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Ourr bodln [email protected] ficld that help: woud: heal Dors changing the feld strength sox healine? Fn experients wth netts Inuostigated thtr qutstion Tnc dan belowActho...

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Ourr bodln [email protected] ficld that help: woud: heal Dors changing the feld strength sox healine? Fn experients wth netts Inuostigated thtr qutstion Tnc dan belowActhonialina ratc ot cutt (mucrom ctc Thont matched Dabe @ienmeni_Throzntethelohind Iinb the Hml rE_with Ine body { nannlfd In One Wlnt Kcontrollandmalananaturaealu - Ine Otn umb lexnenmentall Results sre pwven in the tabla_Ditercnce (Control ExpcrimcntllHnnEx erin entalControlHean ofdiferenceStundurd DewationDifterences3) (3 pts) Here Dot

Ourr bodln [email protected] ficld that help: woud: heal Dors changing the feld strength sox healine? Fn experients wth netts Inuostigated thtr qutstion Tnc dan belowActhonialina ratc ot cutt (mucrom ctc Thont matched Dabe @ienmeni_Throzntethelohind Iinb the Hml rE_with Ine body { nannlfd In One Wlnt Kcontrollandmalananaturaealu - Ine Otn umb lexnenmentall Results sre pwven in the tabla_ Ditercnce (Control Expcrimcntll Hnn Ex erin ental Control Hean ofdiference Stundurd Dewation Difterences 3) (3 pts) Here Dot Plot of the differences; there any evidence that the normality aesumplton not met? Explain your answer. Dotplot of Differenc Dltturenaa b) (4 pts) Calculate any missing needed statistics for the test: c)1(3 pts) Order conduct the test state tne null and alternative hypotheses dl (7 pts) Calculate the statistic; using Table € find the p-value for the test using the appropriate number degrees of freedom: At a level of significance . a =Olstate your conclusion:



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Focus Problem: Meteorology The Focus Problem at the beginning of this chapter asks you to use a sign test with a $5 \%$ level of significance to test the claim that the overall temperature distribution of Madison, Wisconsin, is different (either way) from that of Juneau, Alaska. The monthly average data (in ${ }^{\circ} \mathrm{F}$ ) are as follows. $$ \begin{aligned} &\begin{array}{l|llllll} \hline \text { Month } & \text { Jan. } & \text { Feb. } & \text { March } & \text { April } & \text { May } & \text { June } \\ \hline \text { Madison } & 17.5 & 21.1 & 31.5 & 46.1 & 57.0 & 67.0 \\ \hline \text { Juneau } & 22.2 & 27.3 & 31.9 & 38.4 & 46.4 & 52.8 \\ \hline \end{array}\\ &\text { Juriedu }\\ &\begin{array}{l|cccccc} \hline \text { Month } & \text { July } & \text { Aug. } & \text { Sept. } & \text { Oct. } & \text { Nov. } & \text { Dec. } \\ \hline \text { Madison } & 71.3 & 69.8 & 60.7 & 51.0 & 35.7 & 22.8 \\ \hline \text { Juneau } & 55.5 & 54.1 & 49.0 & 41.5 & 32.0 & 26.9 \\ \hline \end{array}\\ &\text { What is your conclusion? } \end{aligned} $$

We have the following matched pairs and we want to conduct a sign tested matched pairs, testing the claim P does not equal five at alpha equals 50.1 significance. This question is testing an understanding of how to implement statistical non parametric texts. In particular the scientists and matched pairs. We proceed the steps A through D to solve so first and a. We see our alpha and hypotheses alpha is 0.1 H and r S p 0.5 AJ is the claim P does not equal five and B. We compute the test at so your population A to B. We have the following signs thus and equals 20. The total number of pluses and minuses and X is the number of plus over. The total number 10/20. Thus our Zenon is x minus point 5/2 0.0.25 over and a zero because 00.5 minutes 0.50 Thus from a normal distribution, r p value is one because the probability at Z is greater than 0.5 times two is one. Thus, we conclude for this test that we fail to reject because P is greater than alpha, which means that we lack evidence to support the alternative hypothesis.

We have the following matched pairs listed above and we want to conduct a scientist and mashed pears testing acclaimed P greater than 0.5 at alpha equals 0.5 significance. This question assessing your knowledge of how to conduct a scientist and mashed pears to do so. We could see a few cents a 30 below first and a least eight alpha and hypotheses. This is alpha equals 80.5 H. And r P equals 0.5 H. A. The claim P greater than 25. Next computer test at. So we have signs for the the difference from population A to B as follows and is 18. The total number of pluses and minuses. So X. Is the number of pluses over. The total number, which is 14, 18. Thus we have zero equals excellent 00.5 over. Route 0.25 over and equals 2.357 Thus we identify the P value from normal curve. So for normal distribution we have P equals pZ greater than 0.92 From this, we can conclude that we're going to reject H shot because he is the alpha, which means that we have evidence to support the claim that P is greater than 0.5.

All right. So what do we have this time? The types off raw materials used to construct stone tools they were found at an archaeological site called Casa de Rito are shown to us. They're given to us. We have a table and a random sample off 1486 stone tools was obtained from an ongoing excavation site. Now, what we have to do is we have to use a 1% level of significance. Okay, 1%. Level of significance means what are ALPA is 0.1 okay, and one person level of significance to test the claim that the original distribution off raw materials fits the distributions at the current excavation site. So what is going to be a little hypothesis in this case, Arnel hypothesis is that the regional distribution of raw materials that the regional distribution off raw materials fits the distribution of the current excavation site? It's the distribution. It's the distribution off the current excavation site off the current excavation site. This is what we're going to write in an AL hypothesis. And what do we write for our alternative hypothesis? It will be that the regional distribution off raw materials off raw materials does not fit. Or I can say the reason of distribution of raw materials and the distribution off. The current excavation sake are not the same. Our e can say that they are not similar, that they're not similar. Okay, All right. Now what do we have now? We have a table that is given to us. So let's just look at the table closely. We have raw material. Okay? We have raw material. One of the raw materials that we have The first one is Basil. The next one is obsidian. Then we have welded tuff welded tuff. Then we have paternal short. Or let me just write. This is PC. Okay. And then the last one is the other category. The other category. Okay, then we have reasonable percent off stone tools. All right, then. We have the regional percent off stone tools. Okay, on the stone tools. All right. This is 61.3%. 61.3%. Then we have 10.6% 10.6%. Then we have 11 point 4%. Then we have 13 points, 1% and then we have 3.6%. All right, Now, what is the observed number of tools at current excavation site? So let me just write this as the observed values, the observed values. Okay, these are 906. 906. Then these are 1 62 then these are 1 68 then 1 97 and 53. Now, what is the total? I need this total because this is going to be my sample size, my total sample size. And this student is already given to me as 1486 Right. Okay, so this is 1486 Okay, now, the first step in calculating the guys question district is goingto be to find the expected value e. Now, how exactly do I find the expected value? How do I find the expected values? These are going to be for each category. Expected values are given by the sample size in the sample size that is n multiplied by the probability off its category or the proportion that we have. Right. So the probability the probability off each category off each category. Okay. All right. So what is my sample size? My sample sizes. This oil 1 46. Okay, these are going to be the expected values. Okay, these will be the expected values. All right. So what is going to be the expected value for the first category? If I take my calculator, this is going to be in. That is one for 86 multiplied by now. I want 61.3% off 1486 So this is going to be 61% 61.3%. 0.631 This is 9. 37.666 Or let me describe this as nine. 37.67 Okay, the next one stand 10.6% off for 1/4. 861486 multiplied by 10.6. Divided by 100. This is 1 57.516 1, 57.516 Okay, then we have 11.4% off. 1486 multiplied by 0.114 This is 1 69.4041 16, 9 0.404 Then we have 13.1%. 1486 multiplied by 0.131 Right. That in 0.1% this is 1 94.666 Recognized this as 1 94.67 Yeah, and then I have 3.6% or 0.0 36 multiplied by 1486 This is 53.496 This is 53.496 but I could access 53 point Faith. Okay, These are my expected values. Now I want to find the guys question distinct. This column is going to be for individual chi square values, right? For individual chi square values. Okay, Now, what is the formula for Chi Square statistic? The overall case question The stick is going to be first one. Find the individual high school values of the formula for that is observed minus the expected, or we consider the difference of the observed and expected square upon the expected value. We do this for all the categories and in the end of the some of them, all up, and this will give us the highest question to stick for our problem. Okay, so let us do that. Now. The difference is observed and expected. This is 9 37.67 minus 906 I found this. Now I squared the study 1.6 71.67 and I divide this by 9. 37.67 9. 37 point 67 Okay, so this is 1.696 or I could write. This is 1.71 point 07 Okay, moving on to the next category. This is the difference between 1. 62 1. 57.516 We square this 4.484 and divide this by 1. 57 point 516 which is one point. Sorry. This is 0.12764 I can write this as 0.13 Okay, then we have 1 68 and 1 69.404 We square this, okay? And then we divide this by 1. 69.0 40.1 16 So this is 0.11 Okay, let me just write this much. Many of the difference between 1 97 and 1 94 minus 1 94.67 We square this 2.33 and divide this by 1. 90 4.67 which is 0.278 or regulate this as 0.3 Then we have the difference between 53.5 and 53. We square this and we divide this by 53.5. So this is 0.0 467 reconnected says 0.0 05 All right, now I need to some all of these individuals high school values, right? We saw, were you? The submission is going to give us the final chi square statistic. So this is 1.7 Yes. 0.13 less 0.11 plus 0.3 plus 0.5 Now, this is 1.246 on my guys. Question stick is 1.246 All right, so now we have this s one point 1.246 All right, now, what is do we need now? We need the degrees of freedom that is DF now. This is given by number off categories. Number off categories minus one Know how many categories do we have? If you look away? We have basal obsidian welded tuff than PC or spc Paternal short on the other. So these are the five categories, so number of categories is five five minus one is four. So this becomes four. So my diesel fuel it was for okay. Now you can either use a chi square table and with the help of the table, you will get a P value range off the values. You will not be able to find the exactly value in order to find exactly value. You need the guys for statistic, the degrees of freedom. And along with this, you need a statistical software like Excel or SPS or our python or whatever statistical tools that you're using. Like over here. I'm using an online calculator, so this will give me an exactly value. So my thigh square statistic is 1.26 If I'm not mistaking 1.246 1.2 for six and my degrees of freedom is five minus one, which is four. Okay, My significant level. There's al 50.1 Andi, I calculate this and write P value is coming out to be 0.87 my p value. My peopling is 0.87 My Alfa was 0.1 right? Yeah, I also was 0.1 All right, So what I can say over here is that since my P value is much greater than Alfa right, I can say that I will fail to reject that. I feel to reject my null hypothesis. H not right. So I can say that I do not have I do not half enough statistical evidence enough to this critical evidence to suggest to suggest that to suggest that the regional distribution off raw materials that the regional this line is very important. This is the answer the regional distribution, the reasonable distribution off raw materials. We'll draw materials and the distribution and the distribution from the current excavation site from the current excavation site and the current excavation site are different. Okay, so this is my answer. I fail to reject my inner life. But this is aged not And I say that I do not have enough statistical evidence to suggest that the regional distribution off raw materials and the distribution from the Context Commission site are different. So this is an ally prosthesis. We're saying that it fits, and the alternative is that it does not fit right. So we do not have enough evidence to suggest that the distributions don't fit. This is going to be an answer.

Yeah. We have the following matched pairs and we want to conduct a scientist and matched pairs testing the plane P less than 0.5 at alpha equals 0.16 Presents this question assessing our understanding of how to conduct a non parametric test. A scientist and matched pairs for population data. So we have to proceed through such a through D to solve first and a wee state alpha. And our hypotheses this is alpha equals 0.1 H P equals 0.5 Or an H A p is less than 0.5 are clean. Now we can get the test and B we have the signs or rather the change from population would be as follows. There are 16 pluses minuses. So X is 13/16. The number of pluses over total plus minus. Thus we have zero equals excellent 00.5 over 0.25 and routed or 2.5 years in our next identify the P value from a normal pair of this is P equals PZ greater than actually magazine equals 0.62 Thus we conclude indeed, that we reject because P is less than equal to alpha. We have evidence to support our hypothesis are cheering hypothesis. We have evidence to support the less than 0.5.


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