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Running head: ECONOMETRICS
ECONOMETRICS
Student’s name
University Affiliation
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Running head: ECONOMETRICS
Question 1:
1. Calculate and report the mean and standard deviation of the wage variable for this sample.
Sw3
Statistics
wage
Valid
935
Missing
934
N
Mean
957.95
Std. Deviation
404.361
2. The calculated mean serves as an estimate of the true population mean. What is the standard
error of this estimate?
Statistics
wage
Valid
935
N
Missing
Std. Error of Mean
0
13.224
3. Conduct a test of the null hypothesis that the true population mean for wage is equal to 1,000
at the level of alpha=0.05.
Hypothesis
H0: true population mean for wage = 1,000
H1: true population mean for wage ≠ 1,000
H0: µ =1000
H1:µ ≠1000
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Running head: ECONOMETRICS
We first denote the null hypothesis and the alternative hypothesis according to the given question.
In this case we are only dealing with one set of data, wage variable; hence to test the hypothesis
we are going to use t-test. The population standard deviation is not known hence z-test cannot be
used.
One-Sample Statistics
N
wage
Mean
935
957.95
Std. Deviation
Std. Error Mean
404.361
13.224
In this output for testing the hypothesis, we have 935 observations (N), the mean number of the
wage variable is 957.95 while the standard deviation of the mean is 404.361.
The sd was gotten by (957.95/ square root of 935 = 404.361.
One-Sample Test
Test Value = 0
t
df
Sig. (2-tailed)
Mean Difference
95% Confidence Interval of the
Difference
Lower
wage
72.440
934
.000
957.945
931.99
Upper
983.90
The second part of the output gives the value of the statistical test. The second column of the output
gives us the t-test value, 72.440. To get this answer,(404.361-1)/404.361/ square root of 935. The
third column shows the degrees of freedom for the t-test. The fourth column gives us details of a
2-tailed p- value. However, in this case, we are having a on tailed test, hence will look at the critical
t values. The critical vale of t at 934(935-1) degrees of freedom, α=0.05 is .000. To know if we
will reject the null hypothesis, we check if the t value of the two tailed column should be less than
the observed value of t. In this case, the critical t is .000 while the observed t is 72.440, thus we
reject the null hypothesis. There is sufficient evidence to show that the population mean for wage
is not 1000.
Question 2
1. Consider the variables wage, IQ, and educ. For each variable, report the mean value, including the
unit of measurement. For example, “The mean value of wage is 957.95 dollars per month.”
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Running head: ECONOMETRICS
Statistics
wage
Valid
IQ
educ
935
935
935
0
0
0
Mean
957.95
101.28
13.47
Std. Error of Mean
13.224
.492
.072
N
Missing
The mean value of wage is 957.95 dollars per month.”
The mean value of IQ is 101.28 units”
The mean value of educ is 13.47 level of education”
2. Conduct a regression of wage on educ. Report and interpret the estimated coefficient on educ.
Model Summary
Model
R
R Square
.327a
1
Adjusted R
Std. Error of the
Square
Estimate
.107
.106
382.320
a. Predictors: (Constant), educ
ANOVAa
Model
Sum of Squares
Regression
1
df
Mean Square
16340644.545
1
16340644.545
Residual
136375523.673
933
146168.836
Total
152716168.218
934
a. Dependent Variable: wage
b. Predictors: (Constant), educ
F
111.793
Sig.
.000b
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Running head: ECONOMETRICS
Co efficientsa
Model
Unstandardized Coefficients
Standardized
t
Sig.
Coefficients
B
(Constant)
Std. Error
Beta
146.952
77.715
60.214
5.695
1.891
.059
10.573
.000
1
educ
.327
a. Dependent Variable: wage
In this regression analysis, I have ensured I have utmost six assumptions. One of them is
to make sure that the two variables are continuous; they have a linear relationship, no significant
outliers that could be seen. There was also independence of observations together with
homoscedasticity and checked that the residual errors were normally distributed.
In our case, the first table, we have the model summary. The table provides the R and the
R2. The R value gives the correlation between wage and educ variable. The value is 0.37. This
shows a weak linear relationship between the two variables. The R2 shows the variation that exists
in the dependent variable that can be explained by the independent variable. In our case, only
10.7% of the variation of the dependent variable can be explained by the independent variable.
Next table is the ANOVA table. This table predicts if the dependent variable is significant. In
this case the p-value is <0.05 hence indicating that the overall model is statistically significant, hence predicts well for the outcome variable and a good fit for the data. The last table it a coefficients table. Here it helps us predict the wage given the education level as well as determine how the educ variable contributes to the whole model. In our case the coefficient for educ is 60.214 which is significant since it has a p-value<0.05. 3. Conduct a regression of wage on IQ. Report and interpret the estimated coefficient on IQ. What is the predicted wage for a person with the average level of IQ? Model Summary Model 1 R R Square .309a .096 Adjusted R Std. Error of the Square Estimate .095 a. Predictors: (Constant), IQ ANOVAa 384.767 6 Running head: ECONOMETRICS Model Sum of Squares Regression 1 df Mean Square 14589782.589 1 14589782.589 Residual 138126385.629 933 148045.429 Total 152716168.218 934 F Sig. 98.549 .000b a. Dependent Variable: wage b. Predictors: (Constant), IQ Coefficientsa Model Unstandardized Coefficients Standardized t Sig. Coefficients B (Constant) Std. Error 116.992 85.642 8.303 .836 Beta 1.366 .172 9.927 .000 1 IQ .309 a. Dependent Variable: wage In our case, the first table, we have the model summary. The table provides the R and the R2. The R value gives the correlation between wage and educ variable. The value is 0.309. This shows a weak linear relationship between the two variables. The R2 shows the variation that exists in the dependent variable that can be explained by the independent variable. In our case, only 9.6 % of the variation of the dependent variable can be explained by the independent variable. Next table is the ANOVA table. This table predicts if the dependent variable is significant. In this case the p-value is <0.05 hence indicating that the overall model is statistically significant, hence predicts well for the outcome variable and a good fit for the data. The last table it a coefficients table. Here it helps us predict the wage given the education level as well as determine how the educ variable contributes to the whole model. In our case the coefficient for IQ is 8.303 which is significant since it has a p-value<0.05. 4. Compare the previous two regressions. Which of the two independent variables has greater explanatory power for monthly earnings? How do you know? In the two regressions output, the educ variable has a greater explanatory power for monthly earnings. This is because it has a coefficient of determination of 10.7%, higher than that of 7 Running head: ECONOMETRICS IQ which is 9.6. This means that the educ variable has a higher capacity of explaining the variation in the wage variable. Question 3: 1. Estimate the following model: = 0 + 1 + 2 ℎ + 3 + Report the coefficient estimates and standard errors for = −190.881 + 76.736 − 1.995 ℎ + 17.552 + Model Summary Model R R Square .370a 1 Adjusted R Std. Error of the Square Estimate .137 .134 376.222 a. Predictors: (Constant), exper, hours, educ ANOVAa Model Sum of Squares Regression 1 df Mean Square 20939351.246 3 6979783.749 Residual 131776816.973 931 141543.305 Total 152716168.218 934 F Sig. 49.312 .000b a. Dependent Variable: wage b. Predictors: (Constant), exper, hours, educ Coefficientsa Model Unstandardized Coefficients Standardized t Sig. Coefficients B (Constant) Std. Error -190.881 128.089 76.736 6.311 Beta -1.490 .137 12.159 .000 1 educ .417 8 Running head: ECONOMETRICS hours -1.995 1.712 -.036 -1.166 .244 exper 17.552 3.162 .190 5.551 .000 a. Dependent Variable: wage 2. Interpret the coefficient estimates for 1 , 2 , 3 . Consider both the point estimates (actual number estimated) and the standard error for each coefficient. Include the units of measurement in your interpretations. The coefficient of educ is 76.736. for every change in 1 unit of IQ there is corresponding change of 76.736 The coefficient of hours is -1.995 for every change in 1 unit of Educ there is corresponding change of -1.995. The coefficient of exper is 17.552 for every change in 1 unit of Educ there is corresponding change of 17.552 3. Report the average values of education, hours, and work experience for this sample. What is the expected wage for a person with average levels of education, hours, and work experience? = −190.881 + 76.736 − 1.995 ℎ + 17.552 + given µ=0 =17,435.86 4. What is the expected wage for a highly educated person with no experience who does not work? Predict the value of wage when educ is at the maximum observed value, hours = 0, and exper = 0. Is this a reasonable prediction? Explain. 5. = −190.881 + 76.736 − 1.995 ℎ + 17.552 + = -114.145 This scenario is impossible since the wage is coming to a negative number. In real life, you cannot work 0 hours and get a wage, in return, you are going to incur some expenditure money. ... Purchase answer to see full attachment