29 Matching Annotations
  1. Nov 2024
    1. Both tests reject the hypothesis of nonstationarity of the term spread series at the 1%1%1\% level of significance, which is strong evidence in favor of the hypothesis that the term spread is stationary, implying cointegration of long- and short-term interest rates.

      It would be useful to mention that summary() will provide all the relevant information for deciding whether to accept or reject the null hypothesis

    1. This is, however, not the appropriate critical value for the ADF test when an intercept and a time trend are included in the Dickey-Fuller regression: the asymptotic distributions of both test statistics differ and so do their critical values!

      So which ones are the appropriate critical values?

  2. Oct 2024
    1. Things are different for model (3). The coefficient on the level of the logarithm of the dividend yield is different from zero at the 5%5%5\% level and the FFF-test rejects, too.

      It would be useful to explain how to read that information from the table provided by Stargazer

    1. forecast <- c("2013:Q1" = coef(GDPGR_AR2) %*% c(1, GDPGR_level[N-1], GDPGR_level[N-2]))

      A brief explanation about this command would be useful.

    1. stargazer(lpm_HMDA, logit_HMDA, probit_HMDA1, probit_HMDA2, probit_HMDA3, probit_HMDA4, digits = 3, type = "latex", header = FALSE, se = rob_se, model.numbers = FALSE, column.labels = c("(1)", "(2)", "(3)", "(4)", "(5)", "(6)"))

      This command may not work depending on the versions of stargazer and R. If that's the case, just create a list containing all the models first and use that as your first argument:

      allmodels <- list(lpm_HMDA, logit_HMDA, probit_HMDA1, probit_HMDA2, probit_HMDA3, probit_HMDA4)

      stargazer(allmodels,<br /> digits = 3, type = "text", header = FALSE, se = rob_se, model.numbers = FALSE, column.labels = c("(1)", "(2)", "(3)", "(4)", "(5)", "(6)"))

    1. stargazer(fat_mod1, fat_mod2, fat_mod3, fat_mod4, fat_mod5, fat_mod6, fat_mod7, digits = 3, header = FALSE, type = "latex", se = rob_se, title = "Linear Panel Regression Models of Traffic Fatalities due to Drunk Driving", model.numbers = FALSE, column.labels = c("(1)", "(2)", "(3)", "(4)", "(5)", "(6)", "(7)"))

      With R version 4.2.2 or greater and Stargazer version 5.2.3, this code doesn't work. You have to put all the models in a list first and use the list as the first argument of the stargazer command. all_models <- list(fat_mod1,fat_mod2,fat_mod3,fat_mod4,fat_mod5,fat_mod6,fat_mod7)

      stargazer(all_models, digits = 3, header = FALSE, type = "text", se = rob_se, title = "Linear Panel Regression Models of Traffic Fatalities due to Drunk Driving", model.numbers = FALSE, column.labels = c("(1)", "(2)", "(3)", "(4)", "(5)", "(6)", "(7)"))

    2. Fatalities$punish <- with(Fatalities, factor(jail == "yes" | service == "yes", labels = c("no", "yes")))

      The logic of this command is not clear. if either "jail" or "service" is "yes", why is the label for punish "no"?

    1. fatal_fe_lm_mod <- lm(fatal_rate ~ beertax + state - 1, data = Fatalities)

      It would be useful to explain the meaning of " -1" in the formula.

    1. stargazer(TSMA_mod1, TSMA_mod2, TSMA_mod3, TSMA_mod4, TSMA_mod5, TSMA_mod6, title = "Regressions Using Massachusetts Test Score Data", type = "latex", digits = 3, header = FALSE, se = rob_se, object.names = TRUE, model.numbers = FALSE, column.labels = c("(I)", "(II)", "(III)", "(IV)", "(V)", "(VI)"))

      Got an error message when running this code. The code below worked for me:

      stargazer(TSMA_mod1, TSMA_mod2, TSMA_mod3, TSMA_mod4, TSMA_mod5, TSMA_mod6, title = "Regressions Using Massachusetts Test Score Data", type = "text", digits = 3, se = rob_se, dep.var.caption = "Dependent Variable: Test Score", column.labels = c("(I)", "(II)", "(III)", "(IV)", "(V)", "(VI)"))

    1. If including additional variables to mitigate the bias is not an option because there are no adequate controls

      An example for this kind of situation would be very useful

    1. the predicted decrease in test scores amounts to 1.9+3.3=5.21.9+3.3=5.21.9 + 3.3 = 5.2 points

      Why is 18.3 not included in the calculation if HiEL is 1?

  3. Sep 2024
    1. p-value=PH0[|¯¯¯¯Y−μY,0|>|¯¯¯¯Yact−μY,0|]

      This is very difficult to understand. It would be better to add a paragraph explaining the meaning of the formula