How do I have a p-value of 1 in the left tail from a two-sample t-test? I have a sample with a mean of 383.45, a standard deviation of 48.878, and a sample size of 30, and a second sample with a mean of 167.78, a standard deviation of 72.368, and a sample size of 30. When I perform a one-tailed two-sample t-test using a TI-84 Plus CE I get a t statistic of 13.52684957 and a p-value of 8.742353334e-19 for the right tail (μ1 > μ2) and 1 for the left (μ1 < μ2). As you might expect, My null hypothesis is that the populations for both samples have the same mean. My alternative hypothesis is that the first sample's population has a greater mean (μ1 > μ2). So, I plan on using the right tail but I'm confused as to why the left tail has a probability of 1 even though when I do a two-tailed test (μ1 ≠ μ2) I get a p-value of 1.748470667e-18 which is around double the p-value for the right tail.

Let us define $\mu_1= 383$ and $\mu_2= 168$ with reasonable small standard deviations. The differences between the means are over 4 standard deviations apart. By eye one can judge, with a large difference, that it is very very unlikely to occur by chance alone.
With regard to a discussion of the formal analysis:
The right tail t-test is comparing if $\mu_1(383) > \mu_2(168)$ is clearly TRUE: thus a p-value of nearly 0.
The left tail t-test is testing whether $\mu_1(383) < \mu_2(168)$ which we can see is clearly FALSE: thus a very high p-value of nearly 1.
The null hypothesis $\mu_1(383) = \mu_2(168)$  we can also clearly see is thus FALSE given the relatively small standard deviations. Since we are allowing $\mu_1$ to be greater than or less than $\mu_2$ the probability is split on both sides of the normal distribution; thus the p-value is greater than for the right tail test.
Now if the means were closer or if the standard deviations were sufficiently large, the results would not be so clear cut and thus the recommendation to sketch out the distributions to verify the calculations makes sense.
Hope this answered your question.
