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Unformatted text preview: STAT 350 – Spring 2009 Homework #11 Solution
covers through Lecture Q 1. Probability Plots. For this problem, use the dataset given below. It may help to refer to the
lecture notes for Lab #7.
60.8 61.9 28.8 20.4 76.5 43.4 42.4 3.4 69.7 20.8 76.9 21.9 a. For each observed value, calculate the sample percentile this observation represents.
The observations first need to be ordered from smallest to largest. The smallest of the 12
observations would be the 100(10.5)/12 = 4.1667th percentile of the sample.
i 1 2 3 4 5 6 7 8 9 10 11 12 Sample 3.4 20.4 20.8 21.9 28.8 42.4 43.4 60.8 61.9 69.7 76.5 76.9 Percentile 4.1667 12.5 20.833 29.167 37.5 45.833 54.167 62.5 70.833 79.167 87.5 95.833 b. For each percentile from part (a), find the equivalent percentile from the standard normal
distribution (to the nearest 2 decimal places).
Percentile 4.1667 12.5 20.833 29.167
z ‐1.73 ‐1.15 ‐0.81 37.5 45.833 54.167 ‐0.55 ‐0.32 ‐0.10 62.5 70.833 79.167 87.5 95.833 +0.10 +0.32 +0.55 +0.81 +1.15 +1.73 c. Using Excel, make a Normal Probability plot for this sample. Do NOT use SAS – I want to
make sure you understand how these plots are made!
These are the 12 points you should have plotted:
y 3.4 x ‐1.73 20.4 20.8 21.9 28.8 42.4 43.4 60.8 61.9 69.7 76.5 76.9 ‐1.15 ‐0.81 ‐0.55 ‐0.32 ‐0.10 +0.10 +0.32 +0.55 +0.81 +1.15 +1.73 Homework #11 – Solution Page 1 of 1 2. Probability Plots. Refer to Problem 9.24(a) in your textbook (page 415). Using SAS, give a
single QQplot to test the assumption of normality. It may help to refer to the Lecture for Lab#7
(p. 911) for this problem. Use SAS to make the QQplot. You do not need to show your SAS
code. Is the assumption of normality met?
The answer is given below – the explanation follows on the next page.
8 6 4 d
e
v
i
a
t
i
o
n 2 0 2 4
2 1 0
Nm
or al 1 2 Q
uant i l es The assumption of normality appears to NOT be met! Homework #11 – Solution Page 2 of 2 From the following table, we want a QQplot of the column “deviation”, where each value is the
observed value – the mean from that group (e.g., 17.7 – 18.925= 1.225)
group
0 deg
0 deg
0 deg
0 deg
0 deg
0 deg
0 deg
0 deg
45 deg
45 deg
45 deg
45 deg
45 deg
45 deg
45 deg
45 deg
90 deg
90 deg
90 deg
90 deg
90 deg
90 deg
90 deg
90 deg stress
17.7
17.4
17.1
17.3
16.8
22.4
22.3
20.4
22.0
18.7
20.5
19.5
17.4
22.0
19.4
18.3
19.3
20.8
27.5
19.6
19.3
22.3
22.9
19.6 group mean
18.925
18.925
18.925
18.925
18.925
18.925
18.925
18.925
19.725
19.725
19.725
19.725
19.725
19.725
19.725
19.725
21.4125
21.4125
21.4125
21.4125
21.4125
21.4125
21.4125
21.4125 deviation
1.225
1.525
1.825
1.625
2.125
3.475
3.375
1.475
2.275
1.025
0.775
0.225
2.325
2.275
0.325
1.425
2.1125
0.6125
6.0875
1.8125
2.1125
0.8875
1.4875
1.8125 We want a QQplot of the deviations. The SAS code is given below:
data prob924;
input deviation @@;
cards;
1.225
1.525
1.825
1.625
2.125
3.475 3.375 1.475
2.275 1.025
0.775 0.225
2.325
2.275 0.325
1.425
2.1125
0.6125
6.0875
1.8125
2.1125
0.8875
1.4875
1.8125
;
proc univariate data=prob924 noprint;
var deviation;
qqplot deviation/normal (mu=est sigma=est);
run; Homework #11 – Solution Page 3 of 3 3. X is a discrete random variable with the following PMF:
2 1 0
1
2
x
pX(x) 0.1 0.2 0.4 0.2 0.1
2
a. Let Y = X . Make a table giving the Joint PMF of X and Y.
X
1
2
Y 2 1 0
00
0 0.4 0
0 0.4
1 0 0.2 0 0.2 0 0.4
4 0.1 0
0
0 0.1 0.2
0.1 0.2 0.4 0.2 0.1 1.0
b. Find the Correlation of X and Y.
E(X) = 2(0.1) – 1(0.2) + 0(0.4) + 1(0.2) + 2(0.1) = 0
E(Y) = 0(0.4) + 1(0.4) + 4(0.2) = 0.4 + 0.8 = 1.2
Cov(X,Y) = 0 + 0 + (0 – 0)(0 – 1.2)(0.4) + 0 + 0
+ 0 + (1 – 0)(1 – 1.2)(0.2) + 0 + (1 – 0)(1 – 1.2)(0.2) + 0
+ (2 – 0)(4 – 1.2)(0.1) + 0 + 0 +0 + (2 – 0)(4  1.2)(0.1)
=0
Because Cov(X,Y) = 0, Corr(X,Y) also is 0
c. Are X and Y independent random variables? Be sure to support/justify your answer. (Hint:
what would the joint PMF look like if they were independent?)
No, X and Y are NOT independent random variables. If they were, the joint PMF would be the
product of the marginal PMFs and would look like the following table:
X
1
0
1
2
Y 2
0 0.04 0.08 0.16 0.08 0.04 0.4
1 0.04 0.08 0.16 0.08 0.04 0.4
4 0.02 0.04 0.08 0.04 0.02 0.2
0.1 0.2 0.4 0.2 0.1 1.0
d. Find P(Y = 1  X = 1) Hint: this is a conditional probability. Recall the definition of P(AB).
P ( X = 1 ∩ Y = 1) 0.2
=1
=
=
P ( X = 1)
0.2 e. Find P(X = 1  Y = 1)
P ( X = 1 ∩ Y = 1) 0.2
=
=
= 0.5
P (Y = 1)
0.4 Homework #11 – Solution Page 4 of 4 4. X and Y are discrete random variables with the following joint PMF.
X
Y 1 2 3 4 5 6 0 0.001 0.014 0.015 0.028 0.029 0.036 1 0.002 0.013 0.016 0.027 0.030 0.037 2 0.003 0.012 0.017 0.026 0.031 0.038 3 0.004 0.011 0.018 0.025 0.032 0.039 4 0.005 0.010 0.019 0.024 0.033 0.040 5 0.006 0.009 0.020 0.023 0.034 0.041 6 0.007 0.008 0.021 0.022 0.035 0.139 a. Find the marginal PMF of X
X
Y 1 2 3 4 5 6 0 0.001 0.014 0.015 0.028 0.029 0.036 1 0.002 0.013 0.016 0.027 0.030 0.037 2 0.003 0.012 0.017 0.026 0.031 0.038 3 0.004 0.011 0.018 0.025 0.032 0.039 4 0.005 0.010 0.019 0.024 0.033 0.040 5 0.006 0.009 0.020 0.023 0.034 0.041 6 0.007 0.008 0.021 0.022 0.035 0.139 TOTAL 0.028 0.077 0.126 0.175 0.224 0.370 x 1 2 3 4 5 6 p(x) 0.028 0.077 0.126 0.175 0.224 0.370 Answer: Homework #11 – Solution Page 5 of 5 b. Find P(X < Y)
In the following table, I have highlighted all the cases (cells) in which the value for X is less
than the value for Y. The probability is found by summing the probabilities in the shaded cells.
P(X < Y) = 0.003+0.004+0.005+0.006+0.007
+0.011+0.010+0.009+0.008
+0.019+0.020+0.021
+0.023+0.022
+0.035
= 0.203
X
Y 1 2 3 4 5 6 0 0.001 0.014 0.015 0.028 0.029 0.036 1 0.002 0.013 0.016 0.027 0.030 0.037 2 0.003 0.012 0.017 0.026 0.031 0.038 3 0.004 0.011 0.018 0.025 0.032 0.039 4 0.005 0.010 0.019 0.024 0.033 0.040 5 0.006 0.009 0.020 0.023 0.034 0.041 6 0.007 0.008 0.021 0.022 0.035 0.139 c. Find P(X + Y = 4)
In the following table, I have highlighted all the cases (cells) in which the value of X plus the
value of Y is equal to 4. The probability is found by summing the probabilities in the shaded
cells.
Find P(X + Y = 4) = 0.004+0.012+0.016+0.028 = 0.060
X
Y 2 3 4 5 6 0 0.001 0.014 0.015 0.028 0.029 0.036 1 0.002 0.013 0.016 0.027 0.030 0.037 2 0.003 0.012 0.017 0.026 0.031 0.038 3 0.004 0.011 0.018 0.025 0.032 0.039 4 0.005 0.010 0.019 0.024 0.033 0.040 5 0.006 0.009 0.020 0.023 0.034 0.041 6 Homework #11 – Solution 1 0.007 0.008 0.021 0.022 0.035 0.139 Page 6 of 6 d. Find P(Y = 4  X < 4)
P (Y = 4 ∩ X < 4 ) 0.005 + 0.010 + 0.019 0.034
P (Y = 4  X < 4 ) =
=
=
= 0.1471861472
0.028 + 0.077 + 0.126 0.231
P ( X < 4)
X
Y 1 2 3 4 5 6 0 0.001 0.014 0.015 0.028 0.029 0.036 1 0.002 0.013 0.016 0.027 0.030 0.037 2 0.003 0.012 0.017 0.026 0.031 0.038 3 0.004 0.011 0.018 0.025 0.032 0.039 4 0.005 0.010 0.019 0.024 0.033 0.040 5 0.006 0.009 0.020 0.023 0.034 0.041 6 0.007 0.008 0.021 0.022 0.035 0.139 TOTAL 0.028 0.077 0.126 0.175 0.224 0.370 Homework #11 – Solution Page 7 of 7 5. The joint PMF of X and Y and the joint PMF of X and W are given below.
X
Y
1
2
3
total
0.1 0
0
0.1
1
0
0.4 0
0.4
2
0
0
0.5 0.5
6
total 0.1 0.4 0.5 1.0
X
W
1
2
0.1 0
10
0
0.4
20
0
0
60
total 0.1 0.4 3
0
0
0.5
0.5 total
0.1
0.4
0.5
1.0 a. Find P(X = 3  Y = 6)
=0.50/0.50 = 1
b. Find P(X = 2  Y = 6)
= 0/0.50 = 0
c. Find Cov(X, Y)
E(X) = 1(0.1) + 2(0.4) + 3(0.5) = 0.1 + 0.8 + 1.5 = 2.4
E(Y) = 1(0.1) + 2(0.4) + 6(0.5) = 0.1 + 0.8 + 3.0 = 3.9
Cov(X,Y) = (1 – 2.4)(1 – 3.9)(.1) + (2 – 2.4)(2 – 3.9)(0.4) + (3 – 2.4)(6 – 3.9)(0.5) = 1.34
d. Find Cov(X, W)
E(W) = 10(0.1) + 20(0.4) + 60(0.5) = 1 + 8 + 30 = 39
Cov(X,Y) = (1 – 2.4)(10 – 39)(.1) + (2 – 2.4)(20 – 39)(0.4) + (3 – 2.4)(60 – 39)(0.5) = 13.4
e. Find Corr(X,Y)
Var(X) = (1 – 2.4)2(0.1) + (2 – 2.4)2(0.4) + (3 – 2.4)2(0.5) = 0.44
Var(Y) = (1 – 3.9)2(0.1) + (2 – 3.9)2(0.4) + (6 – 3.9)2(0.5) = 4.49
Corr(X,Y) = 1.34/sqrt(0.44×4.49) = 0.9533564071
f. Find Corr(X,W)
Var(W) = (10 – 39)2(0.1) + (20 – 39)2(0.4) + (60 – 39)2(0.5) = 449
Corr(X,W) = 13.4/sqrt(0.44×449) = 0.9533564071
g. If you know the value of X, you know the value of Y with absolute certainty (and visa versa),
but [you should have found that] the Correlation between X and Y is not 1 (or 1). Explain
why.
Correlation is a measure of the linear relationship between X and Y. For correlation to be 1 (or
1), Y must be a linear function of X. There are no values m, b, such that Y = mX + b.
Homework #11 – Solution Page 8 of 8 6. Answer the following questions based on the following summary data:
n = 20, Σxi = 823, Σyi = 806, Σxi2 = 43722, Σyi2 = 79702, Σxiyi = 43337.
a. Calculate SSxy
= 43337(823)(806)/20 = 10170.1
b. Calculate SSxx
= 43722 – 8232/20 = 9855.55
c. Calculate SSyy
= 79702 – 8062/20 = 47220.2
d. Find the sample variance and standard deviation of x.
SS
2
sx = xx = 9855.55/19 = 518.7131579
n −1
sx = 22.77527514 e. Find the sample variance and standard deviation of y.
SS
2
s y = yy = 47220.2/19 = 2485.273684
n −1
sy = 49.85251934
f. Give a 90% CI for the true mean of y (μy)
s
806
49.85251934
y ± tcrit y
⇒
± 1.729
20
20
n ⇒ (21.0262, 59.5738) g. Give a 90% Prediction Interval for the value of y from a new observation.
1
806
1
y ± tcrit s y 1 +
⇒
± 1.729 ( 49.85251934 ) 1 +
⇒
20
20
n
(48.0236, 128.6236)
h. Find the correlation between x and y (Pearson's).
SS xy
10170.1
r=
=
= 0.471434
SS xx SS yy
9855.55 47220.2 Homework #11 – Solution Page 9 of 9 ...
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This note was uploaded on 02/16/2010 for the course STAT 350 taught by Professor Staff during the Spring '08 term at Purdue UniversityWest Lafayette.
 Spring '08
 Staff
 Probability

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