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You wish to test the following claim ( H a ) at a significancelevel of α = 0.10 . For the context of this problem, μ d = μ 2 − μ1 where the first ...

Question

You wish to test the following claim ( H a ) at a significancelevel of α = 0.10 . For the context of this problem, μ d = μ 2 − μ1 where the first data set represents a pre-test and the seconddata set represents a post-test. H o : μ d = 0 H a : μ d < 0 Youbelieve the population of difference scores is normallydistributed, but you do not know the standard deviation. You obtainpre-test and post-test samples for n = 8 subjects. The averagedifference (post - pre) is ¯ d = − 15.1 with a

You wish to test the following claim ( H a ) at a significance level of α = 0.10 . For the context of this problem, μ d = μ 2 − μ 1 where the first data set represents a pre-test and the second data set represents a post-test. H o : μ d = 0 H a : μ d < 0 You believe the population of difference scores is normally distributed, but you do not know the standard deviation. You obtain pre-test and post-test samples for n = 8 subjects. The average difference (post - pre) is ¯ d = − 15.1 with a standard deviation of the differences of s d = 47.4 . What is the critical value for this test? (Report answers accurate to two decimal places.) critical value = What is the test statistic for this sample? (Report answers accurate to two decimal places.) test statistic = The test statistic is... in the critical region not in the critical region This test statistic leads to a decision to... reject the null accept the null fail to reject the null As such, the final conclusion is that... There is sufficient evidence to warrant rejection of the claim that the mean difference of post-test from pre-test is less than 0. There is not sufficient evidence to warrant rejection of the claim that the mean difference of post-test from pre-test is less than 0. The sample data support the claim that the mean difference of post-test from pre-test is less than 0. There is not sufficient sample evidence to support the claim that the mean difference of post-test from pre-test is less than 0.



Answers

Bacteria Colonies: Poisson Distribution A pathologist has been studying the frequency of bacterial colonies within the field of a microscope using samples of throat cultures from healthy adults. Long-term history indicates that there is an average of $2.80$ bacteria colonies per field. Let $r$ be a random variable that represents the number of bacteria colonies per field. Let $O$ represent the number of observed bacteria colonies per field for throat cultures from healthy adults. A random sample of 100 healthy adults gave the following information. (a) The pathologist wants to use a Poisson distribution to represent the probability of $r$, the number of bacteria colonies per field. The Poisson distribution is $$ P(r)=\frac{e^{-\lambda} \lambda^{r}}{r !} $$ where $\lambda=2.80$ is the average number of bacteria colonies per field. Compute $P(r)$ for $r=0,1,2,3,4$, and 5 or more. (b) Compute the expected number of colonies $E=100 P(r)$ for $r=0,1,2,3,4$, and 5 or more. (c) Compute the sample statistic $\chi^{2}=\Sigma \frac{(O-E)^{2}}{E}$ and the degrees of freedom. (d) Test the statement that the Poisson distribution fits the sample data. Use a 5\% level of significance.

18. So it's no if that new one is equal to mean to, and each one is that anyone is not equal commute. So the degree of freedom, the minimum off in one way, this one, which is 17 and 18 to minus one, which is 19th. So the video is 17, using all for April toe 0, 00.11 teen. So for table five, the critical, very equal toe postive or negative 1.74 So the rectory, then they contain all value smaller than negative 1.74 and all various larger than 1.74 So that is the statistic, if that makes one minus x two bar. So it's 56900 minus five 7800 minus me one minus me, too over square root off this one squared over in one, which is one toe. Once you lose your square over and once with the 18 lost, it's too square. So it's 8000 square over into approximately negative 4.267 So it's a very off statistics. Even in the final hypotheses rejected so and this value realize between the booze values. So we failed to reject the null hypothesis

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.

Solution number 17. And this is a, an interesting problem with the Poisson distribution. And the Poisson distribution is the type of discrete distribution where the events are random and rare. And uh, this, in this case it's a traffic accidents, Daily traffic accidents, and there is an average, that's what we use this lambda for. So λ,, which is the average of the mean is 1.72 accidents per day. Were asked to find the probability that zero accidents occur and the probability that one occurs To occur three occurs and then greater than four. Curse. Now you can use the formula, but again, I like to use the uh, software. So what I'm gonna do is I'm gonna go to second distribution and then I'm gonna go down to the present pdf and then here it asks from you, or sometimes it lasts for lambda, and that's 1.72 And then the X value, I'm just going to find the probability that zero occurs, And that gives me .1791. I'm going to go in round here some point 1791. So you can do this with any type of software or you can just use the formula. Although the formula can take awhile .1791. And I'm gonna do it one more time. Just show you whether they're the second bars for distribution. And then I went to the Plaza pdf, right? That's the probability density function. So pdf. And the mu the mean is that's the land of 1.72. And this time we're gonna find the distribution or the probability that one occurs and it's about Point Let's say .308 point 308. And that's what you're gonna do for, you know, basically all the rest of them. Until you get to the greater than so zero point 2649, I'll go ahead and give you these answers here and then 0.15 one night. So then uh to get something that's greater than what you're gonna do is you're gonna take one go all the way up to infinity and save some time. We're just going to take one minus the four that we've already found. 0123 So there's another function in the calculator we can use is called the plaza CDF. The cumulative density function. And we're going to go up to three. So the CDF calculates the probability of zero plus the probability one plus probability two plus probability of three whenever you do CDF of three, so one minus that. Or you could just do one minus these four numbers here whichever you like. So one minus. And then 2nd distribution. And I'm going to go to the present CDF. And the main remember was 172 and then the X value. Now I'm not gonna put 0123 I'm just gonna put the three And it automatically calculates 0 1, 2 and three combined. And whenever you do that that should be your answer. So .0962 0962. You might get 61 if you just use these numbers here but you get the same thing if you do one minus and these all added together. Next up we find the expected value and any time you find the expected value just take the sample size times the probabilities. So for zero, remember there were 90 days that we looked at. So 90 times the probability of zero was remember that .1791? That should give you 16.119. And then all of these are gonna be the same 90 times something. And those some things are the probabilities. So .308 that's going to give you 27.72. So that means we can expect about 27.72 days where there are there's one wreck And then 90 times the point 2649. That should give us 23 841 And then 90 times .1519. He was this 13671. And the 90 times 0.0 962 Gives you 8658. Okay, so those are the probabilities. Those are the I'm sorry those are the expected values. And then the part c we find we're gonna find our use the good as fit test. So they observed that was the chart that was given and the expected we actually just found. So we're just gonna copy those expected values down 27.72 23-841 13.671 and then eight 658. So those are the expected values where we take the probabilities times the sample sizes. Okay? So now we can go back to our calculator and do the goodness of fit test. So if you go to stat and then edit you can see that here in L. One. I put the observed values and then L. Two I put the expected values. And then if you go to stat tests it's the chi square goodness of fit test. And the observed is L. One the expected sell to or you can use the formula or any other software. And degrees of freedom was four. So the degrees of freedom, that's actually another answer. The degrees of freedom of four. Since there are five categories there of 0123 and then greater than equal to four. So we can calculate that And that gives us a chi square value of about 12 509 So chi square equals 12.509 And then it also asked for degrees of freedom. The degrees of freedom, like I said, was equal to four because it's five minus one. Okay, so the p value let's look back at the P value, it's about .01. Let's go at some .014 and that is greater than the alpha, barely, but it is greater than the alpha. So whenever the p value is greater than the alpha than we fail to reject to reject H not. Which means in this case, actually, in all these goodness of fit test cases, the null hypothesis is that the distribution does in fact follow whatever we're talking about in this case, we're failing to reject that. So this follows a person distribution so the traffic accidents, and per day in this particular area, it does in fact follow a Poisson distribution.

Problem. 17 8. Each note is that new one is smaller than or equal commute. Each one is equal to me. One is bigger than you, so the degree of freedom is the minimum off. Anyone minus one into my storage is 18 and 14, so it's equal toe food correspondent to critical value with offer. April 2.451 tail so T is equal to 1.761 So the actual reason contain old values. Great around 0.761 So the test statistic is equal to x one bar minus X to bar. So for 97 year zero minus 4 to 000 over square. Note on 8800 square over 19 plus 51 Use your square over 15, which is ableto 3.19 point. So it's a very all the tests statistic in the rectory does not have pointy inject. Three point is smaller and it's bigger than 1.761 So we reject then the hypothesis


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