Prob-Stat I 1998 Practice Final Answers

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Probability and Statistics Name__________________________
Spring 1998 Flex-Mode and Flex-Time 45-733
Practice Final
Keith Poole

(5 Points)

1. The proportion of people that have a certain type of cancer in a large population is known to be .002. If we take a random sample of 2000 people, what is the probability that exactly 5 people have the type of cancer.

Poisson approximation: l = np = 2000*.002 = 4
P(X = 5) = F(5) - F(4) = .785 - .629 = .156

Probability and Statistics Name__________________________
Spring 1998 Flex-Mode and Flex-Time 45-733
Practice Final
Keith Poole

(10 Points)

2. Suppose we have the continuous probability distribution:


                        { (b⁴/6)x³e^-bx
                 f(x) = {             x > 0                                                        
                        { 0 otherwise

Find the maximum likelihood estimator for b.

L = P_i=1,n (b⁴/6)x_i³ e^-bx_i
lnL = 4nlnb - nln6 + 3å_i=1,nlnx_i - b å_i=1,nx_i
¶lnL/¶ b = 4n/b - å_i=1,nx_i
Hence

       _^     _
       b = 4/X_n

Probability and Statistics Name__________________________
Spring 1998 Flex-Mode and Flex-Time 45-733
Practice Final
Keith Poole

(10 Points)

3. Assume that X₁, X₂, and X₃ are random variables, and that:


      E(X₁) = 3   VAR(X₁) = 6    COV(X₁, X₂) = -2
      E(X₂) = 4   VAR(X₂) = 8    COV(X₁, X₃) =  0
      E(X₃) = 5   VAR(X₃) = 8    COV(X₂, X₃) =  1

Find E(2X₁ - 5X₂ + X₃ - 10) and VAR(2X₁ - 5X₂ + X₃ - 10).

E(2X₁ - 5X₂ + X₃ - 10) = 2E(X₁) - 5E(X₂) + E(X₃) - 10 = 2*3 - 5*4 + 5 - 10 = -19
VAR(2X₁ - 5X₂ + X₃ - 10) = 4VAR(X₁) + 25VAR(X₂) + VAR(X₃) - 20COV(X₁, X₂) + 4COV(X₁, X₃) - 10COV(X₂, X₃) =
4*6 + 25*8 + 1*8 - 20*(-2) + 4*0 - 10*1 = 262

Probability and Statistics Name__________________________
Spring 1998 Flex-Mode and Flex-Time 45-733
Practice Final
Keith Poole

(10 Points)

4. We wish to estimate the average height of women first year GSIA MBA students. Suppose the population standard deviation is known to be 3.5 inches. We take a random sample of 36 women. What is the probability that the difference between the sample mean and the true mean will not exceed .4 inches.


                   _
We Want    P(|X_n - m| £ .4) =
           P(-.4 £ X_n - m £ .4) = 
           P[(-.4)/(3.5/6) £ (X_n - m)/(3.5/6) £ (.4)/(3.5/6)] = 
           P(-.69 £ Z £ .69) =
           F(.69) - F(-.69) = 
           1 - F(-.69) - F(-.69) = 
           1 - .2451 - .2451 = .5098

Probability and Statistics Name__________________________
Spring 1998 Flex-Mode and Flex-Time 45-733
Practice Final
Keith Poole

(10 Points)

5. The checkout times for customers at the Squirrel Hill Giant Eagle are independent random variables with a mean of 3.5 minutes and a standard deviation of 1.5 minutes. What is the approximate probability that it will take more than 3 hours to process the orders of 50 people.

E(X_i) = 3.5, VAR(X_i) = 1.5²
Hence, P(å_i=1,50X_i > 180) = P[(å_i=1,50X_i - 3.5*50)/ (50^1/2*1.5) > (180 - 175)/10.607] =
P(Z > .47) = 1 - F(.47) = F(-.47) = .3192

Probability and Statistics Name__________________________
Spring 1998 Flex-Mode and Flex-Time 45-733
Practice Final
Keith Poole

(10 Points)

6. Suppose we have the bivariate discrete probability distribution:


                           { c(3x + y²)
                           {         x = 1,2,3
                  f(x,y) = {         y = 1,2,3
                           {
                           { 0 otherwise

Find c

                               y
                           1   2   3
                         ------------
                       1 | 4   7  12 | 23
                         |           |
                    x  2 | 7  10  15 | 32
                         |           |
                       3 |10  13  18 | 41
                         |           |
                         ---------------
                          21  30  45 | 96
   Hence c = 1/96

Find r(X, Y)

E(X) = (1/96)*(1*23 + 2*32 + 3*41) = 210/96
E(Y) = (1/96)*(1*23 + 2*30 + 3*45) = 216/96
E(X²) = (1/96)*(1*23 + 4*32 + 9*41) = 520/96
E(Y²) = (1/96)*(1*23 + 4*32 + 9*41) = 546/96
E(XY) = (1/96)(1*1*4 + 1*2*7 + 1*3*12 + 2*1*7 + 2*2*10 + 2*3*15 + 3*1*10 + 3*2*13 + 3*3*18)= 468/96
```
r(X, Y) = [(468/96) - (210/96)(216/96)]/{[(520/96) - (210/96)²]
                                         [(546/96) - (216/96)²]}^1/2
```

Probability and Statistics Name__________________________
Spring 1998 Flex-Mode and Flex-Time 45-733
Practice Final
Keith Poole

(10 Points)

7. Suppose we read in Robber Barrons that 85 percent of 200 first year Flex-Time and Flex-Mode students would rather have 10 root canals than retake 45-733. Compute 95 percent confidence limits for the true proportion of first years who would rather have the root canals.

           _^
           p = .85,  n = 200,  a = .05,  Z_.025 = 1.96

Hence, .85 ± 1.96[.85*.15/200]^1/2 = .85 ± .0495
or (.8005, .8995)

Probability and Statistics Name__________________________
Spring 1998 Flex-Mode and Flex-Time 45-733
Practice Final
Keith Poole

(10 Points)

8. Two new drugs were given to patients with ulcers. The first drug lowered the acid concentration of 15 patients by an average of 25 points with a standard deviation of 8 points. The second drug lowered the acid concentration of 14 patients by 19 points with a standard deviation of 7 points. Determine 95 percent confidence limits for the difference in the mean reductions in acid concentration. Assume the measurements are normally distributed with equal variances.

          _
          X_n = 25   s_x = 8    n = 15
          _                            a = .05,  t_{.025, 27} = 2.052
          Y_m = 19   s_y = 7    m = 14

          s² = (14*64 + 13*49)/(15 + 14 - 2) = 56.778
          s = 7.535

Hence, 25 - 19 ± 2.052*7.535*(1/15 + 1/14)^1/2 = 6 ± 5.746
Hence, (.254, 11.746)

Probability and Statistics Name__________________________
Spring 1998 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 measure the depth of each pothole. The measurements in centimeters are:

          135  101   99   201  121   89  199  148   91

Assume that the depth of Pittsburgh potholes is normally distributed. Find 99 percent confidence limits for the true depth of the population of Pittsburgh potholes.

           _
           X_n = 131.556,  s = 43.609,  a = .01,  t_{.005, 8} = 3.355

Hence, 131.566 ± 3.355*(43.609/3) = 131.566 ± 48.796
Or (82.77, 180.362)

Probability and Statistics Name__________________________
Spring 1998 Flex-Mode and Flex-Time 45-733
Practice Final
Keith Poole

(10 Points)

10. Suppose the time that it takes to complete a Prob-Stat Final Exam is normally distributed. We draw a random sample of 14 first year Flex-Time/Flex-Mode students and find that their exam times (in minutes) are:

   155  145  100   98   76  178  159  161  102  132  144  146  179  140

Construct 99 percent confidence limits for population variance of the exam times.

s² = 991.720, n = 14, df = 13, a = .01, a/2 = .005
Hence, C₁ = 3.56503 and C₂ = 29.8194

and the limits are: (13*991.720/29.8194 , 13*991.720/3.56503) = (432.348 , 3616.340)

Probability and Statistics Name__________________________
Spring 1998 Flex-Mode and Flex-Time 45-733
Practice Final
Keith Poole

(10 Points)

11. Suppose we are drawing a random sample of 36 from a normal distribution. Given the hypothesis test:


              H₀: m = 175
                           s² = 2500
              H₁: m = 195

If a = .01, calculate the probability of a Type II error, b.

           _
     a = P(X_n > m₀ + c | m = m₀)
            _
       = P[(X_n - 175)/(50/6) > (175 + c - 175)/(50/6)] = 
       = P(Z > 6c/50) = 1 - F(6c/50) = .01

Now, since Z_.01 = 2.33
Hence, 2.33 = 6c/50 and c = 19.417

           _
     b = P(X_n < m₀ + c | m = m₁)
            _
       = P[(X_n - 195)/(50/6) < (175 + 19.417 - 195)/(50/6)] = 
       = P(Z < -.07) = F(-.07) = .4721

Probability and Statistics Name__________________________
Spring 1998 Flex-Mode and Flex-Time 45-733
Practice Final
Keith Poole

(10 Points)

12. We draw a random sample of size 30 from a normal distribution and we find that s² = 143.5. Compute 95 and 99 percent confidence limits for s².

For a = .05, a/2 = .025, 29df, hence:
C₁ = 16.0471 and C₂ = 45.7222
(n - 1)s²/C₂ = (29*143.5)/45.7222 = 91.017 and (n - 1)s²/C₁ = (29*143.5)/16.0471 = 259.33
so that the limits are (91.017 , 259.33)

For a = .01, a/2 = .005, 29df, hence:
C₁ = 13.1211 and C₂ = 52.3356
(n - 1)s²/C₂ = (29*143.5)/52.3356 = 79.516 and (n - 1)s²/C₁ = (29*143.5)/13.1211 = 317.161
so that the limits are (79.516 , 317.161)

Probability and Statistics Name__________________________
Spring 1998 Flex-Mode and Flex-Time 45-733
Practice Final
Keith Poole

(10 Points)

13. We give identical Prob-Stat I final exams to first year and second year MSIA students. We draw a random sample of 9 first years and find that their exam scores are:

        111  121  131   99  123  125  102  130  119

We draw a random sample of 8 second years and find that their exam scores are:

        127   90   88  132  130  101   97  131

Assume that the exam scores are normally distributed in both populations. 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 = .05).

First years: s_x² = 132.861, df = 8; Second years: s_y² = 388.0, df = 7
a/2 = .025, F_.025,8,7 = 4.90, 1/F_.025,7,8 = 1/4.53 = .221

Test statistic: s_x²/s_y² = 132.861/388.0 = .342

Since .221 < .342 < 4.90, Do Not Reject H₀:
****************************
or, Test statistic: s_y²/s_x² = 388.0/132.861 = 2.920

Since 1/4.90 = .204 < 2.920 < 4.53, Do Not Reject H₀:

Probability and Statistics Name__________________________
Spring 1998 Flex-Mode and Flex-Time 45-733
Practice Final
Keith Poole

(10 Points)

14. We wish to draw a random sample from a normal distribution with mean m and variance 36. What is the minimum sample size necessary such that:


                _
             P[|X_n - m| £ .5] ³ .99


           _
   P[-.5 £ X_n - m £ .5] = 
                    _
   P[-.5/(6/n^1/2) £ (X_n - m)/(6/n^1/2) £ .5/(6/n^1/2)] = 
   P[-.5/(6/n^1/2) £ Z £ .5/(6/n^1/2)] = F(.5/(6/n^1/2) - F(-.5/(6/n^1/2) = 
   2F(.5/(6/n^1/2) - 1 ³ .99
   F(.5/(6/n^1/2) ³ .995
   Now, since F(2.58) = .9951, therefore, 
   .5/(6/n^1/2) ³ 2.58, hence
   n ³ [(2.58*6)/.5]² = 958.52
   and n = 959