Ho: mean(out) - mean(in) = diff = 0
|
Ha: diff < 0 |
Ha: diff ≠ 0 |
Ha: diff > 0 |
|
t = 3.3795 |
t = 3.3795 |
t = 3.3795 |
|
P < t = 0.9996 |
P > |t| = 0.0007 |
P > t = 0.0004 |
Finally, we run an association tests to check for independence between being an innovative firm and
being in or out of sample, to evaluate whether less innovative firms are those kicked out of the final
panel. The Pearson chi-square tests are listed for different types of innovation activity:
|
R&D expenditures in 2001-2003 (yes/no) |
Pearson chi-square(1) = 3.52 p-value = 0.061 |
|
Introducing product innovations (yes/no) |
Pearson chi-square(1) = 7.194 p-value = 0.007 |
|
Introducing process innovations (yes/no) |
Pearson chi-square(1) = 2.189 p-value = 0.139 |
|
Introducing both process and product innovations (yes/no) |
Pearson chi-square(1) = 2.249 p-value = 0.134 |
We reject the hypothesis of independence for R&D expenditure and product innovation only. That
means that firms investing into R&D and introducing product innovations have a (slightly) higher
probability to survive. We cannot reject the null for process innovations or both kinds of
innovations, instead. Introducing process innovations or not provide a firm equal probability to
remain in our sample.
31