The following is a solution to number 12 goodness of fit test. And this looks at in a library, the number of books that are checked out and comparing it to the distribution of uh students and classes are the subject taught or something like that. So here are the observed data values 268 114 115 115 76. And then the expected values they just give you percent. So the way you find the expected value, you take the percent times the total sample size and there are 888 in the sample. So 32% times 888 is 2 84 16, 25% times 8 88-22, You know, and so and so forth. So I did that for the data that we're going to use now we can start answering some questions. So it says that in part a what's the significance level and that's the alpha value. And they tell you that it's a 5% significance level. So alpha equals .05. And then the alternate or the Nolan alternative. The goal is always saying that these two distributions about the same and then the alternative saying that they're not the same. So in the context of this problem, h not will be the subject distribution. The subject distribution of books in the library fits the distribution of books checked out. Okay, that's the first um the null hypothesis and then the alternative, you know, just the same time, I'm just gonna say not, but you would just say the subject distribution of books in the library does not fit the distribution of books that are checked out. So the second part, we need to find the chi square value and we can use technology to find that. But you can certainly use the formula if you wish. And then we need to check to make sure the expected values. All the expected values need to be greater than five. That's one of those conditions for inference. And we can verify but those are all significantly greater than five. So we're good on that. And then because of that we can use the chi square distribution. Okay. But we also need to say how many degrees of freedom there are? There are four degrees of freedom. Okay. The reason why it's four is you just take the categories minus one. So there are five categories here and you subtract one? You get four. Okay so I think that's it now. We just need to go to our calculator. Now I use A. T I. T. For especially whenever the data sets pretty small like this, but you can use whatever software you like or you can use the formula. So if you go to Stapp and I took the liberty to go and just type these in already. But L1 is where I put the observed to 68 to 14 to 15 1 15 and 76. And then L two is where I put the expected so you can see there and then if you go back to stat and then tests it's one of the last one. So I go up first and I go to the Kaiser. This the D option chi square G. O. F. Test that means goodness of fit. That's what that means. And L one is the observed L two is the expected degrees of freedom remember was four. So I can calculate this thing and that gives me everything I need. So the chi square value. Is this this top thing right here? So 11 point 866 is my chi square value, so 11866. And then the p value is also given, it's about .018 .018. And then if you compare that with the alpha value, remember the alpha value is 0.5, that is less than alpha. And the rule is so you explicitly compare the P value with the alpha value, the significance level, and any time it's less than alpha, you're going to always reject the null hypothesis. So we reject H not. Okay. And then the final step is the conclusion. And you can write this however you want, basically just saying that these are not uh the same distribution. But in the context of this problem, I'm going to write there is sufficient evidence to suggest that the subject distribution of books in the library does not fit the distribution of books checked out by students. Okay, so that's the conclusion that I came up with. There is enough evidence to suggest that these two distributions are not the same.