251solngr2-081b 6/18/08 (Open this document in 'Page Layout' view!)
Graded Assignment 2
Name:Key
There will be a penalty for papers that are unstapled. Note that from now on neatness means paper neatly trimmed on the left side if it has been torn, multiple pages stapled and paper written on only one side. The stapling is for your protection – putting your name on every page helps too, I still have some unclaimed pages from an old exam (as well as an old exam with no name on it that the perp will not admit responsibility for).Note problem 3 at the end to see how risk fits into this.
and
1) Use the joint probability tables above.
For these joint probability tables (i) check for independence, (ii) Compute and , (iii) Compute or and or, (iv) Compute and from the results in (ii) and (iii), (v) Compute and using the formulas in section K4 of 251v2out or section C1 of251var2. Note that .
Solution:
(i) Check for independence: First you need to find and . Look at the upper left hand probability below. Its value is a).10 or b) 0 and it represents . If and are independent, we would have . We need to find out what these probabilities are so we add the rows and columns to get marginal or total probabilities.
a) b)
Thus we have for a) . Since these are not equal to in a) , and cannot be independent. Even one place where the joint probability is not the product of the marginal probabilities is enough to show that x and y are not independent. For b) we have . But this is not equal to If this one is not enough to convince you, how about, for both a) and b), . Actually the fastest way to prove non-independence is to look for zeroes. If in both a) and b) and x and y are independent, then it must be true that or . Notice that the second row is not proportional to the first row or any other row. This is also evidence of non-independence. If variables are independent all rows must be proportional to one another and all columns must also be proportional to one another.
A zero covariance or correlation would be the consequence of independence, but it is not true that a zero correlation or covariance would prove independence. We have already seen one example where there is a zero correlation, but no independence(Downing and Clark, pg. 219, Computational Problem 3).
Let’s finish the job we did in (i) by computing (a check for a valid distribution), ,,,and .
The easiest way to do this is to multiply the items in the column by the items in the column to get the column and then to multiply the items in the column by the items in the column to get the column. Then multiply the items in the row by the items in the row to get the row and then multiply the items in the row by the items in the row to get the row. Then add up all the rows and columns outside the original table.
a)
b)
To summarize a) (a check),,,, and .
b) (a check),, , , and .
(ii) Compute and . Remember that variances and standard deviations are never negative. We actually need means and variances for both and . From the above
a), (), and ();
b) , (), and ().
(iii) Compute or and or. a) We must now compute by multiplying each pair of values of and by their joint probabilities. We had , (), , () and
.
To complete what we have done, write .
So that .
b) We must now compute by multiplying each pair of values of and by their joint probabilities. We had , (), , () and .
To complete what we have done, write .
So that . In general, joint probability tables with only the diagonals filled produce correlations close to +1 or -1. A northwest to southwest diagonal produces a positive correlation and a southwest to northeast diagonal produces a negative correlation. The two tables here have dominant diagonals, each number in the diagonal is larger than other numbers in its row and column and so the correlations are similar to those of tables with only the diagonals filled.
Remember that the correlation must be between -1 and +1!
Note that the strength of a correlation is found by squaring the correlation and measuring the strength on a zero to one scale. In a) we had , so and we can say that there is a relatively strong tendency for to rise as rises. In b) we had , so and we can say that there is a relatively weak tendency for to fall as rises or for to rise as falls.
(iv) Compute and from the results in (ii) and (iii). Compute and from the results in (ii) and (iii). How many of you ignored the instructions and wrote down each value of with its probability. What a great way to waste time!
The formulas that you were given were and
a) We had , (), , () and .
and ()
b) We had , (), , () and .
and ()
(v) Compute and using the formulas in section K4 of 251v2out or section C1 of251var2. Note that .
251v2out says and , where has the value or depending on whether the product of and is negative or positive. and .
a) We had and . So and .
b) We had and . So and .
2) The following data represent the scores of a group of students on a math placement test and their grades in a math course. (i) Compute the sample mean and variance of , (ii) Compute or and or , (iii) Compute the sample mean and variance of from the results in (i) and (ii). (iv) Compute and using the formulas in section K4 of 251v2out or section C1 of251var2. Note that .
Test Score Grades
51 / 8052 / 77
59 / 87
45 / 72
61 / 80
54 / 84
56 / 83
67 / 87
63 92
53 77
(i) Compute the sample mean and variance of .
Row
1 51 2601 80 6400 4080
2 52 2704 77 5929 4004
3 59 3481 87 7569 5133
4 45 2025 72 5184 3240
5 61 3721 80 6400 4880
6 54 2916 84 7056 4536
7 56 3136 83 6889 4648
8 67 4489 87 7569 5829
9 63 3969 92 8464 5796
10 53 2809 77 5929 4081
561 31851 819 67389 46227
To summarize the results of these computations , and . Thus and .
.
.
(ii) Compute or and or .
Recall , and
.
(iii) Compute the sample mean and variance of from the results in (i) and (ii).
Recall , and
.
(iv) Compute and using the formulas in section K4 of 251v2out or section C1 of251var2. Note that .
251v2out says
and , where has the value or depending on whether the product of and is negative or positive. and
and
3) The PHLX Gold/Silver SectorSM (XAUSM) is a capitalization-weighted index composed of 16 companies involved in the gold and silver mining industry. XAU was set to an initial value of 100 in January 1979; options commenced trading on December 19, 1983. The Dow-Jones Utility average is an average based on the prices of 16 (I think) utility stocks. Both gold and utilities can attract cautious investors under certain stock market conditions, so it is interesting to look at how they move relative to one another. The values of these two indices for 20 very recent trading days are given on the next page.
251solngr2-081b 6/18/08 (Open this document in 'Page Layout' view!)
Row pick Date PHLXGS DJUT
1 * 03/05 203.32 496.60
2 * 03/04 195.62 489.97
3 * 03/03 202.98 482.76
4 * 02/29 196.58 477.50
5 * 02/28 202.84 492.40
6 * 02/27 197.84 496.04
7 * 02/26 193.13 504.63
8 * 02/25 188.12 500.78
9 * 02/22 189.94 497.54
10 * 02/21 196.56 491.82
11 0 02/20 189.56 499.95
12 1 02/19 186.10 499.85
13 2 02/15 177.32 500.41
14 3 02/14 176.87 498.79
15 4 02/13 179.43 504.05
16 5 02/12 176.65 502.08
17 6 02/11 182.06 497.90
18 7 02/08 181.25 494.39
19 8 02/07 174.88 496.96
20 9 02/06 129.98 498.66
You are expected to work with 11 of the 20 observations shown above. Use the first 10 rows of data and pick one more row by finding the row marked with the second-to-last digit of your student number.(i) Compute the sample mean and standard deviation of , (ii) Compute or and or , (iii) Compute the sample mean and variance of from the results in (i) and (ii).(iv) The coefficient of variation is computed by dividing the standard deviation by the mean. Compute a coefficient of variation for , and and compare the relative safety of investing in precious metal stocks, investing in utilities and doing both.
251solngr2-081b 6/18/08 (Open this document in 'Page Layout' view!)
Solution:Since you were not supposed to do this problem, I am just going to present the answer with the original numbers.
(i) Compute the sample mean and standard deviation of .
Row
1 03/05 203.32 496.60 41339.0 246612 100969
2 03/04 195.62 489.97 38267.2 240071 95848
3 03/03 202.98 482.76 41200.9 233057 97991
4 02/29 196.58 477.50 38643.7 228006 93867
5 02/28 202.84 492.40 41144.1 242458 99878
6 02/27 197.84 496.04 39140.7 246056 98137
7 02/26 193.13 504.63 37299.2 254651 97459
8 02/25 188.12 500.78 35389.1 250781 94207
9 02/22 189.94 497.54 36077.2 247546 94503
10 02/21 196.56 491.82 38635.8 241887 96672
11 02/20 189.56 499.95 35933.0 249950 94771
12 02/19 186.10 499.85 34633.2 249850 93022
13 02/15 177.32 500.41 31442.4 250410 88733
14 02/14 176.87 498.79 31283.0 248791 88221
15 02/13 179.43 504.05 32195.1 254066 90442
16 02/12 176.65 502.08 31205.2 252084 88692
17 02/11 182.06 497.90 33145.8 247904 90648
18 02/08 181.25 494.39 32851.6 244421 89608
19 02/07 174.88 496.96 30583.0 246969 86908
20 02/06 129.98 498.66 16894.8 248662 64816
3721.03 9923.08 697304 4924233 1845390
To summarize the results of these computations , and . Thus and . . Minitab says 263.201. Minitab says 16.2235.
Though the mean and variance of were not requested, we will need them. Minitab says 45.1342. Minitab says 6.7167.
(ii) Compute or and or .
. Minitab says -42.8130
. Minitab says -.3928.
(iii) Compute the sample mean and variance of from the results in (i) and (ii).
We have , , , and .
So and = 222.6374 .
(iv) The coefficient of variation is computed by dividing the standard deviation by the mean. Compute a coefficient of variation for , and and compare the relative safety of investing in precious metal stocks, investing in utilities and doing both.
, and .
, and . This seems to show that investing in precious metals is much more (over 6 times as) risky than either utilities or a 50-50 strategy of doing both. However, because of the negative covariance, the 50-50 strategy is only about 62% riskier than utilities alone.