Statistics Exam Help

Live Exam Helper

Statistics Exam Help For any help regarding Statistics Exam Help visit : - https://www.liveexamhelper.com/, Email : - info@liveexamhelper.com or call us at - +1 678 648 4277

Live Exam Helper

Part I: Concept questions

(58 points)

T h e s e questions are all multiple choice or short answer. You don’t have to show any work. Work through t h e m quickly. E ac h answer is worth 2 points.

Concept 1. Which of t h e following represents a valid probability table ? (i)

outcomes probabili ty

(ii) ou tc om e s probability Circle t h e b e s t choice: A. (i )

1 1/5

2 1/5

3 1/5

4 1/5

5 1/5

3 4 5 1/51/10 1/10 1/10

1 2 1/2 B. (ii)

C. ( i ) and (ii)D. Not e nough information

Concept 2. True or false: Setting t h e prior probability of a h yp ot h e si s to 0 me ans t ha t make t h e posterior probability of t ha t h yp o th e s i s t h e maximum over all h yp oth e s e s .

no amount of data will

Circle one: True False Concept 3. True or false: It is okay to have a prior t ha t d e p e n d s on more than one unknown parameter.

Circle one: True False Concept 4. Data ishy drawn t h e following p oth efrom s e s: a normal H 0 : µdistribution = 1 andwith For unknown mean µ. We make H A : µ > 1. (i)-(iii) circle t h e correct answers: (i) Is H 0 a simple or comp osi te h yp oth e s i s ? (ii) Is H A a simple or com posite h yp o th e s i s? (iii) Is H A a one or two-sided?

One-sided

Simple

Composite Simple Composite Two-

sided

Concept 5. If t h e original data h a s n points the n a b oots tr ap sam pl e shoul d have A.Fewer points than t h e original b e c a u s e th ere is l es s information in t h e sam ple than in t h e underlying distribution.

B.T h e s a m e number of points as t h e original b e c a u s e we want t h e b oo tst r ap statistic to mimic t h e statistic on t h e original data.

C.M a ny more points than t h e original b e c a u s e we have t h e computing power to handle a lot of data.

Circle t h e b e s t answer:

A

B

C.

Concept 6. In 3 t o s s e s of a coin which of following equals t h e event “exactly two h e ad s ”?

A = {THH, H T H , H H T, HHH} B = {THH, H T H , HHT} C = {HTH, THH} Circle t h e b e s t answer:

A

B

C B and C

Live Exam Helper

Concept 7. T h e s e questions all refer to t h e following fi gure. For e ac h one circle t h e b e s t answer.

A1 B1 C1

C2

x

y C1

A2

z

(i) T h e probability x represents

B1

B2 C2

B2

C1

C2

C1

C2

A. P ( A 1 ) B. P (A 1 |B 2 ) C. P (B 2 |A 1 ) D. P (C 1 |B 2 ∩ A 1 ).

(ii) T h e probability y represents

A. P ( B 2 )

B. P (A 1 |B 2 ) C. P (B 2 |A 1 ) D. P (C 1 |B 2 ∩ A 1 ).

(iii) T h e probability z re presents A. P (C 1 ) (iv) T h e circled node represents t h e event ∩ A1 .

B. P (B 2 |C 1 ) C. P (C 1 |B 2 ) D. P (C 1 |B 2 ∩ A 1 ).

A. C1 B. B 2 ∩ C1 C. A 1 ∩ B 2 ∩ C1

D. C1 |B2

Concept 8. T h e g rap h s below give t h e pmf for 3 random variables.

(A)

(B)

1

2

3

4

(C)

x

x

5

1

2

3

4

5

x 1

2

3

4

5

Circle t h e answer th at orders t h e gr ap h s from smallest to bi gge st standard deviation.

ABC

ACB

BAC

BCA

CAB

CBA

Concept 9. S u p p o s e you have $100 and you need $1000 b y tomorrow morning. Your only way to g e t t h e money you nee d is to gamble. If you b e t $k, you either win $k with probability p or lose $k with probability 1 − p. Here are two strategies: Maximal strategy: B e t as much as you can, u p to w hat you need, e ach time.

Minimal strategy: M ake a small be t, say $10, e ach time.

S u p p o s e p = 0.8.

Circle t h e b e tt e r strategy: Maximal Minimal 1 2 3 3 3 A. f (x , y ) = 4x y B. f (x , y ) = (x y +2 xy ).

2.

C. f (x , y ) = 6e−3x−2y

Concept 10. Consider t h e following joint p d f ’ s for t h e random variables X and Y . Circle t h e ones where X and Y are independent and cross out t h e other ones.

Live Exam Helper

Concept 11. S u p p o s e X ∼ Bernoulli(θ) where θ is unknown. Which of t h e following is t h e correct state me nt?

A. T h e random variable is discrete, t h e s p a c e of h y p o t h e s e s is discrete.

B. T h e random variable is discrete, t h e s p a c e of h y p o t h e s e s is continuous.

C. T h e random variable is continuous, t h e s p a c e of h y p o t h e s e s is discrete.

Circle t h e letter of t h e correct statement: A B C D D. T h e random variable is continuous, t h e s p a c e of h y p o t h e s e s is continuous.

Concept 12. Let θ b e t h e probability of h e a d s for a be nt coin. S u p p o s e your prior f ( θ) is Beta( 6, 8). Also s u p p o s e you flip t h e coin 7 times, getting 2 h e ad s and 5 tails. W h at is t h e posterior p d f f (θ|x)? Circle t h e b e s t answer.

A. Beta(2,5) B. Beta(3,6)

C. Beta(6,8) D. Beta(8,13)

E. Not e nough information to say

Concept 13. S u p p o s e t h e prior h a s b e e n set. Let x 1 and x 2 b e two s e t s of data. Circle true or false for e ac h of x 1 and x 2 have h e sa m e likelihood function the n t h e y result in t h e s am e posterior. True False tA. h eIf following statet ments.

B. If x 1 and x 2 result in t h e s am e posterior the n t h e y have t h e s am e likelihood function. True C. If x 1 and x 2 have proportional likelihood functions the n t h e y result in t h e s a m e posterior.

True

False False

Concept 14. E a c h day J a n e arrives X hours late to class, with X ∼ uniform(0, θ). J o n m ode ls his initial belief a bo ut θ b y a prior p d f f (θ). After J a n e arrives x hours late to t h e next class, J o n c om p u te s t h e likelihood function f (x|θ) and t h e posterior p df f (θ|x).

Circle t h e probability computations a frequentist would consider valid. Cross out t h e others.

A. prior

B. posterior C. likelihood

Concept 15. S u p p o s e we run a two-sample t-te st for equal me ans with signifi cance level α = 0.05. If t h e data implies we should reje ct t h e null h ypothe si s, the n t h e o d d s th at t h e two s am p l e s c o m e from distributions with t h e s a m e mean are (circle t h e b e s t answer) A. 19/1

B. 1/19

C. 20/1

D. 1/20

E. unknown

Concept 16. Consider t h e following state me nts a bo ut a 95% confi dence interval for a parameter θ.

A. P (θ 0 is in t h e CI | θ = θ 0 ) ≥ 0.95 B. P (θ0 is in t h e CI ) ≥ 0.95 C.An experiment p rodu c e s t h e CI [−1, 1.5]: P (θ is in [−1, 1.5] | θ = 0) ≥ 0.95 Circle t h e letter of e ach correct statem e nt and c ross out t h e others: A

B

C

Live Exam Helper

Part II: Problems

(325 points)

Problem 1. (20) (a) Let A and B b e two events. S u p p o s e tha t t h e probability th at neither event occ urs is 3/8. W h at is t h e probability th at at least one of t h e events oc c ur s ? (b)Let C and D b e two events. S u p p o s e P ( C ) = 0.5, P ( C∩D) = 0.2 and P ( ( C∪D) c ) = 0.4.

W h at is P ( D ) ?

Problem 2. (20) An urn contains 3 red balls and 2 blue balls. A ball is drawn. If t h e ball is red, it is ke pt out of t h e urn and a se c on d ball is drawn from t h e urn. If t h e ball is blue, the n it is p u t back in t h e urn and a red ball is a dd e d to t h e urn. Th e n a se c on d ball is drawn from t h e urn. (a) W ha t is t h e probability th at b o t h balls drawn are red? (b)If t h e se c ond drawn ball is red, w hat is t h e probability t ha t t h e fi rst drawn ball was blue ?

Problem 3. (15) You roll a fair six sided die re pe ate dl y until t h e s u m of all numbers rolled is greater than 6. Let X b e t h e number of times you roll t h e die. Let F b e t h e cumulative distribution function for X . Comp ute F (1), F (2), and F (7).

Problem 4. (20) A t e s t is graded on t h e scale 0 to 1, with 0.55 n ee de d to p as s .

Student sc ore s are mode le d b y t h e following density:  for 0 ≤ x ≤ 1/2 4x   for 1/2 ≤ x ≤ 1 f (x) = 4 − 4x  otherwise 0

(a) W ha t is t h e probability th at a random student p a s s e s t h e exa m ? (b) W h at score is t h e 87.5 percentile of t h e distribution?

Problem 5. (15) S u p p o s e X is a random variable with cdf  for x < 0   0 F ( x ) = x(2 − x ) for 0 ≤ x ≤ 1  1 for x > 1 (a) Find E ( X ) . (b) Find P ( X < 0.4).

Live Exam Helper

Problem 6. (15) C om pute t h e mean and variance of a random variable w ho se distribution is uniform on t h e interval [a, b].

It is not enough to simply state these values. You must give the details of the computation.

Problem 7. (20) Defaulting on a loan me ans failing to pa y it back on time. T h e default rate among M I T stud en ts on their stude nt loans is 1%. A s a pro j e c t you deve lop a t e s t to predict which stude nts will default. Your t e s t is g o o d b u t not perfect. It gives 4% false positives, i.e. prediciting a student will default who in fact will not. If h a s a 0% false negative rate, i.e. prediciting a student won’t default who in fact will. (a)S u p p o s e a random student t e s t s positive. W h at is t h e probability t ha t h e will truly default. (b)Someone off ers to b e t m e t h e student in part ( a ) won’t default. T h e y want m e to p ay t h e m $100 if t h e stude nt doe sn’t default and they’ll p ay m e $400 if t h e student d o e s default. Is this a g o o d b e t for m e to take?

Live Exam Helper

Problem 8. (30) Data was taken on he ight and weight from t h e entire population of 700 mountain gorillas living in t h e Democratic Republic of Congo: t \ w t light short h170 tall 85 Let X e nc ode t h e weight, taking t h e values of a heavy respectively.

average 70 30heavy 190 155 randomly c hose n gorilla: 0, 1, 2 for light, average, and

Likewise, let Y e n c od e t h e height, taking values 0 and 1 for short and tall respectively. (a) Determine t h e joint pm f of X and Y and t h e marginal p m f ’ s of X and of Y . (b) Are X and Y independe nt? (c) Find t h e covariance of X and Y .

For this part, you need a numerical (no variables) expression, but you can leave it uneval- uated. (d) Find t h e correlation of X and Y .

For thisT h part, you nition need a numerical (no variables) expression, but you can leaveYit)uneval- uated. (d) e defi of correlation is C or( X, Y ) = t h Cov(X, e . So we fi rst ne ed to c o m p u t e σX σY variances of X and Y .

260 1000 10 185 E(X2 ) = + = 4· = 700 700 700 7 T hus, 2 Var(X ) = E ( X ) − E(X )

E(Y 2) =

2

=

10 − 7

43

70

Var(Y ) = E ( Y 2 ) − E ( Y ) 2 =

43 −70

81 433 = 100 700 . Σ 43 70

2

=

43 · 27 702

therefore 113/700 Cor(X, Y ) = √

433/700

√

4 3 · 27/70 2

Note: We would a c c e c p t –even encourage solutions– t ha t left t h e fractions uncomputed, e.g. σY = √

4 3 / 7 0 − (43/70) 2 .

Live Exam Helper Problem 9. (20) A political poll is taken to determine t h e fraction p of t h e population th at referendum requiring all citizens to b e fl uent in t h e language of probability and statistics. (a)A s s u m e p = 0.5. Use t h e central limit the ore m to estimate t h e probability th at in a least 14 p e op l e su p p o r t t h e referendum.

would s u p p or t a

poll of 25 peopl e, at

Your answer to this problem should be a decimal. (b)With p unknown and n t h e number of random p e o pl e polled, let X n b e t h e fraction of t h e polled p e op l e w ho s u p p or t t h e referendum.

W h at is t h e smallest samp le size n in order to have a 90% confi dence th at X n is within 0.01 of t h e true value of p?

Your answer to this problem should be an integer.

Problem 10. (10 p t s ) S u p p o s e a researcher collects x 1 , . . . , x n i.i.d. measure ments of t h e background radiation in Boston. S u p p o s e also t ha t t h e s e observations follow a Rayleigh distribution with paramete r τ , with p df given b y f ( x ) = x τ e − .12τ x 2 Find t h e maximum likelihood estimate for τ .

Problem 11. (15) Bivariate data (4, 10), (−1, 3), (0, 2) is as s u m e d to arise from t h e model y i = b|xi − 3| + e i , whe re b is a constant and e i are independent random variables. (a)W h at assumptions are ne e de d on e i so t ha t it makes s e n s e to do a least squares fit of a curve y = b|x − 3| to t h e data? (b) Given t h e above data, determine t h e least squares estimate for b. are integers.

For this problem we want you to calculate all the way to a fraction b =

r

s

, where r and s

Problem 12. (30) Data is collected on t h e time betwee n arrivals of consecutive taxis at a downtown hotel. We collect a data s e t of size 45 with sam pl e mean x¯ = 5.0 and sam pl e standard deviation s = 4.0. (a) A s s u m e t h e data follows a normal random variable. (i) Find an 80% confi dence interval for t h e mean µ of X . (ii) Find an 80% χ 2 -confi dence interval for t h e variance? (b) Now make no assumptions ab ou t t h e distribution of of t h e data. B y bootstrapping, we generate 500 values for t h e diff erences δ∗ = x¯∗ − x¯. T h e smallest and largest 150 are written in non-decreasing order on t h e next page .

Use this data to find an 80% b oot st ra p confi dence interval for µ. (c)We s u s p e c t th at t h e time betwee n taxis is modeled b y an exponential distribution, not a normal distribution. In this case, are t h e a pp roac he s in t h e earlier parts justifi ed? (d) Whe n might m e t h od ( b ) b e preferable to m e t h od ( a ) ?

Live Exam Helper

T h e 100 smallest and 100 largest values of δ∗ for problem 12.

1- 10 11- 20 21- 30 31- 40 41- 50 51- 60 61- 70 71- 80 81- 90 91-100 101-110 111-120 121-130 131-140 141-150

-0.534 -0.330 -0.271 -0.241 -0.213 -0.200 -0.182 -0.170 -0.159 -0.153 -0.144 -0.133 -0.118 -0.105 -0.097

-0.494 -0.328 -0.269 -0.240 -0.211 -0.200 -0.182 -0.166 -0.159 -0.152 -0.142 -0.131 -0.114 -0.103 -0.096

-0.491 -0.315 -0.262 -0.232 -0.211 -0.195 -0.181 -0.164 -0.158 -0.151 -0.142 -0.129 -0.114 -0.103 -0.095

-0.485 -0.304 -0.262 -0.226 -0.210 -0.193 -0.179 -0.163 -0.157 -0.151 -0.142 -0.128 -0.114 -0.103 -0.095

-0.422 -0.297 -0.260 -0.225 -0.209 -0.192 -0.179 -0.163 -0.156 -0.150 -0.138 -0.124 -0.112 -0.102 -0.095

-0.403 -0.293 -0.257 -0.223 -0.209 -0.192 -0.178 -0.162 -0.156 -0.148 -0.137 -0.124 -0.111 -0.101 -0.093

-0.382 -0.287 -0.256 -0.222 -0.208 -0.189 -0.176 -0.162 -0.155 -0.148 -0.135 -0.124 -0.109 -0.099 -0.093

-0.365 -0.279 -0.255 -0.220 -0.204 -0.188 -0.175 -0.160 -0.155 -0.146 -0.135 -0.123 -0.108 -0.098 -0.093

-0.347 -0.273 -0.249 -0.216 -0.202 -0.188 -0.174 -0.160 -0.154 -0.145 -0.134 -0.123 -0.108 -0.098 -0.093

-0.336 -0.273 -0.248 -0.216 -0.200 -0.183 -0.170 -0.159 -0.154 -0.145 -0.134 -0.119 -0.107 -0.097 -0.093

351-360 361-370 371-380 381-390 391-400 401-410 411-420 421-430 431-440 441-450 451-460 461-470 471-480 481-490 491-500

0.073 0.079 0.087 0.094 0.104 0.110 0.118 0.134 0.143 0.156 0.170 0.182 0.220 0.243 0.274

0.074 0.079 0.087 0.094 0.104 0.111 0.122 0.135 0.145 0.162 0.172 0.186 0.220 0.244 0.288

0.075 0.080 0.088 0.096 0.106 0.112 0.122 0.136 0.146 0.163 0.172 0.195 0.221 0.245 0.288

0.075 0.081 0.091 0.097 0.106 0.112 0.123 0.136 0.147 0.164 0.175 0.202 0.222 0.251 0.291

0.077 0.081 0.091 0.100 0.108 0.112 0.127 0.137 0.147 0.164 0.178 0.202 0.224 0.253 0.307

0.077 0.082 0.091 0.100 0.108 0.112 0.129 0.140 0.148 0.165 0.179 0.205 0.225 0.258 0.312

0.077 0.083 0.092 0.101 0.108 0.113 0.129 0.141 0.151 0.166 0.180 0.206 0.232 0.261 0.314

0.077 0.084 0.092 0.101 0.108 0.114 0.132 0.142 0.151 0.168 0.181 0.210 0.232 0.263 0.316

0.078 0.085 0.093 0.102 0.108 0.114 0.134 0.142 0.154 0.169 0.182 0.216 0.236 0.266 0.348

0.079 0.085 0.093 0.103 0.110 0.115 0.134 0.143 0.155 0.169 0.182 0.219 0.236 0.273 0.488

Problem 13. (15) Note. In this problem t h e geometric( p) distribution is defi ned as t h e total number of trials to t h e fi rst failure ( t h e value includes t h e failure), where p is t h e probabilitiy of s u c c e ss . (a) W h at samp le statistic would you u s e to esti mate p? (b)Describe how you would u se t h e parametric b oo tst ra p to e stimate a 95% confi dence interval for p. You can b e brief, b u t you should give careful s te p- b y- st e p instructions.

Live Exam Helper Problem 14. (30) You independently draw 100 data points from a normal distribution. (a)S u p p o s e you know t h e distribution is N(µ, 4) (4 = σ 2 ) and you want to t e s t t h e null h yp ot h e s i s H 0 : µ = 3 against t h e alternative h yp oth e s i s H A : µ ƒ = 3.

If you want a signifi cance level of α = 0.05. W h at is your rejection region? You m ust clearly s ta te what t e s t statistic you are using. (b)S u p p o s e t h e 100 data points have sam pl e mean 5. W h at is t h e p-value for this data? Should you rejec t H0 ? (c) Determine t h e power of t h e t e s t using t h e alternative H A : µ = 4.

Problem 15. (30) S u p p o s e th at you have molecular t y p e with unknown atomic m a s s θ. You have an atomic scale with normally-distributed error of mean 0 and variance 0.5. (a)S u p p o s e your prior on t h e atomic m a s s is N(80, 4). If t h e scale reads 85, w hat is your posterior p d f for t h e atomic m a s s ? (b)With t h e s a m e prior as in part ( a) , c om p ut e t h e smallest number of measure ments ne e de d so th at t h e posterior variance is l es s than 0.01.

Problem 16. (20) Your friend grab s a die at random from a drawer containing two 6-sided dice, one 8-sided die, and one 12-sided die. S h e rolls t h e die once and re ports th at t h e result is 7. (a)Ma ke a discrete B a ye s table showing t h e prior, likelihood, and posterior for t h e t y p e of die rolled given t h e data. (b) W h at are your posterior o d d s th at t h e die h a s 12 si de s? (c) Given t h e data of t h e fi rst roll, w hat is your probability th at t h e next roll will b e a 7?

Live Exam Helper

Standard normal table of left tail probabilities. z -4.00 -3.95 -3.90 -3.85 -3.80 -3.75 -3.70 -3.65 -3.60 -3.55 -3.50 -3.45 -3.40 -3.35 -3.30 -3.25 -3.20 -3.15 -3.10 -3.05 -3.00 -2.95 -2.90 -2.85 -2.80 -2.75 -2.70 -2.65 -2.60 -2.55 -2.50 -2.45 -2.40 -2.35 -2.30 -2.25 -2.20 -2.15 -2.10 -2.05

Φ(z) 0.0000 0.0000 0.0000 0.0001 0.0001 0.0001 0.0001 0.0001 0.0002 0.0002 0.0002 0.0003 0.0003 0.0004 0.0005 0.0006 0.0007 0.0008 0.0010 0.0011 0.0013 0.0016 0.0019 0.0022 0.0026 0.0030 0.0035 0.0040 0.0047 0.0054 0.0062 0.0071 0.0082 0.0094 0.0107 0.0122 0.0139 0.0158 0.0179 0.0202

z -2.00 -1.95 -1.90 -1.85 -1.80 -1.75 -1.70 -1.65 -1.60 -1.55 -1.50 -1.45 -1.40 -1.35 -1.30 -1.25 -1.20 -1.15 -1.10 -1.05 -1.00 -0.95 -0.90 -0.85 -0.80 -0.75 -0.70 -0.65 -0.60 -0.55 -0.50 -0.45 -0.40 -0.35 -0.30 -0.25 -0.20 -0.15 -0.10 -0.05

Φ (z ) 0.0228 0.0256 0.0287 0.0322 0.0359 0.0401 0.0446 0.0495 0.0548 0.0606 0.0668 0.0735 0.0808 0.0885 0.0968 0.1056 0.1151 0.1251 0.1357 0.1469 0.1587 0.1711 0.1841 0.1977 0.2119 0.2266 0.2420 0.2578 0.2743 0.2912 0.3085 0.3264 0.3446 0.3632 0.3821 0.4013 0.4207 0.4404 0.4602 0.4801

z 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 0.55 0.60 0.65 0.70 0.75 0.80 0.85 0.90 0.95 1.00 1.05 1.10 1.15 1.20 1.25 1.30 1.35 1.40 1.45 1.50 1.55 1.60 1.65 1.70 1.75 1.80 1.85 1.90 1.95

Φ (z ) 0.5000 0.5199 0.5398 0.5596 0.5793 0.5987 0.6179 0.6368 0.6554 0.6736 0.6915 0.7088 0.7257 0.7422 0.7580 0.7734 0.7881 0.8023 0.8159 0.8289 0.8413 0.8531 0.8643 0.8749 0.8849 0.8944 0.9032 0.9115 0.9192 0.9265 0.9332 0.9394 0.9452 0.9505 0.9554 0.9599 0.9641 0.9678 0.9713 0.9744

z 2.00 2.05 2.10 2.15 2.20 2.25 2.30 2.35 2.40 2.45 2.50 2.55 2.60 2.65 2.70 2.75 2.80 2.85 2.90 2.95 3.00 3.05 3.10 3.15 3.20 3.25 3.30 3.35 3.40 3.45 3.50 3.55 3.60 3.65 3.70 3.75 3.80 3.85 3.90 3.95

Φ ( z) 0.9772 0.9798 0.9821 0.9842 0.9861 0.9878 0.9893 0.9906 0.9918 0.9929 0.9938 0.9946 0.9953 0.9960 0.9965 0.9970 0.9974 0.9978 0.9981 0.9984 0.9987 0.9989 0.9990 0.9992 0.9993 0.9994 0.9995 0.9996 0.9997 0.9997 0.9998 0.9998 0.9998 0.9999 0.9999 0.9999 0.9999 0.9999 1.0000 1.0000

Φ(z) = P ( Z ≤ z ) for N(0, 1).

(Use interpolation to estimate z values to a 3rd decimal place.)

Live Exam Helper

t-table of left tail probabilities. t\df 0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 2.2 2.4 2.6 2.8 3.0 3.2 3.4 3.6 3.8 4.0 t0.0 \df

0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 2.2 2.4 2.6 2.8 3.0

1 0.5000 0.5628 0.6211 0.6720 0.7148 0.7500 0.7789 0.8026 0.8222 0.8386 0.8524 0.8642 0.8743 0.8831 0.8908 0.8976 0.9036 0.9089 0.9138 0.9181 0.9220

2 0.5000 0.5700 0.6361 0.6953 0.7462 0.7887 0.8235 0.8518 0.8746 0.8932 0.9082 0.9206 0.9308 0.9392 0.9463 0.9523 0.9573 0.9617 0.9654 0.9686 0.9714

3 0.5000 0.5729 0.6420 0.7046 0.7589 0.8045 0.8419 0.8720 0.8960 0.9152 0.9303 0.9424 0.9521 0.9598 0.9661 0.9712 0.9753 0.9788 0.9816 0.9840 0.9860

( T h e table s show P (T < t ) for T ∼ t(df ) . ) 4 0.5000 0.5744 0.6452 0.7096 0.7657 0.8130 0.8518 0.8829 0.9076 0.9269 0.9419 0.9537 0.9628 0.9700 0.9756 0.9800 0.9835 0.9864 0.9886 0.9904 0.9919

5 0.5000 0.5753 0.6472 0.7127 0.7700 0.8184 0.8581 0.8898 0.9148 0.9341 0.9490 0.9605 0.9692 0.9759 0.9810 0.9850 0.9880 0.9904 0.9922 0.9937 0.9948

6 0.5000 0.5760 0.6485 0.7148 0.7729 0.8220 0.8623 0.8945 0.9196 0.9390 0.9538 0.9649 0.9734 0.9797 0.9844 0.9880 0.9907 0.9928 0.9943 0.9955 0.9964

7 0.5000 0.5764 0.6495 0.7163 0.7750 0.8247 0.8654 0.8979 0.9232 0.9426 0.9572 0.9681 0.9763 0.9823 0.9867 0.9900 0.9925 0.9943 0.9956 0.9966 0.9974

8 0.5000 0.5768 0.6502 0.7174 0.7766 0.8267 0.8678 0.9005 0.9259 0.9452 0.9597 0.9705 0.9784 0.9842 0.9884 0.9915 0.9937 0.9953 0.9965 0.9974 0.9980

0.5000 0.5000 0.5000 0.5000 0.5000 10 0.5000 110.5000 0.5000 12 13 0.5000 14 0.5000 15 16 17 0.5773 0.5774 0.5776 0.5777 0.5778 0.5779 0.5780 0.5781 0.5781 0.5782 0.6512 0.6516 0.6519 0.6522 0.6524 0.6526 0.6528 0.6529 0.6531 0.6532 0.7191 0.7197 0.7202 0.7206 0.7210 0.7213 0.7215 0.7218 0.7220 0.7222 0.7788 0.7797 0.7804 0.7810 0.7815 0.7819 0.7823 0.7826 0.7829 0.7832 0.8296 0.8306 0.8315 0.8322 0.8329 0.8334 0.8339 0.8343 0.8347 0.8351 0.8711 0.8723 0.8734 0.8742 0.8750 0.8756 0.8762 0.8767 0.8772 0.8776 0.9041 0.9055 0.9066 0.9075 0.9084 0.9091 0.9097 0.9103 0.9107 0.9112 0.9297 0.9310 0.9322 0.9332 0.9340 0.9348 0.9354 0.9360 0.9365 0.9370 0.9490 0.9503 0.9515 0.9525 0.9533 0.9540 0.9546 0.9552 0.9557 0.9561 0.9633 0.9646 0.9657 0.9666 0.9674 0.9680 0.9686 0.9691 0.9696 0.9700 0.9738 0.9750 0.9759 0.9768 0.9774 0.9781 0.9786 0.9790 0.9794 0.9798 0.9813 0.9824 0.9832 0.9840 0.9846 0.9851 0.9855 0.9859 0.9863 0.9866 0.9868 0.9877 0.9884 0.9890 0.9895 0.9900 0.9903 0.9907 0.9910 0.9912 0.9906 0.9914 0.9920 0.9925 0.9929 0.9933 0.9936 0.9938 0.9941 0.9943 0.9933 0.9940 0.9945 0.9949 0.9952 0.9955 0.9958 0.9960 0.9962 0.9963

9 0.5000 0.5770 0.6508 0.7183 0.7778 0.8283 0.8696 0.9025 0.9280 0.9473 0.9617 0.9723 0.9801 0.9856 0.9896 0.9925 0.9946 0.9961 0.9971 0.9979 0.9984 18

19

Live Exam Helper

t\df 0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 2.2 2.4 2.6 2.8 3.0

0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 t2.2 \df 2.4 2.6 2.8 3.0

20 0.5000 0.5782 0.6533 0.7224 0.7834 0.8354 0.8779 0.9116 0.9374 0.9565 0.9704 0.9801 0.9869 0.9914 0.9945 0.9965

21 0.5000 0.5783 0.6534 0.7225 0.7837 0.8357 0.8782 0.9119 0.9377 0.9569 0.9707 0.9804 0.9871 0.9916 0.9946 0.9966

22 0.5000 0.5783 0.6535 0.7227 0.7839 0.8359 0.8785 0.9123 0.9381 0.9572 0.9710 0.9807 0.9874 0.9918 0.9948 0.9967

23 0.5000 0.5784 0.6536 0.7228 0.7841 0.8361 0.8788 0.9126 0.9384 0.9575 0.9713 0.9809 0.9876 0.9920 0.9949 0.9968

24 0.5000 0.5784 0.6537 0.7229 0.7842 0.8364 0.8791 0.9128 0.9387 0.9578 0.9715 0.9812 0.9877 0.9921 0.9950 0.9969

25 0.5000 0.5785 0.6537 0.7230 0.7844 0.8366 0.8793 0.9131 0.9389 0.9580 0.9718 0.9814 0.9879 0.9923 0.9951 0.9970

26 0.5000 0.5785 0.6538 0.7231 0.7845 0.8367 0.8795 0.9133 0.9392 0.9583 0.9720 0.9816 0.9881 0.9924 0.9952 0.9971

27 0.5000 0.5785 0.6538 0.7232 0.7847 0.8369 0.8797 0.9136 0.9394 0.9585 0.9722 0.9817 0.9882 0.9925 0.9953 0.9971

0.5000 0.5000 0.5000 0.5000 0.5000 0.5000 0.5000 0.5000 0.5000 0.5000 0.5786 0.5786 0.5786 0.5786 0.5787 0.5787 0.5787 0.5787 0.5787 0.5787 0.6540 0.6541 0.6541 0.6541 0.6542 0.6542 0.6542 0.6543 0.6543 0.6543 0.7235 0.7236 0.7236 0.7237 0.7238 0.7238 0.7239 0.7239 0.7240 0.7240 0.7850 0.7851 0.7852 0.7853 0.7854 0.7854 0.7855 0.7856 0.7857 0.7857 0.8373 0.8375 0.8376 0.8377 0.8378 0.8379 0.8380 0.8381 0.8382 0.8383 0.8802 0.8804 0.8805 0.8807 0.8808 0.8809 0.8810 0.8811 0.8812 0.8813 0.9141 0.9143 0.9144 0.9146 0.9147 0.9148 0.9150 0.9151 0.9152 0.9153 0.9400 0.9401 0.9403 0.9404 0.9406 0.9407 0.9408 0.9409 0.9411 0.9412 0.9590 0.9592 0.9594 0.9595 0.9596 0.9598 0.9599 0.9600 0.9601 0.9602 0.9727 0.9728 0.9730 0.9731 0.9732 0.9733 0.9735 0.9736 0.9737 0.9738 30 0.9823 310.9824 0.9825 32 33 0.9827 34 0.9828 35 36 37 0.9822 0.9826 0.9829 0.9830 0.9831 0.9886 0.9887 0.9888 0.9889 0.9890 0.9891 0.9892 0.9892 0.9893 0.9894 0.9928 0.9929 0.9930 0.9931 0.9932 0.9932 0.9933 0.9933 0.9934 0.9935 0.9956 0.9956 0.9957 0.9958 0.9958 0.9959 0.9959 0.9960 0.9960 0.9960 0.9973 0.9974 0.9974 0.9974 0.9975 0.9975 0.9976 0.9976 0.9976 0.9977

28 0.5000 0.5785 0.6539 0.7233 0.7848 0.8371 0.8799 0.9138 0.9396 0.9587 0.9724 0.9819 0.9884 0.9926 0.9954 0.9972

38

29 0.5000 0.5786 0.6540 0.7234 0.7849 0.8372 0.8801 0.9139 0.9398 0.9589 0.9725 0.9820 0.9885 0.9927 0.9955 0.9973

39

Live Exam Helper

t\ df 0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 2.2 2.4 2.6 2.8 3.0

40 41 42 43 44 45 46 47 0.5000 0.5000 0.5000 0.5000 0.5000 0.5000 0.5000 0.5000 0.5000 0.5000 0.5788 0.5788 0.5788 0.5788 0.5788 0.5788 0.5788 0.5788 0.5788 0.5788 0.6544 0.6544 0.6544 0.6544 0.6545 0.6545 0.6545 0.6545 0.6545 0.6546 0.7241 0.7241 0.7241 0.7242 0.7242 0.7242 0.7243 0.7243 0.7243 0.7244 0.7858 0.7858 0.7859 0.7859 0.7860 0.7860 0.7861 0.7861 0.7862 0.7862 0.8383 0.8384 0.8385 0.8385 0.8386 0.8387 0.8387 0.8388 0.8388 0.8389 0.8814 0.8815 0.8816 0.8816 0.8817 0.8818 0.8819 0.8819 0.8820 0.8820 0.9154 0.9155 0.9156 0.9157 0.9157 0.9158 0.9159 0.9160 0.9160 0.9161 0.9413 0.9414 0.9415 0.9415 0.9416 0.9417 0.9418 0.9419 0.9419 0.9420 0.9603 0.9604 0.9605 0.9606 0.9606 0.9607 0.9608 0.9609 0.9609 0.9610 0.9738 0.9739 0.9740 0.9741 0.9742 0.9742 0.9743 0.9744 0.9744 0.9745 0.9832 0.9833 0.9833 0.9834 0.9834 0.9835 0.9836 0.9836 0.9837 0.9837 0.9894 0.9895 0.9895 0.9896 0.9897 0.9897 0.9898 0.9898 0.9898 0.9899 0.9935 0.9935 0.9936 0.9936 0.9937 0.9937 0.9938 0.9938 0.9938 0.9939 0.9961 0.9961 0.9962 0.9962 0.9962 0.9962 0.9963 0.9963 0.9963 0.9964 0.9977 0.9977 0.9977 0.9978 0.9978 0.9978 0.9978 0.9978 0.9979 0.9979

48

49

Live Exam Helper

Table of χ 2 critical values (right-tail) T h e table sh ows cdf, p = t h e 1 − p quantile of χ 2 (df ).

In R notation cdf, p = qchisq(1-p, df).

df\p 1 2 3 4 5 6 7 8 9 10 16 17 18 19 20 21 22 23 24 25 30 31 32 33 34 35 40 41 42 43 44 45 46 47 48 49

0.010 6.63 9.21 11.34 13.28 15.09 16.81 18.48 20.09 21.67 23.21 32.00 33.41 34.81 36.19 37.57 38.93 40.29 41.64 42.98 44.31 50.89 52.19 53.49 54.78 56.06 57.34 63.69 64.95 66.21 67.46 68.71 69.96 71.20 72.44 73.68 74.92

0.025 5.02 7.38 9.35 11.14 12.83 14.45 16.01 17.53 19.02 20.48 28.85 30.19 31.53 32.85 34.17 35.48 36.78 38.08 39.36 40.65 46.98 48.23 49.48 50.73 51.97 53.20 59.34 60.56 61.78 62.99 64.20 65.41 66.62 67.82 69.02 70.22

0.050 3.84 5.99 7.81 9.49 11.07 12.59 14.07 15.51 16.92 18.31 26.30 27.59 28.87 30.14 31.41 32.67 33.92 35.17 36.42 37.65 43.77 44.99 46.19 47.40 48.60 49.80 55.76 56.94 58.12 59.30 60.48 61.66 62.83 64.00 65.17 66.34

0.100 2.71 4.61 6.25 7.78 9.24 10.64 12.02 13.36 14.68 15.99 23.54 24.77 25.99 27.20 28.41 29.62 30.81 32.01 33.20 34.38 40.26 41.42 42.58 43.75 44.90 46.06 51.81 52.95 54.09 55.23 56.37 57.51 58.64 59.77 60.91 62.04

0.200 1.64 3.22 4.64 5.99 7.29 8.56 9.80 11.03 12.24 13.44 20.47 21.61 22.76 23.90 25.04 26.17 27.30 28.43 29.55 30.68 36.25 37.36 38.47 39.57 40.68 41.78 47.27 48.36 49.46 50.55 51.64 52.73 53.82 54.91 55.99 57.08

0.300 1.07 2.41 3.66 4.88 6.06 7.23 8.38 9.52 10.66 11.78 18.42 19.51 20.60 21.69 22.77 23.86 24.94 26.02 27.10 28.17 33.53 34.60 35.66 36.73 37.80 38.86 44.16 45.22 46.28 47.34 48.40 49.45 50.51 51.56 52.62 53.67

0.500 0.45 1.39 2.37 3.36 4.35 5.35 6.35 7.34 8.34 9.34 15.34 16.34 17.34 18.34 19.34 20.34 21.34 22.34 23.34 24.34 29.34 30.34 31.34 32.34 33.34 34.34 39.34 40.34 41.34 42.34 43.34 44.34 45.34 46.34 47.34 48.33

0.700 0.15 0.71 1.42 2.19 3.00 3.83 4.67 5.53 6.39 7.27 12.62 13.53 14.44 15.35 16.27 17.18 18.10 19.02 19.94 20.87 25.51 26.44 27.37 28.31 29.24 30.18 34.87 35.81 36.75 37.70 38.64 39.58 40.53 41.47 42.42 43.37

0.800 0.06 0.45 1.01 1.65 2.34 3.07 3.82 4.59 5.38 6.18 11.15 12.00 12.86 13.72 14.58 15.44 16.31 17.19 18.06 18.94 23.36 24.26 25.15 26.04 26.94 27.84 32.34 33.25 34.16 35.07 35.97 36.88 37.80 38.71 39.62 40.53

0.900 0.02 0.21 0.58 1.06 1.61 2.20 2.83 3.49 4.17 4.87 9.31 10.09 10.86 11.65 12.44 13.24 14.04 14.85 15.66 16.47 20.60 21.43 22.27 23.11 23.95 24.80 29.05 29.91 30.77 31.63 32.49 33.35 34.22 35.08 35.95 36.82

0.950 0.00 0.10 0.35 0.71 1.15 1.64 2.17 2.73 3.33 3.94 7.96 8.67 9.39 10.12 10.85 11.59 12.34 13.09 13.85 14.61 18.49 19.28 20.07 20.87 21.66 22.47 26.51 27.33 28.14 28.96 29.79 30.61 31.44 32.27 33.10 33.93

0.975 0.00 0.05 0.22 0.48 0.83 1.24 1.69 2.18 2.70 3.25 6.91 7.56 8.23 8.91 9.59 10.28 10.98 11.69 12.40 13.12 16.79 17.54 18.29 19.05 19.81 20.57 24.43 25.21 26.00 26.79 27.57 28.37 29.16 29.96 30.75 31.55

0.990 0.00 0.02 0.11 0.30 0.55 0.87 1.24 1.65 2.09 2.56 5.81 6.41 7.01 7.63 8.26 8.90 9.54 10.20 10.86 11.52 14.95 15.66 16.36 17.07 17.79 18.51 22.16 22.91 23.65 24.40 25.15 25.90 26.66 27.42 28.18 28.94

Live Exam Helper

18.05 Final Exam Solutions Part I: Concept questions

(58

points) T h e s e questions are all multiple choice or short answer. You don’t have to show any work. Work through t h e m quickly. E ac h answer is worth 2 points.

Concept 1.

answer: C. ( i ) and (ii)

Concept 2.

answer: True

Concept 3.

answer: True

Concept 4. answer: ( i ) Simple (ii) Composite (iii) One-sided Concept 5.

answer: B.

Concept 6.

answer: 2. B

Concept 7. (i) (ii)

answer: A. P ( A 1 ) .

answer: C. P (B 2 |A 1 ).

(iii)

answer: D. P (C 1 |B 2 ∩ A 1 ).

(iv)

answer: C. A 1 ∩ B 2 ∩ C 1 .

Concept 8.

answer: BAC.

Concept 9. answer: p = 0.8 u s e minimal strategy.

If you u s e t h e minimal strate gy t h e law of large numbers s a ys your average winnings p e r certainly b e t h e ex p e c t e d winnings of one b et.

b e t will almost

Win -10 10 p 0.2 0.8 T h e e x p e c t e d value when p = 0.8 is 6. Since this is positive you’d like to make a lot of b e t s and let t h e law of large numbers (practically) guarantee you will win an average of $6 p e r be t. So you u s e t h e minimal strategy.

Concept 10. answer: A. Independent. T h e variables can b e separate d: t h e marginal densities are f X ( x ) = ax 2 and f Y ( y ) = by3 for s o m e constants a and b with ab = 4.

1 B. Not independent. X and Y are not independent b e c a u s e the re is no way to factor f (x , y ) into a produ ct f X ( x ) f Y ( y ).

Live Exam Helper

Concept 11. answer: B . A Bernoulli random variable takes values 0 or 1. So X is discrete. T h e parameter θ can b e anywhere in t h e continuous range [0,1]. Therefore t h e s p a c e of h y p o t h e s e s is continuous.

Concept 12. answer: D. B y t h e form of t h e posterior p df we know it is beta(8 , 13).

Concept 13. A. True, B. False C. True Concept 14.

answer: A. Not valid B. not valid C. valid

B o t h t h e prior and posterior measure a belief in t h e distribution of h y p o t h e s e s ab ou t t h e frequentist d o e s not consider t h e m valid.

value of θ. T h e

T h e likelihood f (x|theta) is perfec tly acc e ptab le to t h e frequentist. It re presents t h e p rob ability of data from a repeatable experiment, i.e. measuring how late J a n e is e ac h day. Conditioning on θ is fine. Thi s j us t fi xe s a model parameter θ. It doesn’t require c o m p u t ing probabilities for θ.

Concept 15. answer: E. unknown. Frequentist m e t h o d s only give probabilities for data under an as s u m e d hyp othe si s. T h e y do not give probabilities or o d d s for h yp ot h e se s . So we don’t know t h e o d d s for distribution m eans Concept 16.

A. Correct, Thi s is t h e definition of a confidence interval.

B. Incorrect. Frequentist m e t h o d s do not give probabilities for h yp o th e s e s .

C. Correct. Given θ = 0 t h e probability θ is in [-1, 1.5] is 100%.

A

B

Part II: Problems

Problem 1. (20) (a) P ( ( A ∪ B ) c ) = 3/8 ⇒ P ( A ∪ B ) = 5/8 ..

0.4

C points) D (325 0.3

0.2 0.1

Figure for part ( a) .

Figure for part ( b ) . (b) S e e t h e figure: P ( ( C U D ) c ) = 0.4 ⇒ P ( ( C U D ) = 0.6).

P (C ∪ D ) = P ( C ) + P ( D ) − P (C ∩ D ) ⇒ 0.6 = 0.5 + P ( D ) − 0.2 ⇒

P ( D ) = 0.3

Problem 2. (20) 3/5 R2/4 2

R1

2/5 B2

R1

2/4

R2

4/6

B2

2/6

.

Live Exam Helper

3 2 6 . · = = 0.3 5 4 P (R 2 |B 1 )P ( 20 B1 ) P = 3 5· (R2 )

(a) P (R 1 ∩ R 2 ) = (b) P ( B |R1 ) = 2 Problem 3. (15)

2 ·4 5 6 2 2 4 4+ 5· 6

8/30 = = 17/ 30 17

8 .

F (1): Since you never g e t more than 6 on one roll we have

F (1) = 0 .

F (2 ) = P ( X = 1) + P ( X = 2): P ( X = 1) = 0 P ( X = 2) = P (total on 2 dice = 7,8,9,10,11,12) =

21 7 = .

36 12 F (7): T h e smallest total on 7 rolls is 7, s o F ( 7 ) = 1 .

Problem 4. (20)

P ( X ≥ 0.55) =

(a) Let X = score of a random student.

1

f ( x ) dx = 0.55

1

4 − 4 x d x = 4x − 2x 0.55

1

0.55

2 = 2 − 4 × 0.55 + 2(0.55) = 0.405

(b) Geometric method: We ne ed t h e sh ad e d area in t h e figure to b e 0.125 S h a d e d area = area of triangle = 12 (1 − x ) ( 4 − 4 x ) = 0.125.

Solving for x we g e t 2(1 − x )

2

= 0.125 ⇒ (1 −2 x )

1 ⇒ 16

=

3 x=

2

y

.

x x = q0.875

4 Analytic mehtod: We want a s u c h th at F ( a ) = 7/8. Since f ( x ) is de fined in two p i e c e s we have to c o m p u te F ( a ) in two pieces.

1/ 2

F (1/2) =

0

2 1/ 2 =0 .

4x dx = 2x

(W hic h we knew geometrically already.) For

1 2

a ≥ 1/2 we the n have F (a) = =

1/ 2

4xdx +

0

1 +

a

a

4 − 4xdx 1/ 2

4 − 4xdx 1/ 2 21 [ 2 = + 4x − 2x 2

a 1/ 2 2

= 4a − 2a − 1.

Solving for a s u c h th at F ( a ) = 7/8 we g e t 4a − 2a2 − 1 = 7/8 ⇒ 2a2 − 4a + 15 / 8 = 0 ⇒ a =

√ 4± 1 3 5 4 = 4, 4.

Live Exam Helper

Since4 5 is not in t h e range of X we have a = 3/4 . ( T h e s am e answer as with t h e geometri c m e th od . )Problem 5. (15) (a) f ( x ) = F / ( x ) = 2 − 2x on [0, 1]. Therefore ∫ 1 E(X) = x f ( x ) dx ∫ 01 = 2x − 2x dx2 0

= x2 − = 1 .

3

2 3 3x

1 0

.

(b) P ( X ≤ 0.4) = F (0.4) = 0.4(2 − 0.4) = 0.4(1.6) = 0.64 .

1 Problem 6. (15) Let X ∼ U(a, b). T h e p df of X is f ( x ) = on t h e interval [a, b]. b− a T hus, ∫ b ∫ b x b .. b b2 − a2 b+ a E ( X ) = x f ( x ) dx = dx = x2 = a a − a a 2 = ∫ b ( x − μ ) f ( x2) 2(b dx − a) 2(b − a) a ) ∫ b( a+ b 2 1 = x− dx 2 a b − a b 3 x − a+b 1 = .

2 3 b− a . a = . . . algebra . . .

1 = (b − a) 3 1 12 b− a 2 (b − a) = .

12

Va r ( X ) =

Problem 7. (20) (a) We organize t h e problem in a tree. Here: D + = default, D − = no default T + = t e s t is positive, T − = t e s t is negative

Live Exam Helper

0.01 1

0.99

0

D+

T+

0.04

D−

0.96

T+

T−

T−

0.01 0.01 1 = = = ≈ 0.2 .

0.0496 4.96 0.01 + 0.99 · 0.04 P (D + |T + ) 1/4.96 1 + + (b) Odds(winning) = O dd s( D |T ) = = = .

4 − |T + (Dodd 3.96/4.96 Since t h e payoff ratio is greater than P1/( s )of winning), it is3.96 a g o o d b et.

1 Equivalently we can argue 1 the P (D + |T ) +=

P (T + |D + )P ( D + ) P (T + )

3.96 Y\

X

40

1

0 170/700 E(winnings) = 400 · 1 marginal for X

4.96

85/700 255/700

2

70/700 − 100 ·

4.96

190/700 260/700

= 30/700> 0.

4.96

155/700 185/700

marginal for Y 270/700 430/700 1

A positive x pife cPt(eXd =winnings e an a g o(Y o d= be (b) We cheeck 0,Y = 0)m= P s( Xit’s = 0)P 0).t.

170 ? 255 270 Problem 8. (30) (a) Probability table: 700 700 = 700 .

Cross-multiply and do a little ? ? algebra ⇔ 11900 = ⇔ 11900 = 68850 ? X and Y are not independent.

Since t h e y are not 170 · 700 = 255 · 270 equal (c)

185 630 9 + 2 · = = 700 = 700 700 700 1070 190 155 500 5 E (430 XY ) = + 2 · = 43 = 700 700 700 5 9 43 113 Cov(X, Y ) −7E ( X Y ) − E ( X ) E ( Y ) = − · = 7 10 70 700 Cov(X, Y ) (d) T h e definition of correlation is Cor( X, Y ) = . So we first ne ed to σ σ X Y c om p ut e t h e variances of X and Y .

E(X) = E(Y ) =

260

260

E(X2 ) =

700

185 1000 10 + 4 · = = 700 700 7

Live Exam Helper

T hus, Var( X ) = E ( X )2 − E ( X ) 2 = E(Y 2 ) =

10 − 7

43

70

Var(Y ) = E ( Y 2 ) − E ( Y ) 2 = therefor e

43 − 70

81 = 100 ( ) 43 70

433 700 2

=

43 · 27 702

113/700 Cor(X, Y ) =

J

J

433 /700 4 3 · 27/70 2

Note: We would ac c e c p t –even encourage solutions– th at left t h e fractions uncompute d, e.g. σY =

J

4 3 / 7 0 − (43/70) 2 .

Problem 9. (20) (a) Let X ∼ binomial(25, 0.5) = t h e number supporting t h e referendum.

We know th at Standardizing and using t h e CLT we have Z = X − 12.5 ≈ N(0, 1) Therefore, 1 25 5/2 5 X − 12.5 14 − 12.5 P ( X ≥ 14) = P ( ≥ ≈ )P ( Z ≥ 0.6) = Φ(−0.6) = 0.2743 , 5/ 2 5/ 2 E ( X ) = 12.5, Va r ( X ) = 25 · = , σX = .

4 4 2 whe re t h e last probability was looked u p in t h e Ztable. (b)

T h e rule of thum b CI is z

x ± z0.05 · √ .

2 n 1

√ S o we want 0.05 ≤ 0.01.

From t h e 2table z0.05 = Φ(−0.05) = 1.65. So we want n √ 1.65 165 ⇒ √ ≤ 0.01 ⇒ n≥ 2 2 n answer: n = 6807

n > (82.5) =2 6806.25

Problem 10. (10 p t s ) For a fixe d τ t h e p df for x i is f ( xi | τ ) = x τ e − 12 τ x 2 . Therefore t h e likelihood function of data is � the f ( d ata | τ ) = x1 x2 ··· xn τ e n − 12 τ x 2 i .

T h e log likelihood is 1 � 2 l n ( f (d ata | τ ) ) = ln(x1 x2 ··· xn ) + n ln(τ ) − τ i 2 x .

Live Exam Helper

We find t h e M L E for τ b y taking a derivative of t h e log likelihood with re s p e c t to τ and setting equal to 0. d ln(f ( d a t a | τ ) ) n 1 Σ n Σ 1 2n 2 . = − xi =20 ⇒ = xi ⇒ τ = � dτ x 2i τ 2 τ 2 Problem 11. (15) (a) We a s s u m e t h e random error te rms e i are independent, have mean 0 and all have t h e s am e variance ( homosc ed asti c ). (b) E(b) = s u m of t h e squared errors =

Σ

( y i − b|xi − 3|)2

= (10 − b)2 + (3 − 4b)2 + (2 − 3b)2 d E( b) = −2(10 − b) − 8(3 − 4b) − 6(2 − 3b) = 52b − 56 = 0. db T h e least squares fit is found b y setting t h e derivative (with re s pe c t to b) to 0, 56 14 .

Therefore t h e least squares estimate of ˆb = 52 =13 b is Problem 12. (30) (a) Since σ is unknown we u s e t h e Studentized x −μ mean √ t= ∼ t(44) s / n which follows a t distributions with 44 d e gre e s of freedom. (i ) T h e 80% CI is x ± t0.1 √ = 44 is approximately 1.3.

Th us, (ii) We use the statistic

n

. From t h e t-table we g 4 e t t 0.1 with df 80% CI = 5 ± √ · 1.3 45

( n − 1)s 2 σ2

∼ χ 2 (44). T h e 80% confidence interval for σ 2 is [ ] ( n − 1)s 2( n − 1)s 2 , , c0.9

c0.1

where c0.9 and c0.1 are t h e right critical values from t h e chi-square distribution with 44 de gre e s of freedom. [ ]2 [ ] ( n − 1)s 2 ( n − 1)s 44 · 16 ,44 · 16 80% CI for σ2 , = 56.37 32.49 56.37 32.49 = (b) T h e 80% b oo tst r ap CI is [x − δ0∗.1, x − δ0∗.9], where δ0∗.1 and δ0∗.9 are empirical right tail critical points for δ∗

Live Exam Helper

δ0∗.1 = 450th element = 0.169 δ0∗.9 = 50th element = -0.2 So CI = [5 5 +since 0.2] =it [4.831, (c) tThhee 80% approach in − ( b0.169, ) is fine makes 5.2]. no assumptions about t h e underlying x −μ √ distri bution. T h e approach in ( a ) is more problematic since d o e s not distribution. However for an exponential distribution and n = 45 t h e approximation is follow a Student-t not too bad. s/ n (d) M e t h o d ( b ) is preferable if t h e underlying distribution is highly asymmetric. pˆ = 1 / x .

Problem 13. (15) (a) Since μ = 1/p we should u s e t h e approximation (b) S t e p 1. Approximate p b y pˆ = 1 / x .

S t e p 2. Generate a bo ots tr ap sam pl e x ∗ 1 ,..., xn∗ from geom(pˆ).

S t e p 3. C om pute p∗ = 1/x ∗ and δ∗ = p∗ − pˆ.

Re p e at s t e p s 2 and 3 many times ( s a y 104 times.

S t e p 4. List all t h e δ∗ and find th e critical values.

Le t δ0∗.025 = 0.025 critical value = 0.975 quantile. Let Problem 14. (30) (a) We will u se t h e standardized mean b a s e d on H 0 as a t e s t statistic: δ0∗.975 = 0.975 critical value = 0.025 quantile. x − μ0 x −3 √interval z= = S t e p 5. T h e b oo tst r ap confidence is. = [pˆ5(x − − 3). σ/ n 2/10 δAt∗ α = , pˆ 0.05 − δ ∗we].re ject H 0 if 0 .025

0 .975

z < z0.975 = −1.96

or z > z0.025 = 1.96.

(Or we could have u s e d x a s a t5e2/ s t statistic and g ot t h e corresponding rejection (b) With this data we have z = − 310= 10. T h e rejection region is two sided s o region.) p = P (|Z| > |z|) = P (|Z| > 10) = 0.

Yes, since p < α you should re ject H 0 . (c) Power = P (re j e c t |xμ−=3 4) Our z-statistic is z =2/ 10 and we don’t re ject if x −3 −1.96 ≤ z ≤ 1.96 ⇔ −1.96 ≤ ≤ 1.96 2/10 So,

⇔

Power = P ( re j e c t | μ = 4) = 1 − P (don’t re ject | μ = 4) = 1 − P (2.61 < x < 3.39 | μ = 4)

2.61 ≤ x ≤ 3.39

Live Exam Helper

We standardize using t h e given mean μ = ( ) 4 2.61 − 4 −.61 = 1 −P < Z< 2/ 10 2/ 10 = 1 − P (−6.9 < Z < −3.05) = 1 − Φ(−3.05) + Φ(−6.9) = 1 − 0.0011 + 0 = 0.9989 .

T h e probabilities were looked u p in t h e z-table. We u s e d Φ(−6.9) ≈ 0.

(We could have u s e d much l e ss calculation to find th at t h e non-rejection range is x Hypothesis Prior Likelihood Posterior between N(μ p o s t , σ 2 N(80, 16) f (x|θ) ∼ N(θ, 0.01) θ −7σx and −3σx from t h e mean μ = 4.)

) post

We have 15. (30) (a) Thi s is a normal/normal conjugate prior/likilihood upd ate .

Problem 1 1 1 1 a= = , b= = prior 4 = 2. σ 2 For t h e 0.5 σ2 update post aμ prior + bx a+ b μ = 80/ 4 + 170 760 = = ≈ 84.44 9 1/4 + 2 1 = σ 2post a+ b 1 4 = = ≈ 0.4444 1 / 4 + 2 So, t h e posterior 9 2 is f (θ | x = 84) ∼ N(μ post, σ post ) = N(84.44, 0.4444) (b) In this c a s e a = 1/4, b = n/ 0 . 5 = 2n. We know 1 1 4 = = = σ 2post a + b 1 / 4 + 2n 8n + 1 Now σ 2 post ≤ 0.01 gives u s 4 399 ≤ n answer: n = 50 . ≤ 0.01 ⇒ 400 ≤ 8n + 1 ⇒ 8 8n + 1 Problem 16. (20) (a) Let θ represent t h e number of side s to t h e die. T h e data is x 1 = 7 Hypothesis

prior

likelihood

θ

p (θ)

p(x 1 = 7 | θ)

po ste rior p(θ)p(x 1 = 7 | θ)

unnorm. p o s t . p(θ)p(x 1 = 7 | θ)

p(θ | x 1 = 7 ) =

p(x = 7) 1

θ= 6 θ= 8 θ = 12

1/2 1/4 1/4

0 1/8 1/12

0 1/32 1/48

0 3/5 2/5

Live Exam Helper

p(θ = 12 | x 1 = 7) 2/5 2 . p(θ = 12 | x = 7) 3/5 1= (b) Odds = = 3 (c) We exte nd t h e table in order to c o m p u t e t h e posterior predictive probability. p(θ | x 1 = 7) p(x 2 = 7 | θ) p(θ | x 1 = 7)p(x 2 = 7 | θ) θ θ= 6 θ= 8 θ = 12

Total

0 3/5 2/5

T h e total probability p(x 2 = 7 | x 1 = 7) =

0 1/8 1/12

13 .

120

0 3/40 2/60

13/120