Title | One-sided tests for coefficients | |
Author | Kristin MacDonald, StataCorp | |
Date | July 2006; minor revision May 2015 |
Estimation commands provide a t test or z test for the null hypothesis that a coefficient is equal to zero. The test command can perform Wald tests for simple and composite linear hypotheses on the parameters, but these Wald tests are also limited to tests of equality.
To perform one-sided tests, you can first perform the corresponding two-sided Wald test. Then you can use the results to calculate the test statistic and p-value for the one-sided test. Let’s say that you perform the following regression:
. sysuse auto, clear (1978 Automobile Data) . regress price mpg weight
Source | SS df MS | Number of obs = 74 | |
F( 2, 71) = 14.74 | |||
Model | 186321280 2 93160639.9 | Prob > F = 0.0000 | |
Residual | 448744116 71 6320339.67 | R-squared = 0.2934 | |
Adj R-squared = 0.2735 | |||
Total | 635065396 73 8699525.97 | Root MSE = 2514 |
price | Coef. Std. Err. t P>|t| [95% Conf. Interval] | |
mpg | -49.51222 86.15604 -0.57 0.567 -221.3025 122.278 | |
weight | 1.746559 .6413538 2.72 0.008 .467736 3.025382 | |
_cons | 1946.069 3597.05 0.54 0.590 -5226.245 9118.382 | |
If you wish to test that the coefficient on weight, β_{weight}, is negative (or positive), you can begin by performing the Wald test for the null hypothesis that this coefficient is equal to zero.
. test _b[weight]=0 ( 1) weight = 0 F( 1, 71) = 7.42 Prob > F = 0.0081
The Wald test given here is an F test with 1 numerator degree of freedom and 71 denominator degrees of freedom. The Student’s t distribution is directly related to the F distribution in that the square of the Student’s t distribution with d degrees of freedom is equivalent to the F distribution with 1 numerator degree of freedom and d denominator degrees of freedom.
As long as the F test has 1 numerator degree of freedom, the square root of the F statistic is the absolute value of the t statistic for the one-sided test. To determine whether this t statistic is positive or negative, you need to determine whether the fitted coefficient is positive or negative. To do this, you can use the sign() function.
. local sign_wgt = sign(_b[weight])
Then, using the ttail() function along with the returned results from the test command, you can calculate the p-values for the one-sided tests in the following manner:
. display "Ho: coef <= 0 p-value = " ttail(r(df_r),`sign_wgt'*sqrt(r(F))) Ho: coef <= 0 p-value = .00406491 . display "Ho: coef >= 0 p-value = " 1-ttail(r(df_r),`sign_wgt'*sqrt(r(F))) Ho: coef >= 0 p-value = .99593509
In the special case where you are interested in testing whether a coefficient is greater than, less than, or equal to zero, you can calculate the p-values directly from the regression output. When the estimated coefficient is positive, as for weight, you can do so as follows:
H_{0}: β_{weight} = 0 | p-value = 0.008 (given in regression output) |
H_{0}: β_{weight} <= 0 | p-value = 0.008/2 = 0.004 |
H_{0}: β_{weight} >= 0 | p-value = 1 − (0.008/2) = 0.996 |
When the estimated coefficient is negative, as for mpg, the same code can be used:
. test _b[mpg]=0 ( 1) mpg = 0 F( 1, 71) = 0.33 Prob > F = 0.5673 . local sign_mpg = sign(_b[mpg]) . display "Ho: coef <= 0 p-value = " ttail(r(df_r),`sign_mpg'*sqrt(r(F))) Ho: coef <= 0 p-value = .71633814 . display "Ho: coef >= 0 p-value = " 1-ttail(r(df_r),`sign_mpg'*sqrt(r(F))) Ho: coef >= 0 p-value = .28366186
However, to calculate the p-values from the regression output directly, you use the following formulas:
H_{0}: β_{mpg} = 0 | p-value = 0.567 (given in regression output) |
H_{0}: β_{mpg} <= 0 | p-value = 1 − (0.567/2) = 0.717 |
H_{0}: β_{mpg} >= 0 | p-value = 0.567/2 = 0.284 |
On the other hand, if you want to perform a test such as H_{0}: β_{weight} <= 1, you cannot calculate the p-value directly from the regression results. Here you would have to perform the Wald test first.
. test _b[weight]=1 ( 1) weight = 1 F( 1, 71) = 1.35 Prob > F = 0.2483 . local sign_wgt2 = sign(_b[weight]-1) . display "Ho: coef <= 1 p-value = " ttail(r(df_r),`sign_wgt2'*sqrt(r(F))) Ho: coef <= 1 p-value = .12415299
In the output for certain estimation commands, you will find that z statistics are reported instead of t statistics. In these cases, when you use the test command, you will get a chi-squared test instead of an F test. The relationship between the standard normal distribution and the chi-squared distribution is similar to the relationship between the Student’s t distribution and the F distribution. In fact, the square root of the chi-squared distribution with 1 degree of freedom is the standard normal distribution. Therefore, one-sided z tests can be performed similarly to one-sided t tests. For example,
. webuse union, clear (NLS Women 14-24 in 1968) . xtset id . xtlogit union age grade, nolog Random-effects logistic regression Number of obs = 26,200 Group variable: idcode Number of groups = 4,434 Random effects u_i ~ Gaussian Obs per group: min = 1 avg = 5.9 max = 12 Integration method: mvaghermite Integration pts. = 12 Wald chi2(2) = 69.07 Log likelihood = -10623.006 Prob > chi2 = 0.0000
union | Coef. Std. Err. z P>|z| [95% Conf. Interval] | |
age | .0149252 .0036634 4.07 0.000 .007745 .0221055 | |
grade | .1130089 .0177088 6.38 0.000 .0783002 .1477175 | |
_cons | -4.313313 .2426307 -17.78 0.000 -4.788861 -3.837766 | |
/lnsig2u | 1.800793 .046732 1.7092 1.892386 | |
sigma_u | 2.460578 .0574939 2.350434 2.575884 | |
rho | .6479283 .0106604 .6267624 .6685287 | |
Here the test command returns r(chi2), which can be used along with the normal() function to calculate the appropriate p-values.
. local sign_grade = sign(_b[grade]) . display "H_0: coef<=0 p-value = " 1-normal(`sign_grade'*sqrt(r(chi2))) H_0: coef<=0 p-value = 8.768e-11 . display "H_0: coef>=0 p-value = " normal(`sign_grade'*sqrt(r(chi2))) H_0: coef>=0 p-value = 1
Finally, if you want to perform a test of inequality for two of your coefficients, such as H_{0}: β_{age} >= β_{grade}, you would first perform the following Wald test:
. test age-grade = 0 ( 1) [union]age - [union]grade = 0 chi2( 1) = 27.44 Prob > chi2 = 0.0000
Then calculate the appropriate p-value:
. local sign_ag = sign(_b[age]-_b[grade]) . display "H_0: age coef >= grade coef. p-value = " normal(`sign_ag'*sqrt(r(chi2))) H_0: age coef >= grade coef. p-value = 8.112e-08
Again, this approach (performing a Wald test and using the results to calculate the p-value for a one-sided test) is appropriate only when the Wald F statistic has 1 degree of freedom in the numerator or the Wald chi-squared statistic has 1 degree of freedom. The distributional relationships discussed above are not valid if these degrees of freedom are larger than 1.