All positions are established on Sep. 15 and liquidated on Nov. 10 . Take care w
ID: 3264384 • Letter: A
Question
All positions are established on Sep. 15 and liquidated on Nov. 10. Take care with the decimals by rounding to 6 decimal places for all solutions (e.g. either 0.123456 or 12.3456% is a 6 decimal response). Answer each question as completely as possible and show all your work for full credit, even for an otherwise correct answer. Use the information in the table below for the questions identified. Important Notes:
1. Place all your solutions on the sheet indicated for that problem.
2. Copy/paste from the Excel.
3. When failed to copy/paste where required, will receive a zero for that problem.
Weekly Domestic T-Bond Price Data
Question 11
Interpret the results of the standard deviation calculations, for the 2039 Apr 15 T-bond and the GBM Index by:
a. Stating the 99.7% confidence intervals for each series.
b. Describe, in your own words, the meaning of each confidence intervals.
All positions are established on Sep. 15 and liquidated on Nov. 10. Take care with the decimals by rounding to 6 decimal places for all solutions (e.g. either 0.123456 or 12.3456% is a 6 decimal response). Answer each question as completely as possible and show all your work for full credit, even for an otherwise correct answer. Use the information in the table below for the questions identified. Important Notes:
1. Place all your solutions on the sheet indicated for that problem.
2. Copy/paste from the Excel.
3. When failed to copy/paste where required, will receive a zero for that problem.
Weekly Domestic T-Bond Price Data
Date T-bond ROPCdb Gov't Index ROPCgi 15-Sep 96.6065 1520.14 22-Sep 98.7979 1525.62 29-Sep 99.8031 1513.01 6-Oct 98.7364 1510.33 13-Oct 98.6713 1492.42 20-Oct 99.5946 1488.13 27-Oct 100.9021 1500.08 3-Nov 102.6137 1491.64 10-Nov 102.75 1489.41 3.5% 2039 Apr 15 T-bondExplanation / Answer
Solution
Back-up Theory
Let X = variable under study. We assume X ~ N(µ, 2).
100(1 – ) % confidence interval for 2 is: [{(n - 1)s2/2n-1,/2)}, {(n - 1)s2/2n-1,1- /2)}], where
= population standard deviation,
s = sample standard deviation,
n = sample size and
2n-1,/2 and 2n-1,1- /2 are respectively upper and lower (/2) % point of
Chi-square Distribution with (n - 1) degrees of freedom.
Given, n = 9, = 0.003 , 99.7% Confidence Interval for = sqrt(CI for 2)
Computations: [by Excel]
Date
T-bond
Gov't Index
15-Sep
96.6065
1520.14
22-Sep
98.7979
1525.62
29-Sep
99.8031
1513.01
6-Oct
98.7364
1510.33
13-Oct
98.6713
1492.42
20-Oct
99.5946
1488.13
27-Oct
100.9021
1500.08
3-Nov
102.6137
1491.64
10-Nov
102.75
1489.41
SD (s)
1.98354628
14.199924
s^2
3.93445584
201.63785
n
9
9
n - 1
8
8
0.003
0.003
/2
0.0015
0.0015
1- (/2)
0.9985
0.9985
Upper (/2)% point of 2(n - 1)
25.0915213
25.091521
Lower (/2)% point of 2(n - 1)
0.95801015
0.9580102
Lower Bound for ^2
1.25443357
64.28876
Upper Bound for ^2
32.8552329
1683.8055
Lower Bound for
1.12001499
8.0180272
Upper Bound for
5.73194844
41.0342
Date
T-bond
Gov't Index
15-Sep
96.6065
1520.14
22-Sep
98.7979
1525.62
29-Sep
99.8031
1513.01
6-Oct
98.7364
1510.33
13-Oct
98.6713
1492.42
20-Oct
99.5946
1488.13
27-Oct
100.9021
1500.08
3-Nov
102.6137
1491.64
10-Nov
102.75
1489.41
SD (s)
1.98354628
14.199924
s^2
3.93445584
201.63785
n
9
9
n - 1
8
8
0.003
0.003
/2
0.0015
0.0015
1- (/2)
0.9985
0.9985
Upper (/2)% point of 2(n - 1)
25.0915213
25.091521
Lower (/2)% point of 2(n - 1)
0.95801015
0.9580102
Lower Bound for ^2
1.25443357
64.28876
Upper Bound for ^2
32.8552329
1683.8055
Lower Bound for
1.12001499
8.0180272
Upper Bound for
5.73194844
41.0342