It seems the creators of this particular R command presume one is familiar with the original Dickey-Fuller formulae, so did not provide the relevant documentation for how to interpret the values. I found that Enders was an incredibly helpful resource (Applied Econometric Time Series 3e, 2010, p. 206-209--I imagine other editions would also be fine). Below I'll use data from the URCA package, real income in Denmark as an example.
> income <- ts(denmark$LRY)
It might be useful to first describe the 3 different formulae Dickey-Fuller used to get different hypotheses, since these match the ur.df "type" options. Enders specifies that in all of these 3 cases, the consistent term used is gamma, the coefficient for the previous value of y, the lag term. If gamma=0, then there is a unit root (random walk, nonstationary). Where the null hypothesis is gamma=0, if p<0.05, then we reject the null (at the 95% level), and presume there is no unit root. If we fail to reject the null (p>0.05) then we presume a unit root exists. From here, we can proceed to interpreting the tau's and phi's.
- type="none": $\Delta y_t = \gamma \, y_{t-1} + e_t$ (formula from Enders p. 208)
(where $e_t$ is the error term, presumed to be white noise; $\gamma = a-1$ from $y_t = a \,y_{t-1} + e_t$; $y_{t-1}$ refers to the previous value of $y$, so is the lag term)
For type= "none," tau (or tau1 in R output) is the null hypothesis for gamma = 0. Using the Denmark income example, I get "Value of test-statistic is 0.7944" and the "Critical values for test statistics are: tau1 -2.6 -1.95 -1.61. Given that the test statistic is within the all 3 regions (1%, 5%, 10%) where we fail to reject the null, we should presume the data is a random walk, ie that a unit root is present. In this case, the tau1 refers to the gamma = 0 hypothesis. The "z.lag1" is the gamma term, the coefficient for the lag term (y(t-1)), which is p=0.431, which we fail to reject as significant, simply implying that gamma isn't statistically significant to this model. Here is the output from R
> summary(ur.df(y=income, type = "none",lags=1))
>
> ###############################################
> # Augmented Dickey-Fuller Test Unit Root Test #
> ###############################################
>
> Test regression none
>
>
> Call:
> lm(formula = z.diff ~ z.lag.1 - 1 + z.diff.lag)
>
> Residuals:
> Min 1Q Median 3Q Max
> -0.044067 -0.016747 -0.006596 0.010305 0.085688
>
> Coefficients:
> Estimate Std. Error t value Pr(>|t|)
> z.lag.1 0.0004636 0.0005836 0.794 0.431
> z.diff.lag 0.1724315 0.1362615 1.265 0.211
>
> Residual standard error: 0.0251 on 51 degrees of freedom
> Multiple R-squared: 0.04696, Adjusted R-squared: 0.009589
> F-statistic: 1.257 on 2 and 51 DF, p-value: 0.2933
>
>
> Value of test-statistic is: 0.7944
>
> Critical values for test statistics:
> 1pct 5pct 10pct
> tau1 -2.6 -1.95 -1.61
- type = "drift" (your specific question above): : $\Delta y_t = a_0 + \gamma \, y_{t-1} + e_t$ (formula from Enders p. 208)
(where $a_0$ is "a sub-zero" and refers to the constant, or drift term)
Here is where the output interpretation gets trickier. "tau2" is still the $\gamma=0$ null hypothesis. In this case, where the first test statistic = -1.4462 is within the region of failing to reject the null, we should again presume a unit root, that $\gamma=0$.
The phi1 term refers to the second hypothesis, which is a combined null hypothesis of $a_0 = \gamma = 0$. This means that BOTH of the values are tested to be 0 at the same time. If p<0.05, we reject the null, and presume that AT LEAST one of these is false--i.e. one or both of the terms $a_0$ or $\gamma$ are not 0. Failing to reject this null implies that BOTH $a_0$ AND $\gamma = 0$, implying 1) that $\gamma=0$ therefore a unit root is present, AND 2) $a_0=0$, so there is no drift term. Here is the R output
> summary(ur.df(y=income, type = "drift",lags=1))
>
> ###############################################
> # Augmented Dickey-Fuller Test Unit Root Test #
> ###############################################
>
> Test regression drift
>
>
> Call:
> lm(formula = z.diff ~ z.lag.1 + 1 + z.diff.lag)
>
> Residuals:
> Min 1Q Median 3Q Max
> -0.041910 -0.016484 -0.006994 0.013651 0.074920
>
> Coefficients:
> Estimate Std. Error t value Pr(>|t|)
> (Intercept) 0.43453 0.28995 1.499 0.140
> z.lag.1 -0.07256 0.04873 -1.489 0.143
> z.diff.lag 0.22028 0.13836 1.592 0.118
>
> Residual standard error: 0.0248 on 50 degrees of freedom
> Multiple R-squared: 0.07166, Adjusted R-squared: 0.03452
> F-statistic: 1.93 on 2 and 50 DF, p-value: 0.1559
>
>
> Value of test-statistic is: -1.4891 1.4462
>
> Critical values for test statistics:
> 1pct 5pct 10pct
> tau2 -3.51 -2.89 -2.58
> phi1 6.70 4.71 3.86
- Finally, for the type="trend": $\Delta y_t = a_0 + \gamma * y_{t-1} + a_{2}t + e_t$ (formula from Enders p. 208)
(where $a_{2}t$ is a time trend term)
The hypotheses (from Enders p. 208) are as follows:
tau: $\gamma=0$
phi3: $\gamma = a_2 = 0$
phi2: $a_0 = \gamma = a_2 = 0$
This is similar to the R output. In this case, the test statistics are -2.4216 2.1927 2.9343
In all of these cases, these fall within the "fail to reject the null" zones (see critical values below). What tau3 implies, as above, is that we fail to reject the null of unit root, implying a unit root is present.
Failing to reject phi3 implies two things: 1) $\gamma = 0$ (unit root) AND 2) there is no time trend term, i.e., $a_2=0$. If we rejected this null, it would imply that one or both of these terms was not 0.
Failing to reject phi2 implies 3 things: 1) $\gamma = 0$ AND 2) no time trend term AND 3) no drift term, i.e. that $\gamma =0$, that $a_0 = 0$, and that $a_2 = 0$. Rejecting this null implies that one, two, OR all three of these terms was NOT zero.
Here is the R output
> summary(ur.df(y=income, type = "trend",lags=1))
>
> ###############################################
> # Augmented Dickey-Fuller Test Unit Root Test #
> ###############################################
>
> Test regression trend
>
>
> Call:
> lm(formula = z.diff ~ z.lag.1 + 1 + tt + z.diff.lag)
>
> Residuals:
> Min 1Q Median 3Q Max
> -0.036693 -0.016457 -0.000435 0.014344 0.074299
>
> Coefficients:
> Estimate Std. Error t value Pr(>|t|)
> (Intercept) 1.0369478 0.4272693 2.427 0.0190 *
> z.lag.1 -0.1767666 0.0729961 -2.422 0.0192 *
> tt 0.0006299 0.0003348 1.881 0.0659 .
> z.diff.lag 0.2557788 0.1362896 1.877 0.0665 .
> ---
> Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
>
> Residual standard error: 0.02419 on 49 degrees of freedom
> Multiple R-squared: 0.1342, Adjusted R-squared: 0.08117
> F-statistic: 2.531 on 3 and 49 DF, p-value: 0.06785
>
>
> Value of test-statistic is: -2.4216 2.1927 2.9343
>
> Critical values for test statistics:
> 1pct 5pct 10pct
> tau3 -4.04 -3.45 -3.15
> phi2 6.50 4.88 4.16
> phi3 8.73 6.49 5.47
In your specific example above, for the d.Aus data, since both of the test statistics are inside of the "fail to reject" zone, it implies that $\gamma=0$ AND $a_0 = 0$, meaning that there is a unit root, but no drift term.
The null hypothesis of the (augmented) Dickey Fuller unit root test is that the series being tested has a unit root (that implies nonstationarity).
The basic logic tells you to reject the null hypothesis if the test statistic is sufficiently extreme. The question could be, does it matter whether the test statistic is extremely large and positive or extremely large and negative. Here you are lucky to see which way the extremality of interest goes: 10% critical value is -1.605, 5% critical value (more extreme than 10%) is lower (more negative) at -1.950 and 1% critical value (more extreme than 5%) is even lower at -2.639. So you see that the more negative the test statistic is, the more extreme. In other words, the extremality goes into the negative direction.
Now see what the actual test statistic is: 10.134. That is less extreme than the conventional 5% value (and even the 10% value) because it is less negative than -2.639 (and even -1.605). Therefore, you do not reject the null hypothesis of unit root at any of the conventional significance levels. You data seems to be nonstationary.
This is a generic way of using tests you do not understand well. Of course, it is always better to know more to understand the intuition behind the test, but that is not always the case.
Best Answer
The null hypothesis of the ADF test is the unit root, i.e. the series is nonstationary. To see this, take the simple case of an AR(1) model:
\begin{equation} y_t = c + \phi y_{t-1} + \epsilon_t \tag{1} \end{equation}
and the unit root corresponds to $\phi = 1$, would you agree? So we formulate the hypothesis:
$$H_0 : \phi = 1 \implies \text{unit root} $$
$$H_1 : \phi <1 \implies \text{stationarity} $$
Why is the alternative $\phi <1$ and not the general $\phi \neq 1$? You will have to think about that. It turns out the hypothesis is more easily tested if we subtract $y_{t-1}$ from both sides in 1 getting
\begin{equation} \Delta y_t = c + \gamma y_{t-1} + \epsilon_t \end{equation}
where $\gamma = \phi -1$. This can be estimated by OLS and the unit root corresponds to the usual t-statistic
$$ t = \frac{\widehat{\gamma}}{SE(\widehat{\gamma})}$$
the catch being however that under $H_0$ the statistic does not follow the normal but the Dickey-Fuller distribution whence the critical values come from.
So you tell me, if the null hypothesis is of unit root and the p-value is too large what is the conclusion?