You are basically there. I assume that you are doing a $Z$ test here (although if you have been taught how to perform a $t$-test, then that test, with 24 degrees of freedom, might be more appropriate).
You want to know if your $Z$ statistic of $-2$ is significant. Since you are performing a one tailed test, the critical value for this statistic is the $\alpha$-quantile of the $Z$ distribution (i.e., the standard normal distribution). The critical value you use would depend on the significance level of the test. If you want to perform the test at the $99\%$ significance level, then the critical value is $-2.326$. The rejection region is therefore the set $(-\infty,-2.326)$. If you want to conduct the test at the $95\%$ significance level, the critical value is $-1.645$. So, since the observed value of $Z$ is $-2$, you would retain the null hypothesis if you wanted to work at the $99\%$ significance level, and you'd reject it if you wanted to work at the $95\%$ level.
I got these values from an online normal distribution calculator, but you can get them from any table of normal values too (although maybe not to the same level of precision).
Since your decision to reject or retain the null hypothesis depends on the significance level you choose, it isn't really reasonable to say that you have determined which hypothesis is true. You have merely decided whether the observed data are consistent with the hypothesis of no improvement.
(It should also be noted that you are assuming that the distribution of costs is still normally distributed after the cost cutting measures. You did not state explicitly that this was the case, but it is necessary in order to use either the $t$ or $Z$ test.)