2000 Prob-Stat I Final Answers Page

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45-733 PROBABILITY AND STATISTICS I Final Examination Answers

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

(10 Points)

1. Suppose X has the continuous distribution:


                         æ e^{-(x + q)}  x > -q
                  f(x) = ç   
                         è   0 otherwise

Is f(x) a probability distribution?

YES.

f(x) ³ 0

ò _-q^¥ e^{-(x + q)}dx = -e^-qe^-x| _-q^¥ = -e^-q [0 - e^q] = 1

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

(10 Points)

2. Suppose that X₁ , X₂ , and X₃ are independently distributed and that X₁ ~ N(1, 25), X₂ ~ N(3, 6), and X₃ ~ N(-3, 10). Find P(-2X₁ + 3X₂ + 2X₃ - 5 > -3).

-2X₁ + 3X₂ + 2X₃ - 5 ~ N(-4, 194)

Hence, P[-2X₁ + 3X₂ + 2X₃ - 5 > -3] = P[(-2X₁ + 3X₂ + 2X₃ - 5 + 4)/194^1/2 > (-3 + 4)/194^1/2] =
P[Z > .0718] = 1 - F(.0718) » .4721 (Using F(.07))

Probability and Statistics Name__________________________
Spring 2000 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 (4x + 2y)/3   0 < x < 1
                  f(x,y) = ç               0 < y < 1
                           è      0 otherwise

Are X, Y Independent?

f₁(x) = (1/3)ò₀¹ (4x + 2y)dy = (1/3)[4xy + 2y²/2]|₀¹ = (1/3)(4x + 1)

f₂(y) = (1/3)ò₀¹ (4x + 2y)dx = (1/3)[(4x²)/2 + 2yx]|₀¹ = (1/3)(2 + 2y) = (2/3)(1 + y)

Clearly X and Y are not independent.

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

(10 Points)

4. We are driving through beautiful Iowa during the summer and we see two fields of corn that look very similar. We decide to test our belief that the corn in the two fields are alike so we stop beside the road, walk into the two fields, and measure the heights of a random sample of 11 stalks of corn from one field and 13 from the second field. We obtain the following measurements in feet:


3.0   3.8   2.8   3.7   3.9   3.3   3.2   3.1   3.9   4.8   3.5  and

3.2   4.4   4.1   3.9   3.8   3.3   3.4   3.6   3.7   3.4   3.0   2.9   3.9

respectively. Assume that the height in both fields 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(F_n-1,m-1 > K₂) = a/2 and P(F_n-1,m-1 < K₁) = a/2

K₂ = F_n-1,m-1,
a/2 = F_10,12,.005 = 5.09, and
K₁ = 1/F_m-1,n-1,
a/2 = 1/F_12,10,.005 = 1/5.66 = .1767

s₁² = .31473 and s₂² = .19141

Test Statistic = .31473/.19141 = 1.644 So Do Not Reject H₀:

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

SCALAR PVAL=@FDIST(1.644,10,12)
which produces a value of .2052. Multiply this by 2 to get the result:

P-Value = .4104

For the One Tail Test set K₂ = F_n-1,m-1,
a = F_10,12,.01 = 4.30, and since
Test Statistic = 1.644 we still Do Not Reject H₀:

Note that the P-Value = .2052.

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

(5 Points)

5. The proportion of people in a large population that have a particular stomach disease is known to be .002. We take a random sample of 1000 people. What is the probability that not more than 4 but more than 1 people will have the stomach disease?

l = np = 1000*.002 = 2

P(1 < X £ 4) = F(4) - F(1) = .947 - .406 = .541

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

(10 Points)

6. Two new drugs were given to patients with asthma. The first drug relieved the symptoms of 11 patients by an average of 31.6 pcv with a standard deviation of 12.36 pcv. The second drug relieved the symptoms of 13 patients by an average of 33.1 pcv with a standard deviation of 10.65 pcv. Assume that the two population variances are the same and that the measurements are normally distributed. Test the hypothesis that the two population means are the same against the alternative that they are not the same (a = .01). 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)/[s(1/n + 1/m)^1/2] > t_a/2 then Reject H₀:
    _    _
If (X_n - Y_m)/[s(1/n + 1/m)^1/2] < t_a/2 then Do Not Reject H₀:

where        

s² = [(n - 1)s₁² + (m - 1)s₂²]/(n + m - 2)

   t_a/2 = t_{.005, 22} = 2.819
   _             _
   X_n = 31.6,    Y_m = 33.1,   s₁² = 152.7696,  s₂² = 113.4225

   s² = (10*152.7696 + 12*113.4225)/22 = 131.308

Test Statistic = (31.6 - 33.1)/[131.308*(1/11 + 1/13)]^1/2 = -.3195

DO NOT REJECT

Using the EViews Command:

SCALAR PVAL=@TDIST(-.3195,22)

Produces a two-tailed P-Value of .7524

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

(10 Points)

7. Suppose the arrival of surfers at our website is a Poisson process with an average of 10 surfers every minute. What is the probability that at least 20 but not more than 34 surfers arrive in a period of 2 minutes?

l = 10*2 = 20

P(20 £ X £ 34) = F(34) - F(19) = .999 - .470 = .529

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

(10 Points)

8. You draw a random sample of 13 of our famous Pittsburgh springtime potholes and find that their depths in inches are:


5.9, 9.9, 12.5, 12.9, 6.0, 8.5, 7.1, 6.9, 7.8, 8.1, 7.9, 13.1, 10.0.

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

n = 13, df = n - 1 = 12, a = .01, a/2 = .005, t_.005,12 = 3.055

_ X_n = 8.97, s = 2.523
The Limits are:

8.97 ± (3.055*2.523)/(13)^1/2 = 8.97 ± 2.138, for the interval

(6.832, 11.108)

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

(5 Points)

9. A large shipment of vacuum tubes is delivered to our manufacturing plant. We take a random sample of 95 vacuum tubes and find that 9 are defective. Compute 95% and 99% confidence limits for p, the true proportion of defective vacuum tubes.


Confidence Limits are:
   _Ù        _Ù     _Ù
   p ± z_a/2[p(1 - p)/n]^1/2
   _Ù
   p = 9/95 = .0947,  z_.025 = 1.96, z_.005 = 2.58

  .0947 ± 1.96*[(.0947*.9053)/95]^1/2 = .0947 ± .0589 

  so the 95% limits are:  (.0358, .1536)

  .0947 ± 2.58*[(.0947*.9053)/95]^1/2 = .0947 ± .0775 

  so the 99% limits are:  (.0172, .1722)

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

(10 Points)

10. A manufacturer of sciuroid scintillators claims that her manufacturing process is very accurate and that the mean length of a certain type of sciuroid scintillator is 32cm with a standard deviation of .1cm. You take a random sample of 29 sciuroid scintillators and obtain a sample standard deviation of .15cm. Assume that the length of the sciuroid scintillators is normally distributed. Test the null hypothesis that the true variance is 0.01 against the alternative hypothesis that the true variance is not equal to 0.01. Do you accept or reject her claim? Use a = .05. Compute the P-Value of the test using EVIEWS.

H₀: s² = .01
H₁: s² ¹ .01

The decision rule for this problem is:

If (n-1)s²/ s₀² > C₂ or (n-1)s²/ s₀² < C₁ Then Reject H₀:

Where P(c²_n-1 > C₂) = a/2 and P(c²_n-1 < C₁) = a/2

s² = .0225, 28df, C₁ = 15.3079, C₂ = 44.4607

Test Statistic: (n-1)s²/ s₀² = (28*.0225)/.01 = 63

So Reject H₀:

The P-Value of the Test = 2*.0001662 = .0003324
In EViews, the command:
SCALAR PVALUE=@CHISQ(63,28)
gives us the tail area above 63 which equals .0001662.

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

(10 Points)

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


        H₀:  m = 175

                             s² = 225
        H₁:  m = 190

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


    _
If  X_n < 185 then do not reject H₀:
    _
If  X_n > 185 then reject H₀:


      _                        _
a = P(X_n > 185 | m = 175) = P[(X_n - 175)/15/6 > (185 - 175)/15/6] =
    P(Z > 4) = 1 - F(4) = .0000317
      _                        _
b = P(X_n < 185 | m = 190) = P[(X_n - 190)/15/6 < (185 - 190)/15/6] =
    P(Z < -2) = F(-2) = .0228

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

(10 Points)

12. Suppose we flip a coin 600 times. What is the probability that we observe less than 325 heads?

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

m = .5 and s² = .5²=.25

Hence: P[(å_i=1,600 X_i) < 325] =
P{[(å_i=1,600 X_i) - 600*.5]/[.5*600^1/2] < [(325 - 600*.5)/(.5*600^1/2]} =
P(Z < 2.041) » .9793 using F(2.04)

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

(10 Points)

13. Suppose the bivariate discrete distribution of X and Y is:


                         æ c(2x - y)  x = 0, 1, 2
                  f(x) = ç            y = -2, -1, 0
                         è   0 otherwise

Find c.

                               y
                          -2  -1   0
                         ------------
                       0 | 2   1   0 |  3
                         |           |
                    x  1 | 4   3   2 |  9
                         |           |
                       2 | 6   5   4 | 15
                         |           |
                         ---------------
                          12   9   6 | 27
   Hence c = 1/27

Find P(X > 0 | Y < 0).

P(X > 0 | Y < 0) = P(X > 0 Ç Y < 0)/P(Y < 0) = (18/27)/(21/27) = 18/21 = 6/7

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

(10 Points)

14. We have a computer that contains 20 sciuroid scintillators. Suppose that the probability that any given sciuroid scintillator will fail is .20 and the sciuroid scintillators fail independently. Given that at least 6 sciuroid scintillators have failed, what is the probability that at least 10 sciuroid scintillators have failed?

P[X ³ 10 | X ³ 6] = P[X ³ 10 Ç X ³ 6]/P[X ³ 6] = P[X ³ 10]/P[X ³ 6] =
[1 - F(9)]/[1 - F(5)] = [1 - .997]/[1 - .804] = .003/.196 = 3/196 = .0153

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

(10 Points)

15. Suppose X has the continuous probability distribution:


                         æ qe^-qx  x > 0
                  f(x) = ç       q > 0                        
                         è   0 otherwise

Find the maximum likelihood estimator for q.

f(x₁ , x₂ , ... , x_n | q) = P_i=1,n q e^{- qX_i} = qⁿ e^{- q
S_i=1,nX_i}

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

¶L/ ¶q = n/q - å_i=1,nx_i = 0

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

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

(10 Points)

16. Suppose we randomly select 7 cards from a standard deck of 52 playing cards. What is the probability that we get 4 cards from one denomination and 1 card each from three other denominations?

æ13öæ 4öæ 4öæ 4öæ 4öæ 4ö ç ÷ç ÷ç ÷ç ÷ç ÷ç ÷ è 4øè 1øè 4øè 1øè 1øè 1ø --------------------- æ52ö ç ÷ è 7ø