Answers for Winter 10 Midterm

Part A.
x = bad question.

 1. a    2. d    3. d    4. x    5. d     6. at least one outlier
                                             but no extreme outliers

 7. b    8. b    9. d   10. a   11. a    12. d

13. 32  14. b   15. d

Part B.
a.
    30 +                           +----+
       |                           |    |
       |                 +---------+    |
       |                 |         |    |
    20 +                 |         |    |
       |                 |         |    |
       |                 |         |    |
       |                 |         |    |
    10 +                 |         |    |
       |  +--------------+         |    |
       |  |              |         |    |    
       |  |              |         |    |
     0 +  +----+----+----+----+----+----+
          0    1    2    3    4    5    6

b. Q0=0.0  Q1=3.2  Q3=4.2  Q3=5.1667  IQR=Q3-Q1=5.167-3.2=1.967   
c. 3.95    d. 75%

1. Regression Problem

a.  y = (95-75)/15 = 1.33  
    Look up -1.33 in the standard normal table: 0.0918 = 9%

b. y = 0.45 x + 43.5

c. y^ = 0.45 * 90 + 43.5 = 84

d. Of those persons scoring 90 on the midterm exam, what percentage
   have final exam scores over 95?

   RMSE = SDy * sqrt(1-r^2) = 15 * sqrt(1-0.6^2) = 15 * 0.8 = 12.

   z = (y - y^) / RMSE = (95 - 84) / 12 = 11/12 = 0.92.
   Look up -0.92 in the standard normal table: 0.1788 = 18%.

Part C. Ideal Measurement Example

1. a. Q0=291  b. Q1=295  c. Q2=296  d. Q3=299  e. Q4=334  

   g. xbar = 297.85  h. SD+ - 6.598   i. SEave = 0.929

   j. 95% conf int = (295.99, 299.73)

2. Skewed to the right.  Without the outliers,
   the data are still somewhat skewed to the right.

3. Unbiased, homoskedastic

Part C. Regression Problem

1. r = 0.78

2. regression equation:  LogBrainWeight = 0.45 * LogBodyWeight + 1.112

3. biased

4. fairly normal