute the sampling error in the estimate s of 𝜎 for each of the 3 samples.
Sample 1: | s - σ | = |562.78891 - 1003.002| = 440.2131
Sample 2: | s - σ | = |541.33219 - 1003.002| = 461.6698
Sample 3: | s - σ | = |972.14824 - 1003.002| = 30.85376
(c) Compute the sampling error in the sample median M for each of the 3 samples.
Sample 1: | Q(0.5) - M | = |819 – 833| = 14
Sample 2: | Q(0.5) - M | = |773.5 – 833| = 59.5
Sample 3: | Q(0.5) - M | = |888 – 833| = 55
Problem 5
Oneway Analysis of Rating By Sample
Means and Std Deviations
Level
1
2
3
Number
Mean
Std Dev
12
24
36
987.42
1015.25
1359.25
562.789
541.332
972.148
Std Err
Mean
162.46
110.50
162.02
Lower 95%
Upper 95%
629.8
786.7
1030.3
1345.0
1243.8
1688.2
Problem 6
P(Y = 731) = 3/329
P(Y < 731) = 134/329
P(Y ≥ 731) = 195/329
P(Y = 581) = 0
P(Y < 581) = 82/329
P(Y ≥ 581) = 247/329
P(806 < Y < 1690) = 102/329
P(806 ≤ Y ≤ 1690) = 103/329
P(Y ≤ 900) = 178/329
P(Y > 1010) = 136/329
P(Y > 1950) = 54/329
P(1050 ≤ Y ≤ 1150) =17/329
Problem 7
(a) P (X = 0) =
!"
!
0.06! 0.94!" = 0.2901
(b) P(X ≤ 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)
= 0.2901 +
!"
!
0.06! 0.94!" +
!"
!
0.06! 0.94!" + !"
!
0.06! 0.94!" +
!"
!
0.06! 0.94!"
= 0.9944
(c) P(X = 5) =
!"
!
0.06! 0.94!" = 0.004766
(d) P(X > 5) = 1 − P(X ≤ 5) = 1 – 0.9943669 – 0.004765603 = 0. 0008685
E[Y] = n𝜋 = 1000*0.06 = 60
Problem 8
(a) Week’s production of switches.
(The intended target population might be all switches produced by the manufacturer in which case the
week’s production is the sampled population.)
(b) Length of each bolt.
(c) The random sample of 40 bolts.
(d) Quantitative.
(e) The requirement is that each bolt has the same probability of being selected. Since the lot may be quite
large (may be 1000’s) it may be not convenient to label the bolts, unless they are already numbered. So
it is easier to select the sample one bolt at a time after mixing the remaining lot well. In fact that the bolt
selected is not replaced will not affect the probability of selection of the subsequent bolts if the
population is “large” enough.
(f) 𝜇, population mean of length.
(g) 𝑥, sample mean.
!
(h) Estimate 𝜎 ! by 𝑠 ! , from which you can estimate s.e.(𝑥) by !.