1999 Prob-Stat I Final Answers Page

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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)

1. Suppose X has the discrete probability distribution:


                         æ q(1 - q)^x  x = 0,1,2, ...
                  f(x) = ç   
                         è   0 otherwise

Find the maximum likelihood estimator for q.

f(x₁ , x₂ , ... , x_n | q) = P_i=1,n q (1 - q)^X_i = qⁿ (1 - q) ^S_i=1,nX_i

L(x₁ , x₂ , ... , x_n | q) = ln{f(x₁ , x₂ , ... , x_n | q)} = nlnq + (S_i=1,nX_i) ln(1 - q)

¶L/ ¶q = n/q - å_i=1,nx_i/ (1 - q) = 0

         _Ù                            _
Hence,   q = n/(n + å_i=1,nx_i) = 1/(1 + X_n)

Probability and Statistics Name__________________________
Spring 1999 Flex-Mode and Flex-Time 45-733
Final Exam
Keith Poole

(10 Points)

2. Independent random samples are taken from the output of two machines on a production line to check the weight of the items being manufactured. From the first machine a sample of 36 is taken with a mean weight of 120 grams and an s² of 4. From the second machine a sample of 64 is taken with a mean weight of 130 grams and an s² of 5. Assuming that the weights for both machines are normally distributed and the machines operate independently, find 99 percent confidence limits for the difference in the true means.

The Confidence Interval is:
  _    _                                    _   _
P{X_n - Y_m - z_a/2[s_x²/n + s_y²/m]^1/2 < m_x - m_y < X_n - Y_m + z_a/2[s_x²/n + s_y²/m]^1/2} 
= 1 - a

Our confidence limits are:  

120 - 130 ± 2.33[4/36 + 5/64]^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)

3. Suppose the bivariate continuous distribution of X and Y is:


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

Find P(X > 1 | Y < 1).

P(X > 1 | Y < 1) = P(X > 1 Ç Y < 1)/P(Y < 1)

P(X > 1 Ç Y < 1) = (1/8)ò₁² ò₀¹(x + y)dxdy = (1/8)ò₀¹ [x²/2 + xy]|₁²dy =
(1/8)ò₀¹ [3/2 + y]dy = (1/8)[3y/2 + y²/2]|₀¹ = 1/4

P(Y < 1) = (1/8)ò₀² ò₀¹(x + y)dxdy = (1/8)ò₀¹ [x²/2 + xy]|₀²dy =
(1/8)ò₀¹ [2 + 2y]dy = (1/8)[2y + y²]|₀¹ = 3/8

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

Probability and Statistics Name__________________________
Spring 1999 Flex-Mode and Flex-Time 45-733
Final Exam
Keith Poole

(10 Points)

4. The same examination is given to two classes of students in statistics. The first class of 51 students has a mean score of 75 with an s² of 100. The second class of 41 students has a mean score of 80 with an s² of 121. Test the hypothesis that the true mean scores are the same against the hypothesis that the true mean scores are not the same (a = .05). Compute the P-Value of the test using EViews.

H₀: m₁ - m₂ = 0
H₁: m₁ - m₂ ¹ 0

The decision rule for this problem is:
    _    _
If (X_n - Y_m)/(s₁²/n + s₂²/m)^1/2 > z_a/2 or 
    _    _
   (X_n - Y_m)/(s₁²/n + s₂²/m)^1/2 < -z_a/2 then Reject H_o:
           _    _
If -z_a/2 < (X_n - Y_m)/(s₁²/n + s₂²/m)^1/2 < z_a/2 then Do Not Reject H_o:

   z_a/2 = 1.96
    _    _
   (X_n - Y_m)/(s₁²/n + s₂²/m)^1/2 = (75 - 80)/[100/51 + 121/41]^1/2 = -2.256

-2.256 < -1.96 so Reject H₀:

P-Value = .0265 using TDIST(-2.256,90); or
.0241 using CNORM(-2.256)*2

Probability and Statistics Name__________________________
Spring 1999 Flex-Mode and Flex-Time 45-733
Final Exam
Keith Poole

(10 Points)

5. You are curious about whether or not the claims of a soft-drink maker that their plastic bottles always contain 16oz of soda pop are true. You randomly open 11 bottles and obtain the following measurements (in ounces):

15.2 15.6 15.9 15.5 15.9 15.8 16.0 15.9 16.0 15.6 15.7

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.

H₀: m = 16
H₁: m < 16

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 H₀: P-Value = .00019

Probability and Statistics Name__________________________
Spring 1999 Flex-Mode and Flex-Time 45-733
Final Exam
Keith Poole

(10 Points)

6. Suppose we know that exactly one of the following hypotheses must be true:


        H₀:  m = 115

                             s² = 400
        H₁:  m = 120

The population sampled has a normal distribution and the sample size is 36. Find a and b for the decision rule:


    _
If  X_n < 118.5 then do not reject H₀:
    _
If  X_n > 118.5 then reject H₀:


      _                          _
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

Probability and Statistics Name__________________________
Spring 1999 Flex-Mode and Flex-Time 45-733
Final Exam
Keith Poole

(10 Points)

7. Suppose that X₁ , X₂ , and X₃ are independently distributed and that X₁ ~ N(4, 9), X₂ ~ N(3, 10), and X₃ ~ N(-2, 9). Find P(3X₁ – 5X₂ + X₃ - 5 > -7).

Let W = 3X₁ – 5X₂ + X₃ - 5

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)/340^1/2 > (-7 + 10)/340^1/2] =
P(Z > .16) = 1 - F(.16) = .4364

Probability and Statistics Name__________________________
Spring 1999 Flex-Mode and Flex-Time 45-733
Final Exam
Keith Poole

(10 Points)

8. We conduct a survey of Flex-Time and Flex-Mode students who survived Prob-Stat I in 1997 and 1998. For 1997 61 of 82 randomly chosen students said they would rather have 5 root canals than re-take Prob-Stat I. For 1998 73 of 90 randomly chosen students said they would rather have 5 root canals than re-take Prob-Stat I. Do these results indicate a significant difference in the proportions of students who would rather have the root canals between the two years (a = .01). Compute the P-Value of the test using Eviews.

H₀: p₁ - p₂ = 0
H₁: p₁ - p₂ ¹ 0

                       _Ù    _Ù
                       p₁ - p₂
     Test:      -------------------------   > z_a/2  or  < - z_a/2     Reject H₀:
                _Ù      _Ù        _Ù      _Ù               
               [p₁(1 - p₁)/n₁ + p₂(1 - p₂)/n₂]^1/2

  z_.005 = 2.58
  _Ù                    _Ù
  p₁ = 61/82 = .744,   p₂ = 73/90 = .811
  Test Statistic = (.744 - .811)/[(.744*.256)/82 + (.811*.189)/90]^1/2 = -1.056

Hence, Do Not Reject H₀:

P-Value = .291

Probability and Statistics Name__________________________
Spring 1999 Flex-Mode and Flex-Time 45-733
Final Exam
Keith Poole

(5 Points)

9. Suppose that the proportion of defective SIMMs in a large shipment to our computer manufacturing plant is known to be .009. We take a random sample of 1000 SIMMs. What is the probability that at most 12 are defective but not less than 3 are defective.

l = np = 1000*.009 = 9

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

Probability and Statistics Name__________________________
Spring 1999 Flex-Mode and Flex-Time 45-733
Final Exam
Keith Poole

(10 Points)

10. A large shipment of roller bearings is delivered to your machine shop. You take a random sample of 80 roller bearings and find that 4 are defective. Compute 95 and 99 percent confidence limits for the true proportion of defective roller bearings.

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.

Probability and Statistics Name__________________________
Spring 1999 Flex-Mode and Flex-Time 45-733
Final Exam
Keith Poole

(10 Points)

11. You have a very large greenhouse and you grow a very large number of orchids. You are curious about the variability of the heights of your orchids so you take a random sample and obtain the following heights (in inches):


  12.1   12.3   12.8   12.0   11.9   12.5   13.1   12.9   12.7   12.4   12.5

Assume that orchid height is normally distributed. Construct 98 and 99% confidence limits for the true variance.

s² = .14618

Confidence Limits are:

(n-1)s²/c₂ and
(n-1)s²/c₁

98% Limits:
c₁ = 2.55821
c₂ = 23.2093
(10*.14618)/23.2093 = .0630
and (10*.14618)/2.55821 = .5714

99% Limits:
c₁ = 2.15585
c₂ = 25.1882
(10*.14618)/25.1882 = .0580
and (10*.14618)/2.15585 = .6781

Probability and Statistics Name__________________________
Spring 1999 Flex-Mode and Flex-Time 45-733
Final Exam
Keith Poole

(10 Points)

12. Pitt and GSIA students independently take an identical Prob-Stat I final examination. Assume that the scores in both populations are normally distributed. We draw a random sample 7 GSIA students and find their scores to be:


    116   119   125   121   120   118   110

and we draw a random sample of 10 Pitt students and find their scores to be:


    110   111   109   105   100   111   106   111   112   103

Test the null hypothesis that the variances of the two populations are the same against the alternative hypothesis that variances are not the same (a = .01). Compute the P-Value of the test using Eviews.

H_o: s₁² = s₂²
H₁: s₁² ¹ s₂²

The decision rule for this problem is:

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

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

s₁²/s₂² ~ F_{n-1,m-1 df}

K₂ = F_6,9,.005= 7.13
K₁ = 1/F_9,6,.005= 1/10.39 = .09625
s₁²/s₂² = 21.619/16.622 = 1.3006

Hence Do Not Reject H₀:
The P-Value = P(F_6,9 > 1.3006) = .34682

Alternatively, do the upper tail test:

If s₁²/s₂² > K₂ Then Reject H_o:

K₂ = F_6,9,.01= 5.80

and again Do Not Reject H₀:

Probability and Statistics Name__________________________
Spring 1999 Flex-Mode and Flex-Time 45-733
Final Exam
Keith Poole

(10 Points)

13. In a certain large city here in the U.S. 30% of the people are Republicans, 45% are Democrats, 15% Reform, and 10% are independents. A public opinion poll is taken and it is found that 70% of Republicans, 30% of Democrats, 40% of Reformers, and 50% of Independents favor cutting taxes. If a person is selected at random from the population and it is found that she does not favor cutting taxes, what is the probability that she is a Democrat.

Let A₁ = Republican, A₂ = Democrat, A₃ = Reform, A₄ = Independent and
B = Cutting Taxes

P(A₁) = .30, P(A₂) = .45, P(A₃) = .15, P(A₄) = .10
P(B | A₁) = .7, P(B | A₂) = .3, P(B | A₃) = .4, P(B | A₄) = .5
P(B^c | A₁) = .3, P(B^c | A₂) = .7, P(B^c | A₃) = .6, P(B^c | A₄) = .5

P(A₂|B^c) = [P(A₂)P(B^c|A₂)]/ [å_j=1,4 P(A_j)P(B^c|A_j)] =
(.45*.7)/(.3*.3 + .45*.7 + .15*.6 + .1*.5) = .315/.545 = .5780

Probability and Statistics Name__________________________
Spring 1999 Flex-Mode and Flex-Time 45-733
Final Exam
Keith Poole

(10 Points)

14. Suppose we draw 6 cards randomly from a standard deck of 52 playing cards. What is the probability that we get 3 cards from one suit in consecutive order and 3 cards from a second suit in consecutive order (e.g., 3§ , 4§ , 5§ , 10© , J© , Q© ; or Q¨ , K¨ , A¨ , 7ª , 8ª , 9ª ; etc.).

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

Probability and Statistics Name__________________________
Spring 1999 Flex-Mode and Flex-Time 45-733
Final Exam
Keith Poole

(10 Points)

15. On any given day 200 customers visit our computer store in the mall. The amount spent by each customer is a random variable with a mean of $40 and a standard deviation of $100, and it is independent across customers. What is the probability that our store’s revenue on any given day will be between $7000 and $10,000.

P{[(å_i=1,n X_i) - nm]/ sn^1/2 £ k} » F(k)

m = 40 and s² = 100²

Hence: P[7000 < (å_i=1,200 X_i) < 10000] =
P{[(7000 - 200*40)/(100*200^1/2)] < [(å_i=1,200 X_i) - 200*40]/[100*200^1/2] < [(10000 - 200*40)/(100*200^1/2]} =
P(-.70711 < Z < 1.4142) = F(1.4142) - F(-.70711) = .9213 - .2398 = .6815

Probability and Statistics Name__________________________
Spring 1999 Flex-Mode and Flex-Time 45-733
Final Exam
Keith Poole

(10 Points)

16. We have a hydraulic system that has 20 components. Suppose that the probability that any component will fail is .10 and the components fail independently. Given that at least 4 components have failed, what is the probability that at least 5 components have failed.

P[X ³ 5 | X ³ 4] = P[X ³ 5]/ P[X ³ 4] = [1- F(4)]/[1 - F(3)] =
[1 - .957]/[1 - .867] = .043/.133 = .323