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45-733 PROBABILITY AND STATISTICS I Final Examination Answers

Probability and Statistics
Name__________________________

Spring 1999 Flex-Mode and Flex-Time 45-733

Final Exam

Keith Poole

(10 Points)

Find the maximum likelihood estimator foræ q(1 - q)^{x}x = 0,1,2, ... f(x) = ç è 0 otherwise

L(x

¶L/ ¶q = n/q - å

_{Ù}_ Hence, q = n/(n + å_{i=1,n}x_{i}) = 1/(1 + X_{n})

Probability and Statistics
Name__________________________

Spring 1999 Flex-Mode and Flex-Time 45-733

Final Exam

Keith Poole

(10 Points)

The Confidence Interval is: _ _ _ _P{XOur confidence limits are: 120 - 130 ± 2.33[4/36 + 5/64]_{n}- Y_{m}- z_{a/2}[s_{x}^{2}/n + s_{y}^{2}/m]^{1/2}< m_{x}- m_{y}< X_{n}- Y_{m}+ z_{a/2}[s_{x}^{2}/n + s_{y}^{2}/m]^{1/2}} = 1 - a^{1/2}= -10 ± 1.0136 (-11.0136, -8.9864)

Probability and Statistics
Name__________________________

Spring 1999 Flex-Mode and Flex-Time 45-733

Final Exam

Keith Poole

(10 Points)

Findæ (x + y)/8 0 < x < 2 f(x,y) = ç 0 < y < 2 è 0 otherwise

P(X > 1 Ç Y < 1) = (1/8)ò

(1/8)ò

P(Y < 1) = (1/8)ò

(1/8)ò

Hence: P(X > 1 | Y < 1) = (2/8)/(3/8) = 2/3

Spring 1999 Flex-Mode and Flex-Time 45-733

Final Exam

Keith Poole

(10 Points)

H

-2.256 < -1.96 so Reject HThe decision rule for this problem is: _ _ If(Xor _ __{n}- Y_{m})/(s_{1}^{2}/n + s_{2}^{2}/m)^{1/2}> z_{a/2}(Xthen Reject_{n}- Y_{m})/(s_{1}^{2}/n + s_{2}^{2}/m)^{1/2}< -z_{a/2}H_ _ If_{o}:-zthen Do Not Reject_{a/2}< (X_{n}- Y_{m})/(s_{1}^{2}/n + s_{2}^{2}/m)^{1/2}< z_{a/2}Hz_{o}:_{a/2}= 1.96 _ _(X= (75 - 80)/[100/51 + 121/41]_{n}- Y_{m})/(s_{1}^{2}/n + s_{2}^{2}/m)^{1/2}^{1/2}= -2.256

.0241 using CNORM(-2.256)*2

Spring 1999 Flex-Mode and Flex-Time 45-733

Final Exam

Keith Poole

(10 Points)

Treating this as a sample from a normal distribution, test the hypothesis that the true mean is 16 ounces against the hypothesis that the true mean is less than 16 ounces (a = .01). Compute the P-Value of the test using EViews.

H

The decision rule for this problem is:

_ If (X_{n}- m_{o})/s/n^{1/2}< -t_{a}then Reject H_{o}: _ If -t_{a}< (X_{n}- m_{o})/s/n^{1/2}then Do Not Reject H_{o}: -t_{.01,10}= -2.764

_ (X_{n}- m_{o})/s/n^{1/2}= (15.736 - 16)/.2461/11^{1/2}= -3.5579 Hence, Reject H_{0}: P-Value = .00019

Spring 1999 Flex-Mode and Flex-Time 45-733

Final Exam

Keith Poole

(10 Points)

The population sampled has a normal distribution and the sample size is 36. FindH_{0}: m = 115

s^{2}= 400 H_{1}: m = 120

_ If X_{n}< 118.5 then do not reject H_{0}: _ If X_{n}> 118.5 then reject H_{0}:

_ _ a = P(X_{n}> 118.5 | m = 115) = P[(X_{n}- 115)/20/6 > (118.5 - 115)/20/6] = P(Z > 1.05) = 1 - F(1.05) = .1469 _ _ b = P(X_{n}< 118.5 | m = 120) = P[(X_{n}- 120)/20/6 < (118.5 - 120)/20/6] = P(Z < -.45) = F(-.45) = .3264

Spring 1999 Flex-Mode and Flex-Time 45-733

Final Exam

Keith Poole

(10 Points)

E(W) = 3*4 - 5*3 + 1*(-2) - 5 = -10

VAR(W) = 9*9 + 25*10 + 1*9 = 340

P(W > -7) = P[(W + 10)/340

P(Z > .16) = 1 - F(.16) = .4364

Spring 1999 Flex-Mode and Flex-Time 45-733

Final Exam

Keith Poole

(10 Points)

H

_{Ù}_{Ù}p_{1}- p_{2}Test: ------------------------- > z_{a/2}or < - z_{a/2}Reject H_{0}:_{Ù}_{Ù}_{Ù}_{Ù}[p_{1}(1 - p_{1})/n_{1}+ p_{2}(1 - p_{2})/n_{2}]^{1/2}z_{.005}= 2.58_{Ù}_{Ù}p_{1}= 61/82 = .744, p_{2}= 73/90 = .811 Test Statistic = (.744 - .811)/[(.744*.256)/82 + (.811*.189)/90]^{1/2}= -1.056 Hence, Do Not Reject H_{0}: P-Value = .291

Spring 1999 Flex-Mode and Flex-Time 45-733

Final Exam

Keith Poole

(5 Points)

P(3 £ l £ 12) = F(12) - F(2) = .876 - .006 = .870

Spring 1999 Flex-Mode and Flex-Time 45-733

Final Exam

Keith Poole

(10 Points)

Confidence Limits are:_{Ù}_{Ù}_{Ù}p ± z_{a/2}[p(1 - p)/n]^{1/2}_{Ù}p = 4/80 = .05, z_{.025}= 1.96, z_{.005}= 2.58 .05 ± 1.96*[(.05*.95)/80]^{1/2}= .05 ± .0478 so the 95% limits are: (.0022, .0978) .05 ± 2.58*[(.05*.95)/80]^{1/2}= .05 ± .0629 so the 99% limits are: (-.0129, .1129) or (.0, .1129) since proportions cannot be less than zero.

Spring 1999 Flex-Mode and Flex-Time 45-733

Final Exam

Keith Poole

(10 Points)

Assume that orchid height is normally distributed. Construct 98 and 99% confidence limits for the true variance.12.1 12.3 12.8 12.0 11.9 12.5 13.1 12.9 12.7 12.4 12.5

Confidence Limits are:

(n-1)s

(n-1)s

98% Limits:

c

c

(10*.14618)/23.2093 = .0630

and (10*.14618)/2.55821 = .5714

99% Limits:

c

c

(10*.14618)/25.1882 = .0580

and (10*.14618)/2.15585 = .6781

Spring 1999 Flex-Mode and Flex-Time 45-733

Final Exam

Keith Poole

(10 Points)

and we draw a random sample of 10 Pitt students and find their scores to be:116 119 125 121 120 118 110

Test the null hypothesis that the variances of the two populations are the same against the alternative hypothesis that variances are not the same (110 111 109 105 100 111 106 111 112 103

H

The decision rule for this problem is:

If s

Where P(F

s

K

K

s

Hence Do Not Reject H

The P-Value = P(F

Alternatively, do the upper tail test:

If s

K

and again Do Not Reject H

Spring 1999 Flex-Mode and Flex-Time 45-733

Final Exam

Keith Poole

(10 Points)

Let A

B = Cutting Taxes

P(A

P(B | A

P(B

P(A

(.45*.7)/(.3*.3 + .45*.7 + .15*.6 + .1*.5) = .315/.545 = .5780

Spring 1999 Flex-Mode and Flex-Time 45-733

Final Exam

Keith Poole

(10 Points)

æ4öæ11öæ11ö ç ÷ç ÷ç ÷ è2øè 1øè 1ø P(E) = ---------- æ52ö ç ÷ è 6ø

Spring 1999 Flex-Mode and Flex-Time 45-733

Final Exam

Keith Poole

(10 Points)

m = 40 and s

Hence: P[7000 < (å

P{[(7000 - 200*40)/(100*200

P(-.70711 < Z < 1.4142) = F(1.4142) - F(-.70711) = .9213 - .2398 = .6815

Spring 1999 Flex-Mode and Flex-Time 45-733

Final Exam

Keith Poole

(10 Points)

[1 - .957]/[1 - .867] = .043/.133 = .323