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HW 4
Problem 1
(SW 5.1) Suppose a researcher, using data on class size (CS) and average test scores from 50
third-grade classes, estimates the OLS regression
ˆTestScore = 640.3
(23.5)
− 4.93
(2.02)
× CS R2 = 0.11, SER = 8.7.
(1) Construct a 95% confidence interval for β1, the regression slope coefficient.
(2) Calculate the p-value for the two-sided test of the null hypothesis 0. Do you reject the null
hypothesis at the 5% level? At the 1% level?
(3) Calculate the p-value for the two-sided test of the null hypothesis H0 : β1 = −5.0.
(4) Without doing any additional calculations, determine whether -5.0 is contained in the 95%
confidence interval for β1.
1
Ans:
(1) C.I. is given by:(
− 4.93− 1.96× s.e.(βˆ1), −4.93 + 1.96× s.e.(βˆ1)
)
= (−4.93− 1.96× 2.02, −4.93 + 1.96× 2.02)
= (−8.889, −0.9708)
(2) p-value is computed as:
t− statistic = −4.93− 0
s.e.(βˆ1)
=
−4.93
2.02
= −2.44.
Since sample size is reasonably large (n = 50), we can use normal approximations. Therefore,
p− value = Pr(|z| > |tact|) = Pr(|z| > 2.44) = 2× 0.00734 ≈ 0.015.
Hence, can reject at 5% significance level, but not at 1%.
(3) First, compute the t-statistic:
t− statistic = −4.93− (−5)
s.e.(βˆ1)
=
0.07
2.02
≈ 0.035.
Therefore,
p− value = Pr(|z| > |tact|) = Pr(|z| > 0.035) ≈ 2× 0.48405 = 0.97
Note that this p-value is much larger than 5%, so the null hypothesis cannot be rejected at 5%
significance level.
(4) It is contained in the 95% confidence interval since the H0 : β1 = −5 cannot be rejected at the
5% level (as per (3)).
2
Problem 2
(SW 5.4) Read the box “Economic Value of a Year of Education: Homoskedasticity or Heteroskedas-
ticity?” in Section 5.4 of our Stock and Watson textbook. Use the regression reported in Equation
(5.23), reproduced below, to answer the following:
ˆEarnings = −12.12
(1.36)
+ 2.37
(0.10)
× Y rsEd
(1) A randomly selected 30-year-old worker reports an education level of 16 years. What is the
worker’s expected average hourly earnings?
(2) A high school graduate (12 years of education is contemplating going to a community college
for a 2-year degree. How much are this worker’s average hourly earnings expected to increase?
(3) A high school counselor tells a student that, on average, college graduates earn $10 per hour
more than high school graduates. Is this statement consistent with the regression evidence? What
range of values is consistent with the regression evidence?
3
Ans:
(1) −12.12 + 2.37× 16 = $25.8 per hour.
(2) (−12.12 + 2.37× 14)− (−12.12 + 2.37× 12) = 2.37× 2 = $4.74 per hour.
(3) Going to college adds 4 years of education. Similar to (2), the expected increment in earnings
is 2.37 × 4 = $9.48 per hour. This seems about consistent with the counselor’s claim. We can do
better: the 95% C.I. for the earnings increment is
4×
(
(2.37− 1.96× 0.1), (2.37 + 1.96× 0.1)
)
= 4× (2.174, 2.566) = (8.696, 10.264).
This confidence interval contains the counselor’s claim ($10). Therefore, the counselor’s claim
cannot be rejected under a significance level of 5%.