Following is a solution to number 25 comparing to means for the amount of time lost due to hot temperatures Compared to the mean number need an amount of time lost due to disputes from superiors attitudes in the workforce. And the first part of this, all we're gonna do is just verify that the mean and the standard deviations for the two datasets are in fact 4.86 and 3.18 for the temperatures And then 65 and 288 for the attitudes. And it says use a calculator, so I'm using the T I T four and if you go to stat and then edit, you can see already put those numbers in. So L one represents, I think that's the temperature column and then L two is the attitude problem. And if you go back to stat and then cowpoke and then one bar stats, one variable stats the list, I'm just doing L one first and that is where we get that 4.86. And the standard deviation where this s is that's up 3.18. So that's where I get these numbers here, and then we'll do the same thing, cal one of our stats, but this time we're gonna do second to for L. Two, and we calculate that, and that's where we get the mean as 6.5, this X bar here, and then the standard deviation, we're not looking at signal, we're looking at essence is a sample sample standard deviation about 2.88 So that's where we get this 2.88 So we have verified that those are in fact the numbers, and now we're gonna do the two sample t test with the significance level of point oh five. Before we do that, we need to figure out what the um alternative hypothesis would be, because we already have the data. We can just punch that in after that and it's going to be a two tail not equal to test because it says one way or the other is just as are the two means different and whenever it doesn't specify which one is greater, you just assume that it's a two tail meaning not equal to so not equal to is going to be our alternative. Okay. And then there are no I didn't write it down but the Noles that they're equal to. Okay. So if we go to stat and then go over to tests and it's the two sample T test and since we already have the data, I'm just going to use the data instead of the summary stats because it's more accurate that way list one will be L one list too will be L to the frequencies can just stay as one and then the alternative the new one, let's change that to not equal to the pool is going to be no unless it's otherwise stated. And then we're going to calculate and this gives us everything. So the t. You know if you want you can put that in there. But really all I care about is this P value the P value is about 0.317 So I'm gonna write that down so the p value is equal to 2.317 And what we do is we explicitly compare the P value with our alpha value and this time the p value 0.317 is greater than our alpha value. Which means we fail to reject the null hypothesis. So any time the peabody is greater than alpha you fail to reject. If it's less than alpha than you actually reject. H not so if we fail to reject this null hypothesis, that means there's not enough evidence to say that these two means are different, or in other words, these two means appear to be the same, so I'm gonna write that out, or I'm gonna type it out because it's a little quicker. That way, I'm gonna, right, there is not sufficient evidence to suggest that the meantime lost due to hot temperatures is different from the meantime lost due to disputes from superiors attitudes in the workforce. Okay, so that is our two sample T test for these two population means.