# Why does the ADF test have a higher critical value than its t-statistic?

I did a ADF test to my data set to find unit root using STATA. I get t-test test statistic which is -2.363 from ADF. I think it is much lower than -1.96 so it can reject the null. But actually its critical value is much higher that normal value like -1.96. What happen in this case? Why critical value is much higher that -1.96??

This data is time series and I check ADF 1 lag with trend. Data set is below. 'sejong' is the region and 'price' means house price. I just wonder it has fewer data, only has 24 data sets, so its critical value is higher than I expected. But I still unclear why critical value can be high when its data is smaller or other reason..

• Welcome to Cross Validated, hwjang. Your question leads me to suppose that you misunderstand the nature of a t-statistic. The magic 1.96 number is not relevant to the t-distribution unless the sample size is very large. Perhaps start by searching for an explanation of a t-test on this site. If you want the most helpful answers you should provide more context than you have given. What is the table an output from? What do your data consist of? What are you hoping to test? – Michael Lew Sep 14 '17 at 7:38
• Thank you for your comment. I add more information to be clarify the question. I don't know it will be helpful or not. The purpose of this ADF test is to find unit root in my data set. – hwjang Sep 14 '17 at 8:21

The link between test statistic, critical value, and the resulting p-value is a common source of confusion in unit root testing. As already explained in this answer, obtaining a p-value in this scenario is not as straight-forward as in a standard hypothesis test. If you look at your output carefully, the key words here are "interpolated" and "approximate". Your software uses the test statistic to interpolate critical values based on the tables in Fuller (1996), in order to produce an approximate test p-value (MacKinnon approximate pp-value).

You can cross-check this method in R and observe that it produces very similar results:

tstat = -2.36

# Grabbed from the source code of adf.test():
pval = approx(c(-4.38, -3.95, -3.6, -3.24, -1.14,
-0.8, -0.5, -0.15), c(0.01, 0.025, 0.05, 0.1, 0.9, 0.95,
0.975, 0.99), tstat, rule = 2)$y print(pval) # p-value = 0.4352381  The p-value of 0.43 is very close to your 0.40. Either way, your series is not weakly stationary but with the right transformations it most likely can be. • This does not explain why the p-value of L1 is 0.03 in the table, which is a huge discrepancy from 0.40. – Richard Hardy Sep 14 '17 at 11:06 • Mentioning the approximiation is a very good point in the answer! I always wondered why it is approximated (and then from such a small table). Nowadays, it should not be too computationally expensive to compute an exact p-value? Or is the approximation just good enough so no one implements an exact algorithm? After all, the original table must have been computed in some way... – Helix123 Sep 14 '17 at 11:51 • The p-value of L1 is computed from pt(-2.36, 5) but this is not the actual test p-value for the reasons already explained. – Digio Sep 14 '17 at 13:13 The$p$-value reported in the table with regression coefficients, their standard errors,$t$statistics and confidence intervals appears to be taken from a$t$distribution which does not apply for this particular coefficient (the lag of the original variable) which has a nonstandard distribution under the null hypothesis. This$p$-value is meaningless and should not be reported. The MacKinnon approximate$p\$-value is the relevant one because it uses the correct asymptotic distribution.

• Richard, thanks for your input but I think your answer is stating the obvious. It is being taken for granted that the MacKinnon approximate p-value is the test p-value. What was of interest here was to explain how to derive it. – Digio Sep 14 '17 at 17:56
• @Digio, thanks. Your answer is interesting, but I think it misses the point that the OP is actually confused about. But I cannot speak for the OP, so perhaps you are right. I just posted what I though addresses the real question here, whether the answer is obvious to some or not. – Richard Hardy Sep 14 '17 at 19:05