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1.Distributed via email are data from a survey on concern about global warming and the policy
options available. The survey was conducted over the Internet.
A.Unzip the data; translate it into STATA.
B.Calculate the sample statistics for the variables. There are two variants, unweighted and
weighted. Weighting corrects for the fact that there are biases in the sampling frame. What is
the weighting variable in the data set? You can weight the data by adding the subcommand
[aw=weight].
C.Construct a table that shows the joint density of religiosity and concern about global
warming. Construct a table that shows the joint density of concern about global warming
and what we should do about it?
D.Calculate the conditional probabilities and interpret them.
E.Regress these variables on each other (concern on religiosity, and policy response on
concern). Interpret the coefficients and relate them to the conditional probabilities in D.
2.
A.Generate a joint normal random variable using the following commands.
set obs 100000
gen x = invnorm(uniform())
gen y = 1 + .4*x + invnorm(uniform())
Make a graph displaying the relationship between the variables.
excelhomeworkhelp.com
B.Compute the means, standard deviations, and correlation between the variables. How
does this relate to the values input in the simulation?
C.Choose the highest 5 percent of cases of Y and the lowest 5 percent of cases of Y. Extract the
conditional distributions of X|Y (make a graph and compute statistics using an if statement).
Choose the highest 5 percent of cases of X and the lowest 5 percent of cases of X. Extract the
conditional distributions of Y|X (make a graph and compute statistics using an if statement).
Compare the difference between the means of Y|X and the difference between the means of
X|Y. Why do they differ?
excelhomeworkhelp.com
Question 1. The data already has a weighting variable under the name of “weight.” I believe that
the variable use all different socio-economic or racial background to correct the sampling frame
bias. For the following conditional density and regression problems, I picked question 10 (global
warming concern), question 14b (policy response to global warming) and question 19 (religiosity).
Their sample statistics follows.
. sum q10 q14b q19
Variable | Obs Mean Std. Dev. Min Max
-------------+--------------------------------------------
--------- q10 | 1205 2.633195 1.273246
-1 5
q14b | 1205 .853112 3.142914 -2 7
q19 | 1205 1.880498 .7110843 -1 3
. sum q10 q14b q19 [aw=weight]
Variable | Obs Weight Mean Std. Dev. Min
Max
-------------+----------------------------------------
-------------------------
q10 | 1205 1205.1800 2.683956 1.299613 -1 5
q14b | 1205 1205.1800 .8744669 3.104373 -2 7
q19 | 1205 1205.1800 1.907383 .7080763 -1 3
excelhomeworkhelp.com
-----------+--------------------------------------------+----
------ Total | 11 319 667 208 | 1205
. tab2 q14b q10
-> tabulation of q14b by q10
| Q10
Q14B | -1 1 2 3 4 | Total
-----------+------------------------------------------------------
-+----------
-2 | 1 112 221 142 47 | 606
-1 | 0 0 1 2 0 | 15
1 | 0 1 5 7 2 | 26
2 | 0 27 62 39 5 | 147
3 | 0 11 51 19 3 | 93
4 | 0 7 24 24 4 | 63
5 | 2 38 61 40 3 | 165
6 | 0 15 15 9 2 | 51
7 | 0 0 0 12 16 | 39
-----------+-------------------------------------------------------
+---------- Total | 3 211 440 294 82 | 1205
excelhomeworkhelp.com
| Q10 Q14B | 5 | Total
-----------+-----------+----------
-2 | 83 | 606
-1 | 12 | 15
1 | 11 | 26
2 | 14 | 147
3 | 9 | 93
4 | 4 | 63
5 | 21 | 165
6 | 10 | 51
7 | 11 | 39
-----------+-----------+----------
Total | 175 | 1205
These tables only show frequencies. By dividing the sample size of 1205, you can have a
probability tables.
Joint Density Table for q10 and q19
Q19
Q10 -1 1 2 3 Total
-1 0 0 0.002 0.001 0.002
1 0.001 0.041 0.094 0.040 0.175
2 0.002 0.088 0.214 0.061 0.365
3 0.002 0.079 0.131 0.032 0.244
4 0.001 0.020 0.037 0.011 0.068
5 0.002 0.037 0.076 0.029 0.145
Total 0.009 0.265 0.554 0.173 1
excelhomeworkhelp.com
Joint Density Table for q10 and q14b
Q10
Q14B -1 1 2 3 4 5 Total
-2 0.001 0.093 0.183 0.118 0.039 0.069 0.503
-1 0.000 0.000 0.001 0.002 0.000 0.010 0.012
1 0.000 0.001 0.004 0.006 0.002 0.009 0.022
2 0.000 0.022 0.051 0.032 0.004 0.012 0.122
3 0.000 0.009 0.042 0.016 0.002 0.007 0.077
4 0.000 0.006 0.020 0.020 0.003 0.003 0.052
5 0.002 0.032 0.051 0.033 0.002 0.017 0.137
6 0.000 0.012 0.012 0.007 0.002 0.008 0.042
7 0.000 0.000 0.000 0.010 0.013 0.009 0.032
Total 0.002 0.175 0.365 0.244 0.068 0.145 1.000
The conditional probabilities can be obtained by using the formula of Pr(A|B) = Pr(A & B)/Pr(B).
For example, the conditional probability of policy response given different concerns about global
warming is
Q19
Q10 1 2 3
-1 0 0.002996 0.004797
1 0.153449 0.169271 0.230254
2 0.33195 0.386476 0.350179
3 0.297503 0.236679 0.182285
4 0.075159 0.065911 0.062361
5 0.140922 0.137813 0.167894
excelhomeworkhelp.com
Q10
Q14B -1 1 2 3 4 5
-2 0.414938 0.53112 0.502473 0.48296 0.57359 0.475032
-1 0 0 0.002274 0.006802 0 0.068679
1 0 0.004742 0.011368 0.023808 0.024408 0.062956
2 0 0.128038 0.140965 0.132644 0.06102 0.080126
3 0 0.052164 0.115955 0.064621 0.036612 0.05151
4 0 0.033195 0.054567 0.081627 0.048816 0.022893
5 0.829876 0.180202 0.138692 0.136045 0.036612 0.120189
6 0 0.071132 0.034104 0.03061 0.024408 0.057233
7 0 0 0 0.040814 0.195265 0.062956
From the first table, we see that the conditional probability of the level of concern about global
warming given a particular level of religiosity. For example, if the person is very religious (q19
= 1), the probability that the person says there is a serious global warming problem is .15 which is
least among religiosity category. In the second table, it is a conditional probability of policies
given the level of concern about global warming. This says that if a person believes that “there is
enough evidence that global warming is taking place” (q10-2), the probability that the person
thinks that “we should invest R&D”(q14b-2) is .141. The conditional probability table shows that
as the concern of global warming increases (from 4 to 1), at least less probability for answering
“do nothing(q14b-1).
excelhomeworkhelp.com
-------------+------------------------------ F( 1, 1203) = 3.51
Model | 5.68303187 1
5.68303187
Prob > F = 0.0611
Residual | 1946.18917 1203
1.61777986
R-squared = 0.0029
-------------+------------------------------ Adj R-squared = 0.0021
Total | 1951.8722 1204 1.62115631 Root MSE = 1.2719
q10 | Coef. Std. Err. t P>|t| [95% Conf.
Interval]
-------------+--------------------------------------------
-------------------- q19 | -.0966175 .0515496
-1.87 0.061 -.1977546 .0045196
_cons | 2.814884 .1036326 27.16 0.000 2.611563
3.018205
In the regression result, you can get a little different outcome. From the conditional probability
tables, we have some confidence that the more a person is religious, s/he does not really think
that global warming is an imminent problem. However,
. reg q10 q19
Source | SS df MS Number of obs = 1205
excelhomeworkhelp.com
The regression result does not guarantee any statistically significant causation relationship between
concern about global warming by religiosity. The same story goes for policy responses and global
warming concern.
. reg q14b q10
Source | SS df MS Number of obs = 1205
-------------+------------------------------ F( 1, 1203) = 0.11
Model | 1.040950521
1.04095052
Prob > F = 0.7456
Residual | 11891.9599 1203
9.88525343
R-squared = 0.0001
-------------+------------------------------ Adj R-
squared
= -
0.0007
Total | 11893.0008 1204
9.87790767
Root MSE = 3.1441
q14b |Coef. Std. Err.t P>|t| [95% Conf. Interval]
-------------+---------------------------------------------------
------------- q10 |.0230935 .0711653 0.32 0.746 -
.1165284 .1627154
_cons | .7923024 .2081329 3.81 0.000 .3839585
1.200646
excelhomeworkhelp.com
-------------+------------------------------ F( 1, 1017) = 8.84
Model | 6.61127946 1 6.61127946 Prob > F = 0.0030
Residual | 760.440732 1017
.747729334
R-squared = 0.0086
-------------+------------------------------ Adj R-squared = 0.0076
Total | 767.052012 1018
.753489206
Root MSE = .86471
q10 | Coef. Std. Err. t P>|t| [95% Conf. Interval]
-------------+------------------------------------------------------
---------- q19 | -.1231698 .0414222 -2.97 0.003 -
.2044526 -.0418869
_cons | 2.47248 .0832298 29.71 0.000 2.309158
2.635801
Coding problem may have caused these results. As you have seen in the file, there are a huge
number of “no opinion” or “I don’t know” answers. Also, many of variables are not coded with
ordinal concerns. Therefore, I transformed codings of concern about global warming question.
“No opinion” and “refused” were coded as missing variables (not always recommended. There
are enormous literatures about missing variables and how they can cause problems… anyway,)
and saw the causal relationship between religiosity and global warming concern.
. reg q10 q19
Source | SS df MS Number of obs = 1019
excelhomeworkhelp.com
Summary statistics for truncated data sets given upper 5 percent and lower 5 percent of each
variable follow:
. sum y if x1 !=.
Variable | Obs Mean Std. Dev. Min Max
-------------+--------------------------------------------
--------- y |5000 .1667309 1.01062 -3.127756
3.890636
. sum y if x2 !=.
Variable | Obs Mean Std. Dev. Min Max
-------------+--------------------------------------------
--------- y | 5000 1.843936 1.003058 -
1.397682 5.711002
The result is different from the first regression of q10 on q19. Now, we have a negative
coefficient which is statistically significant. Therefore, as the person becomes more religious,
s/he is less concerned about global warming problem.
Question 2. Please see attached do-file for the process I created variables. Summary statistics
are :
excelhomeworkhelp.com
. sum x if y1 !=.
Variable | Obs Mean Std. Dev. Min Max
-------------+--------------------------------------------
--------- x | 5000 -.7615851 .9547275 -
4.518918 2.23154
. sum x if y2 !=.
Variable | Obs Mean Std. Dev. Min Max
-------------+--------------------------------------------
--------- x | 5000 .7779928 .9445251 -
2.47037 4.448323
As you see, once you have truncated data sets, you have now different means from original
datasets.
Finally, if we regress y on x, we have the result of:
. reg y x
excelhomeworkhelp.com
Variable Obs Mean Std. Dev. Min Max
X 100000 0.004265 1.002622 -4.51892 4.448323
Y 100000 1.006364 1.076699 -3.46255 5.711002
Source| SS df MS Numberof obs = 100000
-------------+------------------------------ F( 1, 99998) =16399.52
Model | 16333.219 1
16333.219
Prob > F = 0.0000
Residual | 99593.7378 99998
.995957297
R-squared = 0.1409
-------------+------------------------------ Adj R-squared = 0.1409
Total | 115926.957 99999
1.15928116
Root MSE = .99798
y | Coef. Std. Err. t P>|t| [95% Conf. Interval]
-------------+-----------------------------------------------
----------------- x | .4030887 .0031476 128.06
0.000 .3969194 .4092581
_cons | 1.004645 .0031559 318.34 0.000 .9984597
1.010831
excelhomeworkhelp.com
mean and 1 variance. Therefore, the variance of y is about .16 + 1 = .16, and standard
deviation is 1.077. Still, standard deviations of two variables are pretty similar and we expect
the correlation coefficient and regression coefficient will be similar, too. And they are.
. corr y x
(obs=100000)
| y x
-------------+------------------
y |1.0000
x |0.3754 1.0000
The coefficient of x is .4 and very close to our artificial coefficient for the creation of y
variable. (yeah, of course.) The standard error is very small, but still have some value.
That’s because we inserted error term at the end of y variable equation which gave some
noise in y and x relations.
The formula for correlation coefficient is Cov(x,y)/St.dv(x)St.dv(x), while the regression
coefficient formula (in bivariate case) is Cov(x,y)/Var(x). Since the variance of x and y are
very similar (almost same as one. Why? Think about the variable property of Var(a + bX) =
b^2Var(X). Here, b is .4, therefore, the squared term is only .16, which results in .16 x 1 =
.16. But, we have an error term at the end of y equation. Error term is also standard normal
with 0
excelhomeworkhelp.com

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Quantitative Methods Assignment Help

  • 1. For any Homework related queries, Call us at:- +1 678 648 4277 You can mail us at:- info@excelhomeworkhelp.com or reach us at:- www.excelhomeworkhelp.com
  • 2. 1.Distributed via email are data from a survey on concern about global warming and the policy options available. The survey was conducted over the Internet. A.Unzip the data; translate it into STATA. B.Calculate the sample statistics for the variables. There are two variants, unweighted and weighted. Weighting corrects for the fact that there are biases in the sampling frame. What is the weighting variable in the data set? You can weight the data by adding the subcommand [aw=weight]. C.Construct a table that shows the joint density of religiosity and concern about global warming. Construct a table that shows the joint density of concern about global warming and what we should do about it? D.Calculate the conditional probabilities and interpret them. E.Regress these variables on each other (concern on religiosity, and policy response on concern). Interpret the coefficients and relate them to the conditional probabilities in D. 2. A.Generate a joint normal random variable using the following commands. set obs 100000 gen x = invnorm(uniform()) gen y = 1 + .4*x + invnorm(uniform()) Make a graph displaying the relationship between the variables. excelhomeworkhelp.com
  • 3. B.Compute the means, standard deviations, and correlation between the variables. How does this relate to the values input in the simulation? C.Choose the highest 5 percent of cases of Y and the lowest 5 percent of cases of Y. Extract the conditional distributions of X|Y (make a graph and compute statistics using an if statement). Choose the highest 5 percent of cases of X and the lowest 5 percent of cases of X. Extract the conditional distributions of Y|X (make a graph and compute statistics using an if statement). Compare the difference between the means of Y|X and the difference between the means of X|Y. Why do they differ? excelhomeworkhelp.com
  • 4. Question 1. The data already has a weighting variable under the name of “weight.” I believe that the variable use all different socio-economic or racial background to correct the sampling frame bias. For the following conditional density and regression problems, I picked question 10 (global warming concern), question 14b (policy response to global warming) and question 19 (religiosity). Their sample statistics follows. . sum q10 q14b q19 Variable | Obs Mean Std. Dev. Min Max -------------+-------------------------------------------- --------- q10 | 1205 2.633195 1.273246 -1 5 q14b | 1205 .853112 3.142914 -2 7 q19 | 1205 1.880498 .7110843 -1 3 . sum q10 q14b q19 [aw=weight] Variable | Obs Weight Mean Std. Dev. Min Max -------------+---------------------------------------- ------------------------- q10 | 1205 1205.1800 2.683956 1.299613 -1 5 q14b | 1205 1205.1800 .8744669 3.104373 -2 7 q19 | 1205 1205.1800 1.907383 .7080763 -1 3 excelhomeworkhelp.com
  • 5. -----------+--------------------------------------------+---- ------ Total | 11 319 667 208 | 1205 . tab2 q14b q10 -> tabulation of q14b by q10 | Q10 Q14B | -1 1 2 3 4 | Total -----------+------------------------------------------------------ -+---------- -2 | 1 112 221 142 47 | 606 -1 | 0 0 1 2 0 | 15 1 | 0 1 5 7 2 | 26 2 | 0 27 62 39 5 | 147 3 | 0 11 51 19 3 | 93 4 | 0 7 24 24 4 | 63 5 | 2 38 61 40 3 | 165 6 | 0 15 15 9 2 | 51 7 | 0 0 0 12 16 | 39 -----------+------------------------------------------------------- +---------- Total | 3 211 440 294 82 | 1205 excelhomeworkhelp.com
  • 6. | Q10 Q14B | 5 | Total -----------+-----------+---------- -2 | 83 | 606 -1 | 12 | 15 1 | 11 | 26 2 | 14 | 147 3 | 9 | 93 4 | 4 | 63 5 | 21 | 165 6 | 10 | 51 7 | 11 | 39 -----------+-----------+---------- Total | 175 | 1205 These tables only show frequencies. By dividing the sample size of 1205, you can have a probability tables. Joint Density Table for q10 and q19 Q19 Q10 -1 1 2 3 Total -1 0 0 0.002 0.001 0.002 1 0.001 0.041 0.094 0.040 0.175 2 0.002 0.088 0.214 0.061 0.365 3 0.002 0.079 0.131 0.032 0.244 4 0.001 0.020 0.037 0.011 0.068 5 0.002 0.037 0.076 0.029 0.145 Total 0.009 0.265 0.554 0.173 1 excelhomeworkhelp.com
  • 7. Joint Density Table for q10 and q14b Q10 Q14B -1 1 2 3 4 5 Total -2 0.001 0.093 0.183 0.118 0.039 0.069 0.503 -1 0.000 0.000 0.001 0.002 0.000 0.010 0.012 1 0.000 0.001 0.004 0.006 0.002 0.009 0.022 2 0.000 0.022 0.051 0.032 0.004 0.012 0.122 3 0.000 0.009 0.042 0.016 0.002 0.007 0.077 4 0.000 0.006 0.020 0.020 0.003 0.003 0.052 5 0.002 0.032 0.051 0.033 0.002 0.017 0.137 6 0.000 0.012 0.012 0.007 0.002 0.008 0.042 7 0.000 0.000 0.000 0.010 0.013 0.009 0.032 Total 0.002 0.175 0.365 0.244 0.068 0.145 1.000 The conditional probabilities can be obtained by using the formula of Pr(A|B) = Pr(A & B)/Pr(B). For example, the conditional probability of policy response given different concerns about global warming is Q19 Q10 1 2 3 -1 0 0.002996 0.004797 1 0.153449 0.169271 0.230254 2 0.33195 0.386476 0.350179 3 0.297503 0.236679 0.182285 4 0.075159 0.065911 0.062361 5 0.140922 0.137813 0.167894 excelhomeworkhelp.com
  • 8. Q10 Q14B -1 1 2 3 4 5 -2 0.414938 0.53112 0.502473 0.48296 0.57359 0.475032 -1 0 0 0.002274 0.006802 0 0.068679 1 0 0.004742 0.011368 0.023808 0.024408 0.062956 2 0 0.128038 0.140965 0.132644 0.06102 0.080126 3 0 0.052164 0.115955 0.064621 0.036612 0.05151 4 0 0.033195 0.054567 0.081627 0.048816 0.022893 5 0.829876 0.180202 0.138692 0.136045 0.036612 0.120189 6 0 0.071132 0.034104 0.03061 0.024408 0.057233 7 0 0 0 0.040814 0.195265 0.062956 From the first table, we see that the conditional probability of the level of concern about global warming given a particular level of religiosity. For example, if the person is very religious (q19 = 1), the probability that the person says there is a serious global warming problem is .15 which is least among religiosity category. In the second table, it is a conditional probability of policies given the level of concern about global warming. This says that if a person believes that “there is enough evidence that global warming is taking place” (q10-2), the probability that the person thinks that “we should invest R&D”(q14b-2) is .141. The conditional probability table shows that as the concern of global warming increases (from 4 to 1), at least less probability for answering “do nothing(q14b-1). excelhomeworkhelp.com
  • 9. -------------+------------------------------ F( 1, 1203) = 3.51 Model | 5.68303187 1 5.68303187 Prob > F = 0.0611 Residual | 1946.18917 1203 1.61777986 R-squared = 0.0029 -------------+------------------------------ Adj R-squared = 0.0021 Total | 1951.8722 1204 1.62115631 Root MSE = 1.2719 q10 | Coef. Std. Err. t P>|t| [95% Conf. Interval] -------------+-------------------------------------------- -------------------- q19 | -.0966175 .0515496 -1.87 0.061 -.1977546 .0045196 _cons | 2.814884 .1036326 27.16 0.000 2.611563 3.018205 In the regression result, you can get a little different outcome. From the conditional probability tables, we have some confidence that the more a person is religious, s/he does not really think that global warming is an imminent problem. However, . reg q10 q19 Source | SS df MS Number of obs = 1205 excelhomeworkhelp.com
  • 10. The regression result does not guarantee any statistically significant causation relationship between concern about global warming by religiosity. The same story goes for policy responses and global warming concern. . reg q14b q10 Source | SS df MS Number of obs = 1205 -------------+------------------------------ F( 1, 1203) = 0.11 Model | 1.040950521 1.04095052 Prob > F = 0.7456 Residual | 11891.9599 1203 9.88525343 R-squared = 0.0001 -------------+------------------------------ Adj R- squared = - 0.0007 Total | 11893.0008 1204 9.87790767 Root MSE = 3.1441 q14b |Coef. Std. Err.t P>|t| [95% Conf. Interval] -------------+--------------------------------------------------- ------------- q10 |.0230935 .0711653 0.32 0.746 - .1165284 .1627154 _cons | .7923024 .2081329 3.81 0.000 .3839585 1.200646 excelhomeworkhelp.com
  • 11. -------------+------------------------------ F( 1, 1017) = 8.84 Model | 6.61127946 1 6.61127946 Prob > F = 0.0030 Residual | 760.440732 1017 .747729334 R-squared = 0.0086 -------------+------------------------------ Adj R-squared = 0.0076 Total | 767.052012 1018 .753489206 Root MSE = .86471 q10 | Coef. Std. Err. t P>|t| [95% Conf. Interval] -------------+------------------------------------------------------ ---------- q19 | -.1231698 .0414222 -2.97 0.003 - .2044526 -.0418869 _cons | 2.47248 .0832298 29.71 0.000 2.309158 2.635801 Coding problem may have caused these results. As you have seen in the file, there are a huge number of “no opinion” or “I don’t know” answers. Also, many of variables are not coded with ordinal concerns. Therefore, I transformed codings of concern about global warming question. “No opinion” and “refused” were coded as missing variables (not always recommended. There are enormous literatures about missing variables and how they can cause problems… anyway,) and saw the causal relationship between religiosity and global warming concern. . reg q10 q19 Source | SS df MS Number of obs = 1019 excelhomeworkhelp.com
  • 12. Summary statistics for truncated data sets given upper 5 percent and lower 5 percent of each variable follow: . sum y if x1 !=. Variable | Obs Mean Std. Dev. Min Max -------------+-------------------------------------------- --------- y |5000 .1667309 1.01062 -3.127756 3.890636 . sum y if x2 !=. Variable | Obs Mean Std. Dev. Min Max -------------+-------------------------------------------- --------- y | 5000 1.843936 1.003058 - 1.397682 5.711002 The result is different from the first regression of q10 on q19. Now, we have a negative coefficient which is statistically significant. Therefore, as the person becomes more religious, s/he is less concerned about global warming problem. Question 2. Please see attached do-file for the process I created variables. Summary statistics are : excelhomeworkhelp.com
  • 13. . sum x if y1 !=. Variable | Obs Mean Std. Dev. Min Max -------------+-------------------------------------------- --------- x | 5000 -.7615851 .9547275 - 4.518918 2.23154 . sum x if y2 !=. Variable | Obs Mean Std. Dev. Min Max -------------+-------------------------------------------- --------- x | 5000 .7779928 .9445251 - 2.47037 4.448323 As you see, once you have truncated data sets, you have now different means from original datasets. Finally, if we regress y on x, we have the result of: . reg y x excelhomeworkhelp.com
  • 14. Variable Obs Mean Std. Dev. Min Max X 100000 0.004265 1.002622 -4.51892 4.448323 Y 100000 1.006364 1.076699 -3.46255 5.711002 Source| SS df MS Numberof obs = 100000 -------------+------------------------------ F( 1, 99998) =16399.52 Model | 16333.219 1 16333.219 Prob > F = 0.0000 Residual | 99593.7378 99998 .995957297 R-squared = 0.1409 -------------+------------------------------ Adj R-squared = 0.1409 Total | 115926.957 99999 1.15928116 Root MSE = .99798 y | Coef. Std. Err. t P>|t| [95% Conf. Interval] -------------+----------------------------------------------- ----------------- x | .4030887 .0031476 128.06 0.000 .3969194 .4092581 _cons | 1.004645 .0031559 318.34 0.000 .9984597 1.010831 excelhomeworkhelp.com
  • 15. mean and 1 variance. Therefore, the variance of y is about .16 + 1 = .16, and standard deviation is 1.077. Still, standard deviations of two variables are pretty similar and we expect the correlation coefficient and regression coefficient will be similar, too. And they are. . corr y x (obs=100000) | y x -------------+------------------ y |1.0000 x |0.3754 1.0000 The coefficient of x is .4 and very close to our artificial coefficient for the creation of y variable. (yeah, of course.) The standard error is very small, but still have some value. That’s because we inserted error term at the end of y variable equation which gave some noise in y and x relations. The formula for correlation coefficient is Cov(x,y)/St.dv(x)St.dv(x), while the regression coefficient formula (in bivariate case) is Cov(x,y)/Var(x). Since the variance of x and y are very similar (almost same as one. Why? Think about the variable property of Var(a + bX) = b^2Var(X). Here, b is .4, therefore, the squared term is only .16, which results in .16 x 1 = .16. But, we have an error term at the end of y equation. Error term is also standard normal with 0 excelhomeworkhelp.com