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45-733 PROBABILITY AND STATISTICS I Practice Final


Probability and Statistics                      Name__________________________
Spring 2000 Flex-Mode and Flex-Time 45-733
Practice Final
Keith Poole

(10 Points)

1. A random sample of 500 probable voters in California found that 275 preferred McCain to Bush. Use this sample to perform the hypothesis test

H0: p = .50
H1: p ¹ .50

 with a = .05. Compute the P-Value of the test using EVIEWS.

The decision rule for this problem is:

       Ù    
If    (p - p0)/[p0(1 - p0)/n]1/2 > za/2 or 
       Ù      
      (p - p0)/[p0(1 - p0)/n]1/2 < -za/2 then Reject H0:    
               Ù      
If     -za/2 < (p - p0)/[p0(1 - p0)/n]1/2 < za/2 then Do Not Reject H0:    
                   Ù    
    -z.025 = 1.96,  p = 275/500 = .550    

Test Statistic = (.550 - .5)/[(.5*.5)/500]1/2 = 2.236 > 1.96
Hence, Reject H0:

In EVIEWS, the command:
SCALAR PVALUE=@CNORM(-2.236) (Use the negative to get the lower tail)
yields the value .012676
The P-Value of the Test = 2*.012676 = .025352

Probability and Statistics                      Name__________________________
Spring 2000 Flex-Mode and Flex-Time 45-733
Practice Final
Keith Poole

(10 Points)

2. We have two urns. In the first urn there are 10 white, 10 red, and 10 blue balls. In the second urn there are 5 white, 15 red, and 10 blue balls. We randomly draw one ball from each urn. What is the probability that neither ball is red?



(20/30)*(15/30) = 1/3

Probability and Statistics                      Name__________________________
Spring 2000 Flex-Mode and Flex-Time 45-733
Practice Final
Keith Poole

(10 Points)

3. Suppose we have the discrete bivariate probability distribution: 

                           æ c(x2 + y2)  x = 0, 1, 2
                  f(x,y) = ç              y = -2, -1
                           è   0 otherwise
  1. Find c. 

    
                             y  
                          -2  -1
                         ---------
                       0 | 4   1 |  5
                         |       |
                    x  1 | 5   2 |  7
                         |       |
                       2 | 6   3 |  9
                         |       |
                         -----------
                          15   6 | 21
   Hence c = 1/21
  2. Find g2(y | x = 1). 
    
                                  æ  5/7  y = -2
                                  ç
     g2(y | x=1) = f(1,y)/f1(1) = ç  2/7  y = -1
                                  ç
                                  è   0 otherwise




Probability and Statistics                      Name__________________________
Spring 2000 Flex-Mode and Flex-Time 45-733
Practice Final
Keith Poole

(10 Points)

4. Let X1, X2, ..., Xn be a random sample from a distribution with mean m and variance s2. What is the mean and variance of [(3X1)/2 + X2/2]?
E[(3X1)/2 + X2/2] = (3m)/2 + m/2 = 2m

VAR[(3X1)/2 + X2/2] = 9s2/4 + s2/4 = (10s2)/4

Probability and Statistics                      Name__________________________
Spring 2000 Flex-Mode and Flex-Time 45-733
Practice Final
Keith Poole

(10 Points)

5. We wish to estimate the average life time of some custom lightbulbs in hours that have been shipped to our factory. Suppose the population standard deviation is known to be 30 hours. We take a random sample of 144 lightbulbs. What is the probability that the absolute difference between the sample mean and the true mean will not exceed 5 hours?


     _                     _
  P[|Xn - m| £ 5] = P[-5 £ Xn - m £ 5] = 
                   _
  P[-5/(30/12) £ (Xn - m)/(30/12) £ 5/(30/12)] = 

  P[-2 £ Z £ 2] = F(2) - F(-2) = .9544


Probability and Statistics                      Name__________________________
Spring 2000 Flex-Mode and Flex-Time 45-733
Practice Final
Keith Poole

(10 Points)

6. Suppose X, Y have the continuous bivariate distribution: 

                           æ  2xy   0 < x < 1
                  f(x,y) = ç        0 < y < 1
                           è   0 otherwise
Find E(XY).

E(XY) = ò01 ò01 xy(2xy)dxdy = ò01 ò01 (2x2y2)dxdy =
ò01 {[2(x3/3]y2}|01 dy = (2/3) ò01 [y2]dy = (2/3)[y3/3]|01 = 2/9

Probability and Statistics                      Name__________________________
Spring 2000 Flex-Mode and Flex-Time 45-733
Practice Final
Keith Poole

(10 Points)

7. Suppose we read in Robber Barons that a public opinion poll shows that 60 percent of 60 randomly sampled GSIA MBA students think that Karl Freidrich Gauss is the lead singer in a heavy metal band. Compute 99 percent confidence limits for the true proportion of GSIA MBA students who think that Karl Friedrich Gauss is the lead singer of a heavy metal band. 

Confidence Limits are:
   Ù        Ù     Ù
   p ± za/2[p(1 - p)/n]1/2
   Ù
   p = .60,  z.005 = 2.58

  .60 ± 2.58*[(.60*.40)/60]1/2 = .60 ± .06325 

  so the 99% limits are:  (.53675, .66325)


Probability and Statistics                      Name__________________________
Spring 2000 Flex-Mode and Flex-Time 45-733
Practice Final
Keith Poole

(10 Points)

8. Suppose that 35% of Pennsylvanians are Republicans, 45% are Democrats, and 20% are independents. A recent public opinion poll shows that 40% of Republicans, 20% of Democrats, and 90% of Independents plan on voting for John McCain. If a person is selected at random from the Pennsylvania population and it is found that she is not planning to vote for McCain, what is the probability that she is a Democrat?

Let A1 = Republican, A2 = Democrat, A3 = Independent and
B = Vote For McCain

P(A1) = .35, P(A2) = .45, P(A3) = .20
P(B | A1) = .4, P(B | A2) = .2, P(B | A3) = .9
P(Bc | A1) = .6, P(Bc | A2) = .8, P(Bc | A3) = .1

P(A2|Bc) = [P(A2)P(Bc|A2)]/ [åj=1,3 P(Aj)P(Bc|Aj)] =
(.45*.8)/(.35*.6 + .45*.8 + .2*.1) = .36/.59 = .61

Probability and Statistics                      Name__________________________
Spring 2000 Flex-Mode and Flex-Time 45-733
Practice Final
Keith Poole

(10 Points)

9. We take a random sample of potholes in Pittsburgh city streets and Cleveland city streets and measure the depth of each pothole. The measurements for Pittsburgh in centimeters are: 
          115  101  119  151  162   99  159  138  188
and the measurements for Cleveland in centimeters are: 
          135  131  109  167  143   89  199
Assume that the depth of potholes in both cities is normally distributed. Test the null hypothesis that the variances of the two populations are the same against the alternative hypothesis that the variances are not the same (a = .01 ). Compute the P-Value of the test using EVIEWS.
H0: s12 = s22
H1: s12 ¹ s22

The decision rule for this problem is:

If s12/s22 > K2 or s12/s22 < K1 Then Reject H0:

Where P(Fn-1,m-1 > K2) = a/2 and P(Fn-1,m-1 < K1) = a/2

The Cleveland sample variance is the largest so we make it population 1:

s12 = 1313.33 and s22 = 931.86

K2 = Fn-1,m-1, a/2 = F6,8,.005 = 7.95, and
K1 = 1/Fm-1,n-1, a/2 = 1/F8,6,.005 = 1/10.57 = .095

Test Statistic = 1313.33/931.861 = 1.409 So Do Not Reject H0:

To get the P-Value, use the EViews command:

SCALAR PVAL=@FDIST(1.409,6,8)
which produces a value of .3183. Multiply this by 2 to get the result:

P-Value = .6366

For the One Tail Test set K2 = Fn-1,m-1, a = F8,7,.01 = 6.37, and since
Test Statistic = 1.409 we still Do Not Reject H0:

Note that the P-Value = .3183. This is why that I said in class that I prefer this version of the F-Test. But I will accept either version on the final.

Probability and Statistics                      Name__________________________
Spring 2000 Flex-Mode and Flex-Time 45-733
Practice Final
Keith Poole

(10 Points)

10. Suppose the times that it takes for motorists to get through the toll booth area at the Monroeville exit to the Pennsylvania turnpike are independent random variables with a mean of 45 seconds and a standard deviation of 90 seconds. Approximate the probability that 100 motorists can get through the toll booth area in 80 minutes.

P{[(åi=1,n Xi) - nm]/ sn1/2 £ k} » F(k)

m = 45 and s2 = 902

Hence: P[(åi=1,100 Xi) < 4800] =
P{[(åi=1,100 Xi) - 100*45]/[90*1001/2] < [(4800 - 100*45)/(90*1001/2]} =
P(Z < .33) = F(.33) = .6373

Probability and Statistics                      Name__________________________
Spring 2000 Flex-Mode and Flex-Time 45-733
Practice Final
Keith Poole

(5 Points)

11. The proportion of people having a certain type of cancer in a large population is known to be .008. We take a random sample of 1000 people from this population. What is the probability that at most 10 and at least 3 people have the cancer.

l = np = 1000*.008 = 8

P(3 £ X £ 10) = F(10) - F(2) = .816 - .014 = .802

Probability and Statistics                      Name__________________________
Spring 2000 Flex-Mode and Flex-Time 45-733
Practice Final
Keith Poole

(10 Points)

12. We draw a random sample of size 41 from a normal distribution and we find that s2 = 943.8. Compute 95 and 99 percent confidence limits for s2.

The Confidence Interval is:

P[(n-1)s2/C2 < s2 < (n-1)s2/C1] = 1 - a

95% Confidence Limits:

C1 = 24.4331 and C2 = 59.3417
(n-1)s2/C2 = (40*943.8)/59.3417 = 636.180, and
(n-1)s2/C1 = (40*943.8)/24.4331 = 1545.117

99% Confidence Limits:

C1 = 20.7065 and C2 = 66.7659
(n-1)s2/C2 = (60*943.8)/66.7659 = 848.158, and
(n-1)s2/C1 = (60*943.8)/20.7065 = 2734.793



Probability and Statistics                      Name__________________________
Spring 2000 Flex-Mode and Flex-Time 45-733
Practice Final
Keith Poole

(10 Points)

13. We have a spacecraft with 15 identical computers each of which operates independently of the others. The probability that any given computer will fail during a long space flight is .2. We take our spacecraft out for a long cruise to Venus. At some point in the flight at least 4 computers have failed. What is the probability that before the flight is over that at least 8 computers will have failed?

P[X ³ 8 | X ³ 4] = P[X ³ 8]/ P[X ³ 4] = [1- F(7)]/[1 - F(3)] =
[1 - .996]/[1 - .836] = .004/.164 = .024

Probability and Statistics                      Name__________________________
Spring 2000 Flex-Mode and Flex-Time 45-733
Practice Final
Keith Poole

(10 Points)

14. Suppose we have the continuous probability distribution: 

                           æ  (q + 4)x(q + 3)  0 < x < 1
                    f(x) = ç   
                           è   0 otherwise
Find the maximum likelihood estimator for q.

f(x1 , x2 , ... , xn | q) = Pi=1,n (q + 4)Xi (q + 3) = (q + 4)n Pi=1,n Xi(q + 3)

L(x1 , x2 , ... , xn | q) = ln{f(x1 , x2 , ... , xn | q)} = nln(q + 4) + (q + 3) (Si=1,nlnXi)

L/ q = n/(q + 4) + åi=1,nlnxi = 0

 
         Ù                           
Hence,   q = -n/(åi=1,nlnxi) - 4