54 Chapter 3, Exercise Solutions, Principles of Econometrics, 4e 55
(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
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
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 .
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.
-3 -2 -1 0 1 t 2 3
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,
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
significance with 95% confidence, which relate s to interval estimation procedures, not
hypothesis tests. Chapter 3, Exercise Solutions, Principles of Econometrics, 4e 57
(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.
0 1 2 3 4 5 6 7 8
Figure xr3.2(a) Estimated regression function
(b) Using the value tt 2.074 , the 95% confidence interval for is given by