Connect Hosted by ALEKS Access Card or Elementary Statistics
3rd Edition
ISBN: 9781260373752
Author: William Navidi Prof., Barry Monk Professor
Publisher: McGraw-Hill Education
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Chapter 13.3, Problem 21E
a.
To determine
To find:The regression equation for the data.
a.
Expert Solution
Answer to Problem 21E
The regression equation for the data is
Explanation of Solution
Given information: The data is shown below.
| 31.3 | 50 | 19 | 4.0 | 37.4 | 60 | 30 | 5.0 | 24.0 | 60 | 24 | 4.0 | 51.0 | 80 | 34 | 7.5 |
| 56.9 | 90 | 38 | 8.0 | 43.3 | 60 | 26 | 7.0 | 36.2 | 50 | 21 | 7.0 | 34.3 | 60 | 22 | 2.5 |
| 43.1 | 70 | 28 | 6.5 | 36.3 | 70 | 25 | 7.5 | 26.5 | 50 | 17 | 2.0 | 31.5 | 60 | 24 | 5.0 |
| 41.5 | 70 | 25 | 5.5 | 38.4 | 70 | 31 | 5.5 | 47.1 | 80 | 34 | 8.5 | 33.2 | 60 | 23 | 4.0 |
| 39.0 | 60 | 26 | 6.5 | 41.5 | 60 | 27 | 7.5 | 38.1 | 70 | 27 | 5.5 | 39.2 | 60 | 29 | 6.5 |
| 40.9 | 70 | 29 | 5.0 | 36.l | 60 | 23 | 6.0 | 33.5 | 60 | 24 | 2.5 | 46.7 | 70 | 27 | 7.5 |
| 35.9 | 60 | 23 | 5.5 | 38.5 | 60 | 23 | 6.0 | 43.6 | 70 | 27 | 10.0 | 30.4 | 70 | 32 | 4.0 |
| 43.5 | 70 | 28 | 5.5 | 42.4 | 60 | 24 | 9.0 | 41.0 | 80 | 32 | 6.5 | 43.2 | 60 | 25 | 5.5 |
| 47.9 | 80 | 34 | 6.5 | 46.5 | 70 | 31 | 5.5 | 50.2 | 50 | 26 | 9.0 | 30.6 | 60 | 26 | 3.5 |
| 33.8 | 70 | 26 | 4.5 | 43.1 | 80 | 32 | 6.0 | 34.4 | 50 | 22 | 4.0 | 43.3 | 70 | 28 | 7.5 |
| 41.1 | 70 | 26 | 8.0 | 50.8 | 60 | 26 | 10.0 | 47.9 | 80 | 34 | 6.5 | 32.4 | 60 | 22 | 5.5 |
| 38.7 | 70 | 26 | 8.0 | 44.2 | 70 | 28 | 4.5 | 46.2 | 50 | 21 | 10.0 | 35.5 | 60 | 22 | 4.5 |
Calculation:
The MINITAB is shown below,
Figure-1
From Figure-1 it is clear that the regression equation is
b.
To determine
To find: The value of variable
b.
Expert Solution
Answer to Problem 21E
The value of variable
Explanation of Solution
Given information: The data is shown below.
| 31.3 | 50 | 19 | 4.0 | 37.4 | 60 | 30 | 5.0 | 24.0 | 60 | 24 | 4.0 | 51.0 | 80 | 34 | 7.5 |
| 56.9 | 90 | 38 | 8.0 | 43.3 | 60 | 26 | 7.0 | 36.2 | 50 | 21 | 7.0 | 34.3 | 60 | 22 | 2.5 |
| 43.1 | 70 | 28 | 6.5 | 36.3 | 70 | 25 | 7.5 | 26.5 | 50 | 17 | 2.0 | 31.5 | 60 | 24 | 5.0 |
| 41.5 | 70 | 25 | 5.5 | 38.4 | 70 | 31 | 5.5 | 47.1 | 80 | 34 | 8.5 | 33.2 | 60 | 23 | 4.0 |
| 39.0 | 60 | 26 | 6.5 | 41.5 | 60 | 27 | 7.5 | 38.1 | 70 | 27 | 5.5 | 39.2 | 60 | 29 | 6.5 |
| 40.9 | 70 | 29 | 5.0 | 36.l | 60 | 23 | 6.0 | 33.5 | 60 | 24 | 2.5 | 46.7 | 70 | 27 | 7.5 |
| 35.9 | 60 | 23 | 5.5 | 38.5 | 60 | 23 | 6.0 | 43.6 | 70 | 27 | 10.0 | 30.4 | 70 | 32 | 4.0 |
| 43.5 | 70 | 28 | 5.5 | 42.4 | 60 | 24 | 9.0 | 41.0 | 80 | 32 | 6.5 | 43.2 | 60 | 25 | 5.5 |
| 47.9 | 80 | 34 | 6.5 | 46.5 | 70 | 31 | 5.5 | 50.2 | 50 | 26 | 9.0 | 30.6 | 60 | 26 | 3.5 |
| 33.8 | 70 | 26 | 4.5 | 43.1 | 80 | 32 | 6.0 | 34.4 | 50 | 22 | 4.0 | 43.3 | 70 | 28 | 7.5 |
| 41.1 | 70 | 26 | 8.0 | 50.8 | 60 | 26 | 10.0 | 47.9 | 80 | 34 | 6.5 | 32.4 | 60 | 22 | 5.5 |
| 38.7 | 70 | 26 | 8.0 | 44.2 | 70 | 28 | 4.5 | 46.2 | 50 | 21 | 10.0 | 35.5 | 60 | 22 | 4.5 |
Calculation:
From Figure-1 it is clear that the regression equation is
Substitute the values in above equation.
Thus, the value of variable
c.
To determine
To find: The confidence interval.
c.
Expert Solution
Answer to Problem 21E
The confidence intervalis
Explanation of Solution
Given information: The data is shown below.
| 31.3 | 50 | 19 | 4.0 | 37.4 | 60 | 30 | 5.0 | 24.0 | 60 | 24 | 4.0 | 51.0 | 80 | 34 | 7.5 |
| 56.9 | 90 | 38 | 8.0 | 43.3 | 60 | 26 | 7.0 | 36.2 | 50 | 21 | 7.0 | 34.3 | 60 | 22 | 2.5 |
| 43.1 | 70 | 28 | 6.5 | 36.3 | 70 | 25 | 7.5 | 26.5 | 50 | 17 | 2.0 | 31.5 | 60 | 24 | 5.0 |
| 41.5 | 70 | 25 | 5.5 | 38.4 | 70 | 31 | 5.5 | 47.1 | 80 | 34 | 8.5 | 33.2 | 60 | 23 | 4.0 |
| 39.0 | 60 | 26 | 6.5 | 41.5 | 60 | 27 | 7.5 | 38.1 | 70 | 27 | 5.5 | 39.2 | 60 | 29 | 6.5 |
| 40.9 | 70 | 29 | 5.0 | 36.l | 60 | 23 | 6.0 | 33.5 | 60 | 24 | 2.5 | 46.7 | 70 | 27 | 7.5 |
| 35.9 | 60 | 23 | 5.5 | 38.5 | 60 | 23 | 6.0 | 43.6 | 70 | 27 | 10.0 | 30.4 | 70 | 32 | 4.0 |
| 43.5 | 70 | 28 | 5.5 | 42.4 | 60 | 24 | 9.0 | 41.0 | 80 | 32 | 6.5 | 43.2 | 60 | 25 | 5.5 |
| 47.9 | 80 | 34 | 6.5 | 46.5 | 70 | 31 | 5.5 | 50.2 | 50 | 26 | 9.0 | 30.6 | 60 | 26 | 3.5 |
| 33.8 | 70 | 26 | 4.5 | 43.1 | 80 | 32 | 6.0 | 34.4 | 50 | 22 | 4.0 | 43.3 | 70 | 28 | 7.5 |
| 41.1 | 70 | 26 | 8.0 | 50.8 | 60 | 26 | 10.0 | 47.9 | 80 | 34 | 6.5 | 32.4 | 60 | 22 | 5.5 |
| 38.7 | 70 | 26 | 8.0 | 44.2 | 70 | 28 | 4.5 | 46.2 | 50 | 21 | 10.0 | 35.5 | 60 | 22 | 4.5 |
Calculation:
The MINITAB output is shown below.
Figure-2
From Figure-2 it is clear that the confidence interval is
d.
To determine
To find: The prediction interval.
d.
Expert Solution
Answer to Problem 21E
The prediction intervalis
Explanation of Solution
Given information: The data is shown below.
| 31.3 | 50 | 19 | 4.0 | 37.4 | 60 | 30 | 5.0 | 24.0 | 60 | 24 | 4.0 | 51.0 | 80 | 34 | 7.5 |
| 56.9 | 90 | 38 | 8.0 | 43.3 | 60 | 26 | 7.0 | 36.2 | 50 | 21 | 7.0 | 34.3 | 60 | 22 | 2.5 |
| 43.1 | 70 | 28 | 6.5 | 36.3 | 70 | 25 | 7.5 | 26.5 | 50 | 17 | 2.0 | 31.5 | 60 | 24 | 5.0 |
| 41.5 | 70 | 25 | 5.5 | 38.4 | 70 | 31 | 5.5 | 47.1 | 80 | 34 | 8.5 | 33.2 | 60 | 23 | 4.0 |
| 39.0 | 60 | 26 | 6.5 | 41.5 | 60 | 27 | 7.5 | 38.1 | 70 | 27 | 5.5 | 39.2 | 60 | 29 | 6.5 |
| 40.9 | 70 | 29 | 5.0 | 36.l | 60 | 23 | 6.0 | 33.5 | 60 | 24 | 2.5 | 46.7 | 70 | 27 | 7.5 |
| 35.9 | 60 | 23 | 5.5 | 38.5 | 60 | 23 | 6.0 | 43.6 | 70 | 27 | 10.0 | 30.4 | 70 | 32 | 4.0 |
| 43.5 | 70 | 28 | 5.5 | 42.4 | 60 | 24 | 9.0 | 41.0 | 80 | 32 | 6.5 | 43.2 | 60 | 25 | 5.5 |
| 47.9 | 80 | 34 | 6.5 | 46.5 | 70 | 31 | 5.5 | 50.2 | 50 | 26 | 9.0 | 30.6 | 60 | 26 | 3.5 |
| 33.8 | 70 | 26 | 4.5 | 43.1 | 80 | 32 | 6.0 | 34.4 | 50 | 22 | 4.0 | 43.3 | 70 | 28 | 7.5 |
| 41.1 | 70 | 26 | 8.0 | 50.8 | 60 | 26 | 10.0 | 47.9 | 80 | 34 | 6.5 | 32.4 | 60 | 22 | 5.5 |
| 38.7 | 70 | 26 | 8.0 | 44.2 | 70 | 28 | 4.5 | 46.2 | 50 | 21 | 10.0 | 35.5 | 60 | 22 | 4.5 |
Calculation:
From Figure-12 it is clear that the
e.
To determine
To find: The percentage of variation in variable
e.
Expert Solution
Answer to Problem 21E
The percentage of variation in variable
Explanation of Solution
Given information: The data is shown below.
| 31.3 | 50 | 19 | 4.0 | 37.4 | 60 | 30 | 5.0 | 24.0 | 60 | 24 | 4.0 | 51.0 | 80 | 34 | 7.5 |
| 56.9 | 90 | 38 | 8.0 | 43.3 | 60 | 26 | 7.0 | 36.2 | 50 | 21 | 7.0 | 34.3 | 60 | 22 | 2.5 |
| 43.1 | 70 | 28 | 6.5 | 36.3 | 70 | 25 | 7.5 | 26.5 | 50 | 17 | 2.0 | 31.5 | 60 | 24 | 5.0 |
| 41.5 | 70 | 25 | 5.5 | 38.4 | 70 | 31 | 5.5 | 47.1 | 80 | 34 | 8.5 | 33.2 | 60 | 23 | 4.0 |
| 39.0 | 60 | 26 | 6.5 | 41.5 | 60 | 27 | 7.5 | 38.1 | 70 | 27 | 5.5 | 39.2 | 60 | 29 | 6.5 |
| 40.9 | 70 | 29 | 5.0 | 36.l | 60 | 23 | 6.0 | 33.5 | 60 | 24 | 2.5 | 46.7 | 70 | 27 | 7.5 |
| 35.9 | 60 | 23 | 5.5 | 38.5 | 60 | 23 | 6.0 | 43.6 | 70 | 27 | 10.0 | 30.4 | 70 | 32 | 4.0 |
| 43.5 | 70 | 28 | 5.5 | 42.4 | 60 | 24 | 9.0 | 41.0 | 80 | 32 | 6.5 | 43.2 | 60 | 25 | 5.5 |
| 47.9 | 80 | 34 | 6.5 | 46.5 | 70 | 31 | 5.5 | 50.2 | 50 | 26 | 9.0 | 30.6 | 60 | 26 | 3.5 |
| 33.8 | 70 | 26 | 4.5 | 43.1 | 80 | 32 | 6.0 | 34.4 | 50 | 22 | 4.0 | 43.3 | 70 | 28 | 7.5 |
| 41.1 | 70 | 26 | 8.0 | 50.8 | 60 | 26 | 10.0 | 47.9 | 80 | 34 | 6.5 | 32.4 | 60 | 22 | 5.5 |
| 38.7 | 70 | 26 | 8.0 | 44.2 | 70 | 28 | 4.5 | 46.2 | 50 | 21 | 10.0 | 35.5 | 60 | 22 | 4.5 |
Calculation:
From Figure-1 it is clear that the percentage of variation in variable
f.
To determine
To find:Whether the given model is useful for prediction.
f.
Expert Solution
Answer to Problem 21E
The model is useful in prediction.
Explanation of Solution
Given information: The data is shown below.
| 31.3 | 50 | 19 | 4.0 | 37.4 | 60 | 30 | 5.0 | 24.0 | 60 | 24 | 4.0 | 51.0 | 80 | 34 | 7.5 |
| 56.9 | 90 | 38 | 8.0 | 43.3 | 60 | 26 | 7.0 | 36.2 | 50 | 21 | 7.0 | 34.3 | 60 | 22 | 2.5 |
| 43.1 | 70 | 28 | 6.5 | 36.3 | 70 | 25 | 7.5 | 26.5 | 50 | 17 | 2.0 | 31.5 | 60 | 24 | 5.0 |
| 41.5 | 70 | 25 | 5.5 | 38.4 | 70 | 31 | 5.5 | 47.1 | 80 | 34 | 8.5 | 33.2 | 60 | 23 | 4.0 |
| 39.0 | 60 | 26 | 6.5 | 41.5 | 60 | 27 | 7.5 | 38.1 | 70 | 27 | 5.5 | 39.2 | 60 | 29 | 6.5 |
| 40.9 | 70 | 29 | 5.0 | 36.l | 60 | 23 | 6.0 | 33.5 | 60 | 24 | 2.5 | 46.7 | 70 | 27 | 7.5 |
| 35.9 | 60 | 23 | 5.5 | 38.5 | 60 | 23 | 6.0 | 43.6 | 70 | 27 | 10.0 | 30.4 | 70 | 32 | 4.0 |
| 43.5 | 70 | 28 | 5.5 | 42.4 | 60 | 24 | 9.0 | 41.0 | 80 | 32 | 6.5 | 43.2 | 60 | 25 | 5.5 |
| 47.9 | 80 | 34 | 6.5 | 46.5 | 70 | 31 | 5.5 | 50.2 | 50 | 26 | 9.0 | 30.6 | 60 | 26 | 3.5 |
| 33.8 | 70 | 26 | 4.5 | 43.1 | 80 | 32 | 6.0 | 34.4 | 50 | 22 | 4.0 | 43.3 | 70 | 28 | 7.5 |
| 41.1 | 70 | 26 | 8.0 | 50.8 | 60 | 26 | 10.0 | 47.9 | 80 | 34 | 6.5 | 32.4 | 60 | 22 | 5.5 |
| 38.7 | 70 | 26 | 8.0 | 44.2 | 70 | 28 | 4.5 | 46.2 | 50 | 21 | 10.0 | 35.5 | 60 | 22 | 4.5 |
Calculation:
The null hypothesis is, the model is not useful for prediction and the alternative hypothesis is, the model is useful in prediction.
From Figure-1 it is clear that the p value is less than the level of significance of
Hence, the null hypothesis is rejected.
Thus, the model is useful in prediction.
g.
To determine
To explain:The test for the hypothesis
g.
Expert Solution
Explanation of Solution
Given information: The data is shown below.
| 31.3 | 50 | 19 | 4.0 | 37.4 | 60 | 30 | 5.0 | 24.0 | 60 | 24 | 4.0 | 51.0 | 80 | 34 | 7.5 |
| 56.9 | 90 | 38 | 8.0 | 43.3 | 60 | 26 | 7.0 | 36.2 | 50 | 21 | 7.0 | 34.3 | 60 | 22 | 2.5 |
| 43.1 | 70 | 28 | 6.5 | 36.3 | 70 | 25 | 7.5 | 26.5 | 50 | 17 | 2.0 | 31.5 | 60 | 24 | 5.0 |
| 41.5 | 70 | 25 | 5.5 | 38.4 | 70 | 31 | 5.5 | 47.1 | 80 | 34 | 8.5 | 33.2 | 60 | 23 | 4.0 |
| 39.0 | 60 | 26 | 6.5 | 41.5 | 60 | 27 | 7.5 | 38.1 | 70 | 27 | 5.5 | 39.2 | 60 | 29 | 6.5 |
| 40.9 | 70 | 29 | 5.0 | 36.l | 60 | 23 | 6.0 | 33.5 | 60 | 24 | 2.5 | 46.7 | 70 | 27 | 7.5 |
| 35.9 | 60 | 23 | 5.5 | 38.5 | 60 | 23 | 6.0 | 43.6 | 70 | 27 | 10.0 | 30.4 | 70 | 32 | 4.0 |
| 43.5 | 70 | 28 | 5.5 | 42.4 | 60 | 24 | 9.0 | 41.0 | 80 | 32 | 6.5 | 43.2 | 60 | 25 | 5.5 |
| 47.9 | 80 | 34 | 6.5 | 46.5 | 70 | 31 | 5.5 | 50.2 | 50 | 26 | 9.0 | 30.6 | 60 | 26 | 3.5 |
| 33.8 | 70 | 26 | 4.5 | 43.1 | 80 | 32 | 6.0 | 34.4 | 50 | 22 | 4.0 | 43.3 | 70 | 28 | 7.5 |
| 41.1 | 70 | 26 | 8.0 | 50.8 | 60 | 26 | 10.0 | 47.9 | 80 | 34 | 6.5 | 32.4 | 60 | 22 | 5.5 |
| 38.7 | 70 | 26 | 8.0 | 44.2 | 70 | 28 | 4.5 | 46.2 | 50 | 21 | 10.0 | 35.5 | 60 | 22 | 4.5 |
The null hypothesis is, there is no relationship between
For the variable
From Figure-1 it is clear that the p value is
Hence, the null hypothesis is not rejected.
Thus, there is no linear relationship between
For the variable
From Figure-1 it is clear that the p value is
Hence, the null hypothesis is rejected.
Thus, there is alinear relationship between
For the variable
From Figure-1 it is clear that the p value is
Hence, the null hypothesis is rejected.
Thus, there is a linear relationship between
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Chapter 13 Solutions
Connect Hosted by ALEKS Access Card or Elementary Statistics
Ch. 13.1 - Prob. 7ECh. 13.1 - Prob. 8ECh. 13.1 - In Exercises 9 and 10, determine whether the...Ch. 13.1 - Prob. 10ECh. 13.1 - Prob. 11ECh. 13.1 - Prob. 12ECh. 13.1 - Prob. 13ECh. 13.1 - Prob. 14ECh. 13.1 - Prob. 15ECh. 13.1 - Prob. 16E
Ch. 13.1 - Prob. 17ECh. 13.1 - Prob. 18ECh. 13.1 - Prob. 19ECh. 13.1 - Prob. 20ECh. 13.1 - Prob. 21ECh. 13.1 - Prob. 22ECh. 13.1 - Prob. 23ECh. 13.1 - Prob. 24ECh. 13.1 - Prob. 25ECh. 13.1 - Prob. 26ECh. 13.1 - Prob. 27ECh. 13.1 - Prob. 28ECh. 13.1 - Prob. 26aECh. 13.1 - Calculator display: The following TI-84 Plus...Ch. 13.1 - Prob. 28aECh. 13.1 - Prob. 29ECh. 13.1 - Prob. 30ECh. 13.1 - Confidence interval for the conditional mean: In...Ch. 13.2 - Prob. 3ECh. 13.2 - Prob. 4ECh. 13.2 - Prob. 5ECh. 13.2 - Prob. 6ECh. 13.2 - Prob. 7ECh. 13.2 - Prob. 8ECh. 13.2 - Prob. 9ECh. 13.2 - Prob. 10ECh. 13.2 - Prob. 11ECh. 13.2 - Prob. 12ECh. 13.2 - Prob. 13ECh. 13.2 - Prob. 14ECh. 13.2 - Prob. 15ECh. 13.2 - Prob. 16ECh. 13.2 - Prob. 17ECh. 13.2 - Dry up: Use the data in Exercise 26 in Section...Ch. 13.2 - Prob. 19ECh. 13.2 - Prob. 20ECh. 13.2 - Prob. 21ECh. 13.3 - Prob. 7ECh. 13.3 - Prob. 8ECh. 13.3 - Prob. 9ECh. 13.3 - In Exercises 9 and 10, determine whether the...Ch. 13.3 - Prob. 11ECh. 13.3 - Prob. 12ECh. 13.3 - Prob. 13ECh. 13.3 - For the following data set: Construct the multiple...Ch. 13.3 - Engine emissions: In a laboratory test of a new...Ch. 13.3 - Prob. 16ECh. 13.3 - Prob. 17ECh. 13.3 - Prob. 18ECh. 13.3 - Prob. 19ECh. 13.3 - Prob. 20ECh. 13.3 - Prob. 21ECh. 13.3 - Prob. 22ECh. 13.3 - Prob. 23ECh. 13 - A confidence interval for 1 is to be constructed...Ch. 13 - A confidence interval for a mean response and a...Ch. 13 - Prob. 3CQCh. 13 - Prob. 4CQCh. 13 - Prob. 5CQCh. 13 - Prob. 6CQCh. 13 - Construct a 95% confidence interval for 1.Ch. 13 - Prob. 8CQCh. 13 - Prob. 9CQCh. 13 - Prob. 10CQCh. 13 - Prob. 11CQCh. 13 - Prob. 12CQCh. 13 - Prob. 13CQCh. 13 - Prob. 14CQCh. 13 - Prob. 15CQCh. 13 - Prob. 1RECh. 13 - Prob. 2RECh. 13 - Prob. 3RECh. 13 - Prob. 4RECh. 13 - Prob. 5RECh. 13 - Prob. 6RECh. 13 - Prob. 7RECh. 13 - Prob. 8RECh. 13 - Prob. 9RECh. 13 - Prob. 10RECh. 13 - Air pollution: Following are measurements of...Ch. 13 - Icy lakes: Following are data on maximum ice...Ch. 13 - Prob. 13RECh. 13 - Prob. 14RECh. 13 - Prob. 15RECh. 13 - Prob. 1WAICh. 13 - Prob. 2WAICh. 13 - Prob. 1CSCh. 13 - Prob. 2CSCh. 13 - Prob. 3CSCh. 13 - Prob. 4CSCh. 13 - Prob. 5CSCh. 13 - Prob. 6CSCh. 13 - Prob. 7CS
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- Question 3. We want to price a put option with strike price K and expiration T. Two financial advisors estimate the parameters with two different statistical methods: they obtain the same return rate μ, the same volatility σ, but the first advisor has interest r₁ and the second advisor has interest rate r2 (r1>r2). They both use a CRR model with the same number of periods to price the option. Which advisor will get the larger price? (Explain your answer.)arrow_forwardQuestion 5. We consider a put option with strike price K and expiration T. This option is priced using a 1-period CRR model. We consider r > 0, and σ > 0 very large. What is the approximate price of the option? In other words, what is the limit of the price of the option as σ∞. (Briefly justify your answer.)arrow_forwardQuestion 6. You collect daily data for the stock of a company Z over the past 4 months (i.e. 80 days) and calculate the log-returns (yk)/(-1. You want to build a CRR model for the evolution of the stock. The expected value and standard deviation of the log-returns are y = 0.06 and Sy 0.1. The money market interest rate is r = 0.04. Determine the risk-neutral probability of the model.arrow_forward
- Several markets (Japan, Switzerland) introduced negative interest rates on their money market. In this problem, we will consider an annual interest rate r < 0. We consider a stock modeled by an N-period CRR model where each period is 1 year (At = 1) and the up and down factors are u and d. (a) We consider an American put option with strike price K and expiration T. Prove that if <0, the optimal strategy is to wait until expiration T to exercise.arrow_forwardWe consider an N-period CRR model where each period is 1 year (At = 1), the up factor is u = 0.1, the down factor is d = e−0.3 and r = 0. We remind you that in the CRR model, the stock price at time tn is modeled (under P) by Sta = So exp (μtn + σ√AtZn), where (Zn) is a simple symmetric random walk. (a) Find the parameters μ and σ for the CRR model described above. (b) Find P Ste So 55/50 € > 1). StN (c) Find lim P 804-N (d) Determine q. (You can use e- 1 x.) Ste (e) Find Q So (f) Find lim Q 004-N StN Soarrow_forwardIn this problem, we consider a 3-period stock market model with evolution given in Fig. 1 below. Each period corresponds to one year. The interest rate is r = 0%. 16 22 28 12 16 12 8 4 2 time Figure 1: Stock evolution for Problem 1. (a) A colleague notices that in the model above, a movement up-down leads to the same value as a movement down-up. He concludes that the model is a CRR model. Is your colleague correct? (Explain your answer.) (b) We consider a European put with strike price K = 10 and expiration T = 3 years. Find the price of this option at time 0. Provide the replicating portfolio for the first period. (c) In addition to the call above, we also consider a European call with strike price K = 10 and expiration T = 3 years. Which one has the highest price? (It is not necessary to provide the price of the call.) (d) We now assume a yearly interest rate r = 25%. We consider a Bermudan put option with strike price K = 10. It works like a standard put, but you can exercise it…arrow_forward
- In this problem, we consider a 2-period stock market model with evolution given in Fig. 1 below. Each period corresponds to one year (At = 1). The yearly interest rate is r = 1/3 = 33%. This model is a CRR model. 25 15 9 10 6 4 time Figure 1: Stock evolution for Problem 1. (a) Find the values of up and down factors u and d, and the risk-neutral probability q. (b) We consider a European put with strike price K the price of this option at time 0. == 16 and expiration T = 2 years. Find (c) Provide the number of shares of stock that the replicating portfolio contains at each pos- sible position. (d) You find this option available on the market for $2. What do you do? (Short answer.) (e) We consider an American put with strike price K = 16 and expiration T = 2 years. Find the price of this option at time 0 and describe the optimal exercising strategy. (f) We consider an American call with strike price K ○ = 16 and expiration T = 2 years. Find the price of this option at time 0 and describe…arrow_forward2.2, 13.2-13.3) question: 5 point(s) possible ubmit test The accompanying table contains the data for the amounts (in oz) in cans of a certain soda. The cans are labeled to indicate that the contents are 20 oz of soda. Use the sign test and 0.05 significance level to test the claim that cans of this soda are filled so that the median amount is 20 oz. If the median is not 20 oz, are consumers being cheated? Click the icon to view the data. What are the null and alternative hypotheses? OA. Ho: Medi More Info H₁: Medi OC. Ho: Medi H₁: Medi Volume (in ounces) 20.3 20.1 20.4 Find the test stat 20.1 20.5 20.1 20.1 19.9 20.1 Test statistic = 20.2 20.3 20.3 20.1 20.4 20.5 Find the P-value 19.7 20.2 20.4 20.1 20.2 20.2 P-value= (R 19.9 20.1 20.5 20.4 20.1 20.4 Determine the p 20.1 20.3 20.4 20.2 20.3 20.4 Since the P-valu 19.9 20.2 19.9 Print Done 20 oz 20 oz 20 oz 20 oz ce that the consumers are being cheated.arrow_forwardT Teenage obesity (O), and weekly fast-food meals (F), among some selected Mississippi teenagers are: Name Obesity (lbs) # of Fast-foods per week Josh 185 10 Karl 172 8 Terry 168 9 Kamie Andy 204 154 12 6 (a) Compute the variance of Obesity, s²o, and the variance of fast-food meals, s², of this data. [Must show full work]. (b) Compute the Correlation Coefficient between O and F. [Must show full work]. (c) Find the Coefficient of Determination between O and F. [Must show full work]. (d) Obtain the Regression equation of this data. [Must show full work]. (e) Interpret your answers in (b), (c), and (d). (Full explanations required). Edit View Insert Format Tools Tablearrow_forward
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