What are the hypothesis and results explaination of joint null test?

When trying to satisfying the paralell assumption in Difference-in-Differences setting, I have the equation:

$$y_{kt} = \alpha_k + \lambda_t + + \delta_{-4} d_{k,t-4} + \delta_{-3} d_{k,t-3} + \delta_{-2} d_{k,t-2} + \delta_{-1} d_{k,t-1} + \delta d_{kt} + \delta_{+1} d_{k,t+1} + \delta_{+2} d_{k,t+2} + \delta_{+\bar{3}} d_{k,t+\bar{3}} + \epsilon_{kt},$$

So, to satisfy the parallel equation, as suggested by Miller, 2021, Table 3, we can do a joint test that $$d_{k,t-i}$$ {i=1;4} equal to 0.

What I tried is: From my understanding, if p-value of joint test is higher than 0.1, we can say that there is no significant difference between treatment and control value in average.

However, when using STATA, I have the joint test code andthe result is as below, I do not know how to read the result because there is no t-stat or p-value here.

    . test (dkt_4=0) (dkt_3=0) (dkt_2 = 0) (dkt_1 = 0)

( 1)  dkt_4 = 0
( 2)  dkt_3 = 0
( 3)  dkt_2 = 0
( 4)  dkt_1 = 0

F(  4, 25334) =    4.48
Prob > F =    0.0013


There is no t-statistics because you are using F-test, with F-statistics. Also you do get p-value, p-value is the probability value of obtaining test results at least as large as the results actually observed, under the assumption that the $$H_0$$ is correct. So in your case 0.0013.
• @NoviceMindset 1. yes, that’s the p value. No if p-value is smaller than 0.01 you can not only reject the null but you already reject it at 1% level (e.g. equivalent of $^{***}$). So the null is rejected. Also as I understood from your text when null is rejected parallel trend is violated