Economics 345
Applied Econometrics
Problem Set 5
Prof: Martin Farnham
Problem sets in this course are ungraded. An answer key will be posted on the course
website within a few days of the release of each problem set. As noted in class, it is
highly recommended that you make every effort to complete these problems before
viewing the answer key.
Introduction: p-values for t-tests
You might wish to supplement what follows by browsing pages 133-135 of your text.
There are a couple reasons p-values can be useful to the econometrician (or someone
reading their research results). First, the p-value contains information about where in the
relevant t-distribution the t-statistic calculated for a particular null hypothesis lies. This
can eliminate the need to look up critical t-values for the chosen significance level of
your test. Second, if you just report that you reject the null at the 5% significance level
(for example), your reader doesn’t know whether this was a close call or not, unless they
take your t-stat and reference the t-table. If you give them a p-value, this tells them
whether your rejection of the null was a close call or whether it was a slam dunk (or
somewhere in between). I find the first reason particularly compelling.
Here’s an example. Suppose you’re testing the following hypothesis:
H 0 : β1 = 0, against the alternative
H 1 : β1 ≠ 0
If you’re performing this test at the 5% significance level, then you will reject the null if
your t-statistic lies at the 97.5 percentile or above, or if it lies at the 2.5 percentile or
below. In other words, the rejection regions sum up to an area of 0.05 under the tdistribution, so it’s the top 0.025 and the bottom 0.025 of the distribution (because the
alternative is two-sided).
The p-value can be interpreted as telling us the lowest significance level at which the null
can be rejected, given the t-statistic we have calculated under the null. So if we just
happened to get a p-value in this case of 0.05, this would tell us that our t-statistic lies
either at the 2.5 percentile of the 97.5 percentile (and we could tell which, by looking at
whether the t-stat is positive or negative).
Suppose now that we obtained a p-value of 0.08. What would this tell us about where
our t-statistic lies in the relevant t-distribution? Well, for a two-sided alternative, if we’re
just on the border of rejection at the 8% significance level, this tells us our rejection
region would have to start at the 4th percentile on the LHS of the distribution, and at the
96th percentile on the RHS of the distribution. Note that this gives us a total rejection
area of 0.08. The fact that our p-value is 0.08 tells us that our t-stat lies at either the 4th or
96th percentile of the t-distribution. With a p-value of 0.08 we fail to reject the null at the
5% level, but we reject the null at the 10% significance level.
Note that EViews always assumes with t-tests that our alternative hypothesis is twosided. This means interpretation of the p-value is straightforward when we have a twosided alternative, but that we need to be careful interpreting the p-value when our test is
one-sided.
Consider now obtaining a p-value of 0.08 with a 1-sided alternative. Consider the
following setup.
H 0 : β1 = 0, against the alternative
H 1 : β1 > 0
Here, you know that your only rejection region lies in the right tail of the distribution.
Since EViews calculates p-values for a two-sided alternative, it is assuming that the top
4% AND bottom 4% of the t distribution comprise the rejection region. But you know
that if you calculated a t-stat that lay at the 4th percentile (or below), you would fail to
reject. So in fact, the p-value of 0.08 really corresponds to a rejection region that
comprises only the top 4% of the distribution (i.e. the 96th percentile and above). If your
rejection region comprises only 4% of the distribution, we say that you are testing at the
4% significance level.
Therefore, for one-sided alternatives of the t-test, you need to take EViews’ 2-sided pvalue and divide by two, to get the one-sided p-value. So you’d take your two-sided pvalue of 0.08, and divide by 2 to get 0.04. This would tell you that you could reject the
null at a minimum significance level of 4%. In other words, now you would reject the
null at both the 10% and 5% levels. You would fail to reject at the 1% level. You would
just barely reject at the 4% level.
Note that for an F-test this problem of one- versus two-sided p-values doesn’t exist. All
F-tests are one-sided (because the F-distribution doesn’t take on negative values), and so
EViews calculates a 1-sided p-value for all F-tests. Therefore you never need to worry
about dividing the p-value by 2 when doing a hypothesis test involving the F-statistic.
1) Suppose you are testing the hypothesis that beta1 equals zero, against the alternative
hypothesis that beta1 is less than zero.
a) State the setup of the hypothesis test formally.
b) Now suppose that when you estimate your model using your sample of data, you
obtain a p-value of 0.12 for your estimate of beta1 (and suppose that estimate is
negative). What is the lowest level of significance at which you can reject the null?
c) Will you reject the null at the 1% level? At the 5% level? At the 10% level?
d) Now suppose you obtain a p-value of 0.12 and your coefficient estimate is positive. At
what percentile in the t-distribution does your t-statistic lie? Will you reject the null at
the 10% level? What about at the 50% level? What is the lowest significance level at
which you could reject the null, in this case?
2) Suppose you are testing the null that beta1 equals 1, in a model you have just
estimated. Your alternative hypothesis is that beta1 is not equal to 1. You obtain a pvalue of 0.02 for the t-statistic generated for this test.
a) Can you reject this null at the 1% level? At the 5% level? At the 10% level?
b) Supposing your coefficient estimate is greater than 1, at what percentile in the tdistribution does your t-statistic lie?
c) Supposing your coefficient estimate is less than 1, at what percentile in the tdistribution does your t-statistic lie?
3) This question deals with how to calculate a p-value by hand, given that you’ve
calculated a t-statistic. I will assume that we have enough degrees of freedom so that we
can approximate the t-distribution with the standard normal distribution.
a) At what percentile in the standard normal distribution, does the value 1.96 lie?
b) At what percentile in the standard normal distribution, does the value -1.96 lie?
c) For a sufficiently large sample size, what would be the p-value associated with a tstatistic of -1.96 be, assuming a 2-sided alternative?
d) For a two-sided alternative, what would be the p-value associated with a t-stat of 1.6?
e) For a one-sided (positive) alternative, what would be the p-value associated with a tstat of 1.6?
4) Problems 4.5, 4.6, and 4.9 from text
Note for Problem 4.6 that housing assessments are estimated values of homes generated
by the folks who collect property taxes. So when the problem refers to “testing the
rationality of assessments…” it’s referring to asking whether official property
assessments track actual market prices.
5) Book Problems 5.2-5.3
6) Book Problem 6.3-6.4. Note: Problem 6.3 is looking at whether research and
development activity by a firm is affected by the level of sales of the firm.