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POST Hut NOw that You rejected the null hypothesis; you need treatments are more effective; runan Matmeni _ Maybe one analya Lnealanen deletirune maybe tWO treatmen...

Question

POST Hut NOw that You rejected the null hypothesis; you need treatments are more effective; runan Matmeni _ Maybe one analya Lnealanen deletirune maybe tWO treatments are equally effective but both are beuol bcttet Itun which than Ihc allthe 4hjsis will determine which of these possibilities occurred thutd uer05 could - WCWM While Therc Hlollx up follow-up [ests WC only cam ont this text many the KSD; duleter @fetence follow-up test This test produces mean difference that you honestly = signifi

POST Hut NOw that You rejected the null hypothesis; you need treatments are more effective; runan Matmeni _ Maybe one analya Lnealanen deletirune maybe tWO treatments are equally effective but both are beuol bcttet Itun which than Ihc allthe 4hjsis will determine which of these possibilities occurred thutd uer05 could - WCWM While Therc Hlollx up follow-up [ests WC only cam ont this text many the KSD; duleter @fetence follow-up test This test produces mean difference that you honestly = significantly different from each other significant ol mcans. Any differences daictmine the difterence larger than the HSD valuc: belxicen two means palrs snficant If 4 Ito Ican? an (or more) , the difference aflercnt by = 4notnt unlikely [0 have resulted from chance = probably created by the treatment differences You compute the HSD as Rhitntet can lock upq Appendix by' using the number TreJUNEn conditions (g) the dfierer' and_ alpha Iihue Your the number of participants each condition. The MS_ comes from the Fratio calculation you did earlier Compute the HSD for this analysis now: Aaameemt KSD = Cczipute the differences berween each pair of treatment conditions The first one cone for you: 4-book vs. 20-book: 30 40-book = book: #book book: Based 0n the HSD you computed Question 33 and the condition mean differences You (cmputed for Question 34. which of the three group mean differences are so large that they are Etlcly have been caused by chance? Choose all that apply: The difference between the 40-book group and the book group diference = between the 40-book group and the 20-book group diflerence between the 20-book group and the book group Na that yoU " Qton - identified which group differences are larger than the HSD you computed in 4105 Siaie, which ones are unlikely to result from sampling error), you need to compute Fnthe { each pairwise comparison (. those you computed in Question 35). This done following ! Kdcompated : formula which the numerator the difference berween the Ewo conditions UAaons and the denominator is the square root of the pooled variance for those two M-M SD;



Answers

Assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. (Note: Answers in Appendix D include technology answers based on Formula 9 - 1 along with "Table" answers based on Table $A$ - 3 with df equal to the smaller of $\boldsymbol{n}_{I}-\boldsymbol{I}$ and $\boldsymbol{n}_{2}-\boldsymbol{I} .$ ) Blanking Out on Tests Many students have had the unpleasant experience of panicking on a test because the first question was exceptionally difficult. The arrangement of test items was studied for its effect on anxiety. The following scores are measures of "debilitating test anxiety." which most of us call panic or blanking out (based on data from "Item Arrangement, Cognitive Entry Characteristics, Sex and Test Anxiety as Predictors of Achievement in Examination Performance," by Klimko, Journal of Experimental Education, Vol. 52, No. 4.) Is there sufficient evidence to support the claim that the two populations of scores have different means? Is there sufficient evidence to support the claim that the arrangement of the test items has an effect on the score? Is the conclusion affected by whether the significance level is 0.05 or 0.01? $$\begin{array}{|c|c|c|c|c|} \hline {}{} {\text { Questions Arranged from Easy to Difficult }} \\ \hline 24.64 & 39.29 & 16.32 & 32.83 & 28.02 \\ \hline 33.31 & 20.60 & 21.13 & 26.69 & 28.90 \\ \hline 26.43 & 24.23 & 7.10 & 32.86 & 21.06 \\ \hline 28.89 & 28.71 & 31.73 & 30.02 & 21.96 \\ \hline 25.49 & 38.81 & 27.85 & 30.29 & 30.72 \\ \hline\end{array}$$ $$\begin{array}{|c|c|c|c|} \hline {}{} {\text { Questions Arranged from Difficult to Easy }} \\ \hline 33.62 & 34.02 & 26.63 & 30.26 \\ \hline 35.91 & 26.68 & 29.49 & 35.32 \\ \hline 27.24 & 32.34 & 29.34 & 33.53 \\ \hline 27.62 & 42.91 & 30.20 & 32.54 \\ \hline\end{array}$$

Following is a solution video to number 24 and this is where we compare to means uh for the soil, water content for field a compared to feel b. And the first part is just to verify that the mean and the standard deviations are in fact these numbers, So the mean for field day is 12.53 and the standard deviation is 2.39 And then the mean for field B was 10.77 with the standard deviation of 2.4, and it says use a calculator something use this T I 84 I went ahead and typed in the means, or sorry, the data values L one and L two, so L one represents fuel day and then L two represents, you'll be and if you go to stat falcon and it's one of our stats and we're going to change that to L one and calculate and that gives us everything we need. So the X bar, is that 12.53 So that's verified. And then we're looking at the s the standard of the sample standard deviation is about 2.39 So for the field A. That is correct. And then let's just go and double check field be. So we're gonna change it to L. Two now because that's where field B. Is, and then that verifies its 10.77 for the mean, then about 2.40 for the standard deviation. So the first part is done, that's verified. And then the second part it says conduct not formally, but we're basically just going to conduct a uh uh two sample T. Test with an alpha value of point oh five. And we're gonna go back to the calculator because doing it by hand can be pretty cumbersome. So if you go to stat and then test now, since we already have, we're gonna go to two sample t. Test since we already have the data in there. I'm just gonna use the data instead of the summary stats. So list one is L. One that's the field A. List to is L. Two. That's field B. And then these frequencies just keep them the same and then we have the the alternative hypothesis and you kinda have to use your context clues here. But this one actually explicitly states is field A. Is the soil content or water content higher or greater than. So I'm going to change this to greater than So μ one is greater than you to warm you. A. is greater than YouTube pulled is usually zero or no unless they tell you otherwise. So we can go ahead and calculate and that's gonna give us everything we need. So the T. Value if you want you can put it down there. But really it's just this P. Value that I want to see. So 00.27 is the p value. So P value equals 0.27 And then we compare that with the alpha value and it's less than the alpha value. So any time it's less than the alpha value. That means we reject the no hypothesis. So we're rejecting the statement that these means are the same. And we are I guess you could say accepting the fact that these two means are in fact the water content and field A. Is greater than the water content and field be on average. So we can type this out as there is enough evidence to suggest that field A. Has on average a higher soil content soil. Sorry, water content. Banfield beat. Okay so this field A. Has on average a higher soil water content on field because since we are rejecting that null hypothesis and accepting the alternative hypothesis.

We have muse equal to 1300 as a known mean of the population. We want to calculate the sample mean exploring the sample standard deviation s for the randomly sampled data that follows. So to do so, we simply have to remember the definition of expire to ask for a sample. That's why, for example, in some of the data to buy buy em in this case, 12 68 S is some of the deviations about the moon squared, all divided by n minus one. In this case. 37.29 Next, what we want to do is implement a two tailed test. That is we want to test whether or not this sample suggests the actual meaning this population differs either higher or lower. Either direction from the no mean 1300. We're going to do so with an alcohol level of 0.1 and we're gonna know at this point that X is approximately normally distributed. So to implement this test, we have to follow the following procedures, one by one 1st. What is the significance of hypotheses? Alpha equals 0.1 The null hypothesis H not is musical 1300. And the alternative hypothesis H is that it is not equal 1300. Next distribution We're going to use compute the associated testes, cystic. We're going to use a student's t distribution because the population standard deviation sigma is not known. We only have a sample standard deviation S It's appropriate to use the distribution because X is approximately normally distributed, meaning it's both symmetric and round shape necessary to use students T. Six reasons to distribution. Let's calculate the T statistic. It's defined by this formula, expire minus mu divided by s over root end, which in this case equates the negative 2.714 next let's compute the p interval and sketch the results. So, since we have a degree of freedom of n minus one equals nine, no, from a two tailed T table, which we can find in google and the stats textbook, we find that the T statistic corresponds to a p value range between 20.0 to 1.5 We can think of this also as the area underneath the T distribution outside of negative 2.714 and 2.714 The T statistic as is graft on the right. Finally, given this p interval range, we can make a conclusion about this test. Since P is greater than alpha, we have statistically insignificant bindings and we cannot reject h shot. We fail to a general hypothesis. Therefore, we interpret this to mean that we lack evidence suggesting that the population mean differs from its no mean of 1300.

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.

The following is solution video to number 26 which is two sample T tests comparing meantime loss in the workforce with stressors and intimidators. And you just got to be really careful here. The only kind of weird thing is They kind of switch up the order on you so X one represents the intimidators. In the next two represents the stressors. And we want to see if the stressors, so I'm actually gonna write that down. So this is intimidators and this is stressors. Okay, So it says is there enough evidence to suggest that the stressor is greater than the intimidator? So keeping that in mind, we're going to write less than some you won Is less than you two, Even though it says greater than but the orders just switched, if that makes sense? So the greater than in this case means less than for our alternative, since the order is opposite. Okay, so the first part of this is just verifying that these means. And standard deviations are four and 2.38 for intimidators, and then 5.5 and 2.784 stressors. So I went ahead and it says to use a calculator, so I'm using a T 84 if you go to stat and then edit here are the list, so L one is the intimidator and then L two is the stressor. So if you go back to stat over to couch and then it's one of our stats, we can change that to L one and this will be the four and the 238 that was over here. So four and 238 And then go back to stat couch. One bar stats, change that to L two calculate, and that's where we get to 55 and the 27855 to 78 So we verified that that is the way to do it. So then the second part of it is to kind of sort of not formally conduct a hypothesis test but essentially come up with the conclusion here and it is gonna be a two sample T test. We're gonna use the calculator since we already have the data input anyway. So it's two sample T test for the alpha is point oh five. And the alternative we already talked about, the alternative is less than not greater than but less than so. Um if you go back to stat and then we're gonna air over two tests and it's the fourth one down is the T. Test. Two sample T test because we don't know the standard deviation for the population, We don't know sigma, we only know as um so sincerity of the data, I'm just going to keep it as data instead of doing the summary stats. L one is the first list. L two is the second list and you can keep those frequencies as one and then we're gonna change this alternative to less than you two. And then pulled is usually no unless they tell you otherwise and then we're gonna go and calculate and you can have that test statistic there if you want but you really don't need to because the p value is all you need. So it's 0.14. So let's write that down. So the p values 0.14 and what we do is we explicitly compare the P value with the alpha. In this case 0.14 is greater than point oh five. So that means whenever the p values greater than alpha, we fail to reject H not so we failed to reject h. Not whenever the p values greater than alpha at the p values less than alpha, then we will reject the null hypothesis. So keep in mind that the null hypothesis is generally always saying that these two means are equal. We can we're failing to reject that, so we're accepting that to be the truth. So let's go ahead and complete this hypothesis test. And we're gonna say there is not sufficient evidence to suggest that the mean time lost due to stressors is greater than the meantime lost due to intimidators. That's spelled right? I have no idea. Looks like it's not so intimidators. Okay, that's why math teachers aren't spelling enthusiast. So intimidators um now you could also say there is not sufficient evidence to suggest that the meantime lost due to intimidators is less than the meantime loss due to stressors. If you want to keep to that less than thing. I just wrote it like this because that's the way the book had it. So there is not enough evidence to say that these two means are different or that one means more than the other. So we're accepting the truth that these two means are probably about the same.


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