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Chapter 3

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CHAPTER 3
Exercise Solutions
54 Chapter 3, Exercise Solutions, Principles of Econometrics, 4e 55
EXERCISE 3.1
(a) The required interval estimator is b1 1 ce( ) . When b183.416, tc (0.975,38)2.024
and se(b1) 43.410, we get the interval estimate:
83.416 2.024 43.410 = (4.46, 171.30)
We estimate that 1 lies between 4.46 and 171.30. In repeated samples, 95% of similarly
constructed intervals would contain the true 1.
(b) test H 0 10 against H 1 1 we compute the t-value
b1 1 83.416 0
t1 1.92
se(b1) 43.410
Stiece t = 1.92 value does not exceed the 5% critical value tt c (0.975,38)2.024 , we do
not reject H 0. The data does not reject the zero-intercept hypothesis.
(c) The p-value 0.0622 represents the sum of the areas under the t distribution to the left of t =
1.92 and to the right of t = 1.92. Since the t distribution is symmetric, each of the tail
areas that make up the p-value are p/2 0.0622 2 0.0311. The level of significance, ,
is given by the sum of the areas under the PDF for ||t|t,c so the area under the curve for
tt is /2 .025 and likewise for tt . Therefore not rejecting the null hypothesis
c c
because /2 p/2, or p, is the same as not rejecting the null hypothesis because
tc c . From Figure xr3.1(c) we can see that having a p-value > 0.05 is equivalent to
having ttt .
c c
Figure xr3.1(c) Critical and observed t values
Chapter 3, Exercise Solutions, Principles of Econometrics, 4e 56
Exercise 3.1 (continued)
(d) Testing H 0 10 against H,1 10 uses the same t-value as in part (b), t = 1.92.
Because it is a one-tailed test, the critical value is chosen such that there is a probability of
0.05 in the right tail. That is, ct (0.95,38).686 . Since t = 1.92 > tc = 1.69, H 0 is
rejected, the alternative is accepted, and we conc lude that the intercept is positive. In this
case p-value = P(t > 1.92) = 0.0311. We see from Figure xr3.1(d) that having the p-value
< 0.05 is equivalent to having t > 1.69.
0.4
Rejection Region
0.3
PDF 0.2
0.1 1.92
0.0
-3 -2 -1 0 1 t 2 3
c
T
Figure xr3.1(d) Rejection region and observed t value
(e) The term "level of significance" is used to describe the probability of rejecting a true null
hypothesis when carrying out a hypothesis test. The term "level of confidence" refers to
the probability of an interval estimator yi elding an interval that includes the true
parameter. When carrying out a two-tailed test of the form H c: versus H c:,
0 k 1 k
non-rejection of H implies c lies within the confidence interval, and vice versa,
0
providing the level of significance is equal to one minus the level of confidence.
(f) False. The test in (d) uses the level of significance 5%, which is the probability of a Type
I error. That is, in repeated samples we ha ve a 5% chance of rejecting the null hypothesis
when it is true. The 5% significance is a probability statement about a procedure not a
probability statement about . It is careless and dangerous to equate 5% level of
1
significance with 95% confidence, which relate s to interval estimation procedures, not
hypothesis tests. Chapter 3, Exercise Solutions, Principles of Econometrics, 4e 57
EXERCISE 3.2
(a) The coefficient of EXPER indicates that, on average, a technical artist's quality rating goes
up by 0.076 for every additional year of experience.
3.9
3.8
3.7
3.6
3.5
RA3.4G
3.3
3.2
3.1
0 1 2 3 4 5 6 7 8
EXPER
Figure xr3.2(a) Estimated regression function
(b) Using the value tt 2.074 , the 95% confidence interval for is given by
c (0.975,2

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