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Problems with solutions -MS Exam: Option I
Probability and Statistics- Fall 2007
[1] Math 310 -MS Exam, Fall 2007
If A is a real symmetric matrix then show that its range (column space) R(A) and null space N (A) have only the null vector in common, i.e.,
R(A) ∩ N (A) = { 0 }.
Solution to Math 310 -MS Exam, Fall 2007-Majumdar
Let y ∈ R(A)
N (A). Then y = Ax, and Ay = 0. i.e AAx = 0 i.e AT^ Ax = 0(sinceA = AT^ ) i.e xT^ AT^ Ax = 0 i.e Ax = 0 i.e y = 0
[2] Math 313 - MS Exam, Fall 2007
Find
xlim→ 0
x − sin x x^3
Solution to Math 313 - MS Exam, Fall 2007-Miescke
We use L’Hospital’s Rule three times:
xlim→ 0
x − sin x x^3
= lim x→ 0
1 − cos x 3 x^2
= lim x→ 0
sin x 6 x
= lim x→ 0
cos x 6
[3] Stat 401 - MS Exam, Fall 2007
If X and Y are independent random variables each uniform on the interval [0, 1] then find
(i). p.d.f. of X + Y.
(ii). E(X + Y )^2
Solution to Math 401 - MS Exam, Fall 2007-Majumdar
(i.)
fX (x) =
1 if 0 < x < 1 , 0 Otherwise
fY (y) =
1 if 0 < y < 1 0 Otherwise
fX,Y (x, y) =
1 if 0 < x, y < 1 0 Otherwise Let Z = X + Y. Let FZ (z) be the cumulative distribution function of Z.
case 1. 0 < z < 1. In this case 0 < x, y < 1. Thus FZ (z) = P (Z ≤ z) =
P (X + Y ≤ z) =
∫ (^) z 0 dx^
∫ (^) (z−x) 0 dy^ =^
z^2
case 2. 1 ≤ z < 2. In This case FZ (z) = P (Z ≤ z) = P (X + Y ≤ z) = 1 − P (X + Y > z) Now P (X + Y > z) for z > 1 is simply the area of the triangle formed by the intersection of the sets A = {(x, y) : x + y > z} and B = {(x, y) :, 0 ≤ x, y ≤ 1 } which is (2−z)
2 2.^ Thus FZ (z) = 1 − (2−z)
2
- Now the density of^ Z^ is given by
fX (x) =
z if 0 ≤ z ≤ 1 , 2 − z if 1 < z ≤ 2 , 0 Otherwise
(ii.)
By symmetry and by the independence of X, Y we have E(X) = E(Y ), E(X^2 ) = E(Y 2 ). Now E(X+Y )^2 = E(X^2 )+E(Y 2 )+2E(XY ) = 2E(X^2 )+2[E(X)]^2 = 2
0 x
(^2) dx + 2
[∫ 1
0 xdx)
] 2
[5] Stat 411 - (Chapter 8,9):-MS Exam, Fall 2007
Let X = (X 1 ,... , Xn)′^ denote a random sample from the distribution N (0, θ) that has the pdf
f (x; θ) =
2 πθ
exp
x^2 2 θ
, −∞ < x < ∞.
The purpose is to test the hypothesis H 0 : θ = 1 against the alternative hypothesis H 1 : θ > 1.
(a) Show that the likelihood ratio L(θ = 1; X)/L(θ = θ 1 ; X) is based upon the statistic Y =
∑n i=1 X i^2 , where^ θ^1 >^ 1 is fixed. (b) If n = 15, find a uniformly most powerful critical region of size α = 0. 05 for the hypothesis test. (Hint: Use the chi-square table attached.)
Solution to Stat 411 - (chapters 8,9):- MS Exam, Fall 2007-Yang
(a) Proof: The likelihood function of θ given x = (x 1 ,... , xn)′^ is
L(θ; x) =
2 πθ
)n/ 2 exp
2 θ
∑^ n
i=
x^2 i
Therefore the likelihood ratio
L(θ = 1; X) L(θ = θ 1 ; X)
= θn/ 1 2 exp
θ 1 − 1 2 θ 1
) (^) ∑n
i=
X i^2
So it is based upon the statistic Y =
∑n i=1 X i^2. (b) For any fixed θ 1 > 1 and for any positive constant k,
L(θ = 1; x) L(θ = θ 1 ; x)
≤ k ⇐⇒
∑^ n
i=
x^2 i ≥
2 θ 1 θ 1 − 1
[n 2
log θ 1 − log k
]
= c
By the Neyman-Pearson Theorem, the best critical region C for testing H 0 : θ = 1 against H 1 : θ = θ 1
takes the form of {(x 1 ,... , xn) :
∑n i=1 x^2 i ≥^ c}, where the constant c is determined by
PH 0 (X ∈ C) = α
If n = 15, Y ∼ χ^2 (15) under the null hypothesis. By the chi-square table attached, c = 24.996 if α = 0.05. Note that the best critical region C does not depend on θ 1. So
C =
(x 1 ,... , xn) :
∑^ n
i=
x^2 i ≥ 24. 996
is a uniformly most powerful critical region of size α = 0. 05 for testing H 0 : θ = 1 against H 1 : θ > 1.
[6] Stat 416 - MS Exam, Fall 2007
In a study of effectiveness of hypnosis the emotions of fear, happiness, depres- sion, and calmness were requested in random order for each of five subjects during hypnosis. The following table gives the resulting measurements of skin potential (adjusted for initial level) in millivolts.
Subject A B C D E Fear 23. 1 54. 6 20. 3 11. 9 30. 5 Happiness 22. 7 47. 1 23. 6 13. 8 29. 7 Depression 22. 5 39. 2 16. 3 13. 7 30. 8 Calmness 22. 6 37. 0 14. 9 13. 3 28. 3
Assuming that the skin potential of a person has different levels in the four types of emotions suppose we want to test whether hypnosis can cause these emotions. (a) Describe the basic model of the Friedman test, give its test statistic S, and compute S. (b) Explain why the Friedman test is here more appropriate than the Kruskal- Wallis test.
(a). Work out first and second order inclusion probabilities of the units in the population, to be denoted as Πj s and Πjks.
(b). Show that Πj Πk > Πjk for all pairs of units (j, k), j 6 = k.
(c). Suggest an unbiased estimate of the population mean Y¯ ,
i. based only on the first unit selected from the population;
ii. based on all the n units selected from the population.
Solution to Stat 431 - MS Exam, Fall 2007-Hedayat
We denote by pj the ”normed” size of j-th population unit, j = 1, 2 ,... , N so that
j pj^ = 1. (a). In standard notations :
(i). Πj = pj + (1 − pj ). (^) ((Nn− −1)1)
(ii). Πjk = pj. (^) ((Nn− −1)1) + pk. (^) ((Nn− −1)1) + (1 − pj − pk). (^) ((Nn− −1)(1)(nN− −2)2)
(b). Easy
(c). y pii serves as an unbiased estimate of the population total. Hence, (^) N pyii serves as an unbiased estimate of Y .¯ We may suggest the Horvitz- Thompson Estimate [HTE] of the population mean Y¯ which is given by (^) N^1
j
yj Πj ”sum” being over all sample units in s.
[8] Stat 461 - MS Exam, Fall 2007
Consider a three dimensional Poisson process of particles in the space with intensity parameter ν. Fix a particle and let D be the distance from this particle to its nearest neighbor. Find
(i). P (D > d).
(ii). E(D).
Solution to Stat 461 - MS Exam, Fall 2007-El-Neweihi
P (D > d) = P (no particles in a sphere with center at chosen particle and radius d)
. =e−^43 πd^3 ν^ , d > 0.
E(D) =
0
e−^
(^43) πνx 3 dx
0
e−^
(^43) πνy y−^
(^23) dy
4 3 πν
[9] Stat 481 Problem 1 - MS Exam, Fall 2007
In studying the thrust force developed by a drill press, it is thought that drilling speed and feed rate of the material are the most important factors. An experiment is conducted with 4 feed rates and two drill speeds. The experiment is conducted by using a completely randomized design with 2 replicates for each level combination. The measurements(thrust force) are presented in the table below:
Drill speed Feed Ratio 0.015 .030 0.045 0. 125 2.70 2.45 2.60 2. 2.78 2.49 2.72 2. 200 2.83 2.85 2.86 2. 2.86 2.80 2.87 2.
(i.) Complete the df and MS column for the following ANOVA table.
based on two replicates, are shown in the following table.
Treatment mean
(1) 22 a 25 b 12 ab 35 c 23 ac 30 bc 15 abc 38
For the contrast corresponding to the main effect of factor A, compute an estimate and the corresponding standard error.
(iii.) It turned out that A and AB were the only significant factorial effects. With this information and the treatment means given in part (ii.) discuss in as much detail as you can at how the three factors A, B, and C affect the response. Also decide which treatment(s) you would recommend if it was desirable to maximize the response. (Hint: An appropriate two- way table might be helpful).
Solution to Stat 481 Problem [9] - MS Exam, Fall 2007
Solution:
Solution to Stat 481 Problem [10] - MS Exam, Fall 2007
Appendix A: Table I – Chi-Square Distribution
The following table presents selected quantiles of chi-square distribution; i.e., the value x such that
P (X ≤ x) =
∫ (^) x
0
Γ(r/2)2r/^2
wr/^2 −^1 e−w/^2 dw ,
for selected degrees of freedom r.
Appendix B: Table II – Normal Distribution
The following table presents the standard normal distribution. The probabilities tabled are
P (X ≤ x) = Φ(x) =
∫ (^) x
−∞
√^1
2 π
e−w
(^2) / 2 dw
Note that only the probabilities for x ≥ 0 are tabled. To obtain the probabilities for x < 0, use the identity Φ(−x) = 1 − Φ(x).
- r 0.010 0.025 0.050 0.100 0.900 0.950 0.975 0.
- 1 0.000 0.001 0.004 0.016 2.706 3.841 5.024 6.
- 2 0.020 0.051 0.103 0.211 4.605 5.991 7.378 9.
- 3 0.115 0.216 0.352 0.584 6.251 7.815 9.348 11.
- 4 0.297 0.484 0.711 1.064 7.779 9.488 11.143 13.
- 5 0.554 0.831 1.145 1.610 9.236 11.070 12.833 15.
- 6 0.872 1.237 1.635 2.204 10.645 12.592 14.449 16.
- 7 1.239 1.690 2.167 2.833 12.017 14.067 16.013 18.
- 8 1.646 2.180 2.733 3.490 13.362 15.507 17.535 20.
- 9 2.088 2.700 3.325 4.168 14.684 16.919 19.023 21.
- 10 2.558 3.247 3.940 4.865 15.987 18.307 20.483 23.
- 11 3.053 3.816 4.575 5.578 17.275 19.675 21.920 24.
- 12 3.571 4.404 5.226 6.304 18.549 21.026 23.337 26.
- 13 4.107 5.009 5.892 7.042 19.812 22.362 24.736 27.
- 14 4.660 5.629 6.571 7.790 21.064 23.685 26.119 29.
- 15 5.229 6.262 7.261 8.547 22.307 24.996 27.488 30.
- 16 5.812 6.908 7.962 9.312 23.542 26.296 28.845 32.
- 17 6.408 7.564 8.672 10.085 24.769 27.587 30.191 33.
- 18 7.015 8.231 9.390 10.865 25.989 28.869 31.526 34.
- 19 7.633 8.907 10.117 11.651 27.204 30.144 32.852 36.
- 20 8.260 9.591 10.851 12.443 28.412 31.410 34.170 37.
- 21 8.897 10.283 11.591 13.240 29.615 32.671 35.479 38.
- 22 9.542 10.982 12.338 14.041 30.813 33.924 36.781 40.
- 23 10.196 11.689 13.091 14.848 32.007 35.172 38.076 41.
- 24 10.856 12.401 13.848 15.659 33.196 36.415 39.364 42.
- 25 11.524 13.120 14.611 16.473 34.382 37.652 40.646 44.
- 26 12.198 13.844 15.379 17.292 35.563 38.885 41.923 45.
- 27 12.879 14.573 16.151 18.114 36.741 40.113 43.195 46.
- 28 13.565 15.308 16.928 18.939 37.916 41.337 44.461 48.
- 29 14.256 16.047 17.708 19.768 39.087 42.557 45.722 49.
- 30 14.953 16.791 18.493 20.599 40.256 43.773 46.979 50.
- x 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.
- 0.0 .5000 .5040 .5080 .5120 .5160 .5199 .5239 .5279 .5319.
- 0.1 .5398 .5438 .5478 .5517 .5557 .5596 .5636 .5675 .5714.
- 0.2 .5793 .5832 .5871 .5910 .5948 .5987 .6026 .6064 .6103.
- 0.3 .6179 .6217 .6255 .6293 .6331 .6368 .6406 .6443 .6480.
- 0.4 .6554 .6591 .6628 .6664 .6700 .6736 .6772 .6808 .6844.
- 0.5 .6915 .6950 .6985 .7019 .7054 .7088 .7123 .7157 .7190.
- 0.6 .7257 .7291 .7324 .7357 .7389 .7422 .7454 .7486 .7517.
- 0.7 .7580 .7611 .7642 .7673 .7704 .7734 .7764 .7794 .7823.
- 0.8 .7881 .7910 .7939 .7967 .7995 .8023 .8051 .8078 .8106.
- 0.9 .8159 .8186 .8212 .8238 .8264 .8289 .8315 .8340 .8365.
- 1.0 .8413 .8438 .8461 .8485 .8508 .8531 .8554 .8577 .8599.
- 1.1 .8643 .8665 .8686 .8708 .8729 .8749 .8770 .8790 .8810.
- 1.2 .8849 .8869 .8888 .8907 .8925 .8944 .8962 .8980 .8997.
- 1.3 .9032 .9049 .9066 .9082 .9099 .9115 .9131 .9147 .9162.
- 1.4 .9192 .9207 .9222 .9236 .9251 .9265 .9279 .9292 .9306.
- 1.5 .9332 .9345 .9357 .9370 .9382 .9394 .9406 .9418 .9429.
- 1.6 .9452 .9463 .9474 .9484 .9495 .9505 .9515 .9525 .9535.
- 1.7 .9554 .9564 .9573 .9582 .9591 .9599 .9608 .9616 .9625.
- 1.8 .9641 .9649 .9656 .9664 .9671 .9678 .9686 .9693 .9699.
- 1.9 .9713 .9719 .9726 .9732 .9738 .9744 .9750 .9756 .9761.
- 2.0 .9772 .9778 .9783 .9788 .9793 .9798 .9803 .9808 .9812.
- 2.1 .9821 .9826 .9830 .9834 .9838 .9842 .9846 .9850 .9854.
- 2.2 .9861 .9864 .9868 .9871 .9875 .9878 .9881 .9884 .9887.
- 2.3 .9893 .9896 .9898 .9901 .9904 .9906 .9909 .9911 .9913.
- 2.4 .9918 .9920 .9922 .9925 .9927 .9929 .9931 .9932 .9934.
- 2.5 .9938 .9940 .9941 .9943 .9945 .9946 .9948 .9949 .9951.
- 2.6 .9953 .9955 .9956 .9957 .9959 .9960 .9961 .9962 .9963.
- 2.7 .9965 .9966 .9967 .9968 .9969 .9970 .9971 .9972 .9973.
- 2.8 .9974 .9975 .9976 .9977 .9977 .9978 .9979 .9979 .9980.
- 2.9 .9981 .9982 .9982 .9983 .9984 .9984 .9985 .9985 .9986.
- 3.0 .9987 .9987 .9987 .9988 .9988 .9989 .9989 .9989 .9990.
- 3.1 .9990 .9991 .9991 .9991 .9992 .9992 .9992 .9992 .9993.
- 3.2 .9993 .9993 .9994 .9994 .9994 .9994 .9994 .9995 .9995.
- 3.3 .9995 .9995 .9995 .9996 .9996 .9996 .9996 .9996 .9996.
- 3.4 .9997 .9997 .9997 .9997 .9997 .9997 .9997 .9997 .9997.
- 3.5 .9998 .9998 .9998 .9998 .9998 .9998 .9998 .9998 .9998.
