Fall 2013: Mercer University
Econ 353: Introduction to Econometrics
Homework 3:Answers
1. Using the data in RDCHEM.RAW for 32 firms in the chemical industry, the following equation would be
estimated:
𝒓𝒅𝒊𝒏𝒕𝒆𝒏𝒔 = 𝜶 + 𝜷𝟏 𝒍𝒐𝒈(𝒔𝒂𝒍𝒆𝒔) + 𝜷𝟐 𝒑𝒓𝒐𝒇𝒎𝒂𝒓𝒈 + 𝝁𝒕
Where, rdintens=R&D spending as a %of sales; sales=firm sales (in million $); profmarg=profit as a % of
sales. Now do the following:
a) Test the null hypothesis that 𝐻0: 𝜷𝟐 = 𝟎 against the alternative hypothesis that𝐻1: 𝜷𝟐 > 𝟎.
Explain what the null hypothesis is testing. Then report the t-statistics associated with the test.
Report the p-values for the test. Do you accept or reject the null hypothesis. Do the test for 5%
and 10% level of significance. Explain your results.
Answer: The null hypothesis is assuming that there is no effect of profit margin on the R&D
expenditures made by firms. The value of the t-test statistics is 1.09 which is less than critical
t-values at 5% level of significance and 10% level of significance which are (with n=32) 1.697
and 1.310 respectively for a one-tail test. Therefore, at the 5% level of significance we accept
the null hypothesis and at the 10% level of significance, we also accept the null hypothesis. ,
The p-value is 0.147. Since, the p-value is greater than both 0.05 and 0.1; we accept the null
hypothesis at both 5% and 10% level of significance.
b) Test the null hypothesis that 𝐻0: 𝜷𝟐 = 𝟎 against the alternative that𝐻1: 𝜷𝟐 #𝟎. Explain what the
null hypothesis is testing. Then report the t-statistics associated with the test. Report the pvalues for the test. Do you accept or reject the null hypothesis. Do the test for 5% and 10% level
of significance. Explain your results.
Answer: The null hypothesis is assuming that there is no effect profit margin on the R&D
expenditures made by firms. The value of the t-test statistics is 1.09 which is less than critical
t-values at 5% level of significance and 10% level of significance which are (with n=32) 2.042
and 1.697 respectively for a two-tail test. Therefore, at the 5% level of significance we accept
the null hypothesis and at the 10% level of significance, we accept the null hypothesis. , The pvalue is 0.147. Since, the p-value is greater than both 0.05 and 0.1; we accept the null
hypothesis at both 5% and 10% level of significance.
c) Test the null hypothesis that 𝐻0: 𝜷𝟐 = 𝟏 against the alternative that𝐻1: 𝜷𝟐 #𝟏. Explain what the
null hypothesis is testing. Then report the t-statistics associated with the test. Report the pvalues for the test. Do you accept or reject the null hypothesis. Do the test for 5% and 10% level
of significance. Explain your results.
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Answer: The null hypothesis is assuming that there is no effect profit margin on the R&D
expenditures made by firms. The estimated value of beta2 and standard error is .0500367 and
.0457768 respectively. Therefore, the value of the t-test statistics is = (0.0500367-1)/
.0457768=-20.752068 whose absolute value is more than critical t-values at 5% level of
significance and 10% level of significance which are (with n=32) 2.042 and 1.697 respectively
for a two-tail test. Therefore, at the 5% level of significance we reject the null hypothesis and
at the 10% level of significance, we reject the null hypothesis. , The p-value is 0.0. Since, the pvalue is less than both 0.05 and 0.1; we reject the null hypothesis at both 5% and 10% level of
significance.
d) Test the Joint null hypothesis that 𝐻0: 𝜷𝟐 = 𝟎, 𝜷𝟏 = 𝟎 against the alternative
that𝐻1: 𝜷𝟏 #𝟎, 𝜷𝟐 #𝟎, . Explain what the null hypothesis is testing. Then report the F-statistics
associated with the test. Report the p-values for the test. Do the test for 5% and 10% level of
significance. Do you accept or reject the null hypothesis. Explain your results.
Answer: the null hypothesis is assuming that there is no effect of either sales or profit margin
on the R&D. This is a joint test. The calculated F-test statistics is 1.58 which is less than critical
value at 5% level of significance and 5% level of significance which are 3.33 and 2.50
respectively. Therefore, we can accept the null hypothesis. The calculated p-value for the test
is 0.2224. Since the calculated p-value is greater than both 0.05 and 0.1, we can also accept
the null hypothesis at both 5 and 10% level of significance.
2. Download the “wage2.dta file from blackboard. Run the following regression:
𝑒𝑑𝑢𝑐𝑡 = 𝛽0 + 𝛽1 𝑏𝑙𝑎𝑐𝑘𝑡 + 𝛽2 𝑚𝑒𝑑𝑢𝑐𝑡 + 𝛽3 𝑓𝑒𝑑𝑢𝑐𝑡 +𝛽4 𝐼𝑄𝑡 + 𝛽5 𝑠𝑖𝑏𝑠𝑡 + 𝑈𝑡
Now, do the following:
a) Test the null hypothesis that 𝐻0: 𝜷𝟐 = 𝟎 against the alternative that𝐻1: 𝜷𝟐 < 𝟎. Explain
what the null hypothesis is testing. Then report the t-statistics associated with the test.
Report the p-values for the test. Do you accept or reject the null hypothesis. Do the test for
5% and 10% level of significance. Explain your results.
Answer: The null hypothesis is assuming no effect of the mothers’ education on the
education level of anyone in the sample. The value of the t-test statistics is 2.70 which is
more than critical t-values at 5% level of significance and 10% level of significance which
are (with n=722) 1.645 and 1.282 respectively for a one-tail test. Therefore, at the 5%
level of significance we reject the null hypothesis and at the 10% level of significance, we
also reject the null hypothesis. , The p-value is 0.007. Since, the p-value is smaller than
both 0.05 and 0.1; we reject the null hypothesis at both 5% and 10% level of significance
b) Test the null hypothesis that 𝐻0: 𝜷𝟒 = 𝟏 against the alternative that 𝐻1: 𝜷𝟒 #𝟏. Explain
what the null hypothesis is testing. Then report the t-statistics associated with the test.
Report the p-values for the test. Do you accept or reject the null hypothesis. Do the test for
5% and 10% level of significance. Explain your results.
2
Answer: The null hypothesis is assuming that there is 1-to-1 effect of IQ on the
EDUCATION. This means that an increase in IQ by one point would increase years of
education by one year. The estimated value of beta4 and standard error is .06758 and
.0051098 respectively. Therefore, the value of the t-test statistics is = (.06758 -1)/
.0051098 =-182.47 whose absolute value is more than critical t-values at 5% level of
significance and 10% level of significance which are (with n=722) 1.960 and 1.645
respectively for a two-tail test. Therefore, at the 5% level of significance we reject the null
hypothesis and at the 10% level of significance, we reject the null hypothesis. , The p-value
is 0.0. Since, the p-value is less than both 0.05 and 0.1; we reject the null hypothesis at
both 5% and 10% level of significance.
c) Test the Joint null hypothesis that 𝐻0: 𝜷𝟐 = 𝜷𝟑 against the alternative that 𝐻1: 𝜷𝟐 # 𝜷𝟑 .
Explain what the null hypothesis is testing. Then report the F-statistics associated with the
test. Report the p-values for the test. Do the test for 5% and 10% level of significance. Do
you accept or reject the null hypothesis. Explain your results.
Answer: The null hypothesis is assuming that equal effect of fathers’ and mothers’ education
on the EDUCATION. This is a joint test. The calculated F-test statistics is 2.34 which are less
than critical value at 5% level of significance and 10% level of significance which are 3.84 and
2.71 respectively. Therefore, we can accept the null hypothesis. The calculated p-value for the
test is 0.1269. Since the calculated p-value is greater than both 0.05 and 0.1, we can also
accept the null hypothesis at both 5 and 10% level of significance.
d) Test the null hypothesis that 𝐻0: 𝜷𝟐 = 𝜷𝟑 = 𝟎, 𝜷𝟒 = 𝟏 against the alternative
that 𝐻1: 𝜷𝟐 # 𝜷𝟑 , 𝜷𝟒 #𝟏 . Explain what the null hypothesis is testing. Then report the Fstatistics associated with the test. Report the p-values for the test. Do the test for 5% and
10% level of significance. Do you accept or reject the null hypothesis. Explain your results.
Answer: The null hypothesis is assuming that equal but zero effect of fathers’ and mothers’
education on the EDUCATION and a 1-to-1 effect of IQ. This is a joint test. The calculated F-test
statistics is 11860.85which are more than critical value at 5% level of significance and 10%
level of significance which are 2.60 and 2.08 respectively. Therefore, we can reject the null
hypothesis. The calculated p-value for the test is 0.0. Since the calculated p-value is smaller
than both 0.05 and 0.1, we can also reject the null hypothesis at both 5 and 10% level of
significance.
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