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Of data using the 4ing collected amc type genetkcut and 7) are the same population Galculations from Questions 2012, (along With rour methods: Their Temulu Table 3....

Question

Of data using the 4ing collected amc type genetkcut and 7) are the same population Galculations from Questions 2012, (along With rour methods: Their Temulu Table 3.1 below: reduce anxiety then terms clarity student thinking penetkcs TABLE %1 Compating popuquidic aqualion- uslng the Hardy-Weinberg Frcquencies Gcnolipe Equitalnt 2001 2012 Phcnotype Btological Term 7042Homozygous Dominant 0 Heicrozykous 5740.12{0ds Reressvc HGIortJorovunand reduce anxiety when penetics temu to clarity student thin

of data using the 4ing collected amc type genetkcut and 7) are the same population Galculations from Questions 2012, (along With rour methods: Their Temulu Table 3.1 below: reduce anxiety then terms clarity student thinking penetkcs TABLE %1 Compating popuquidic aqualion- uslng the Hardy-Weinberg Frcquencies Gcnolipe Equitalnt 2001 2012 Phcnotype Btological Term 7042 Homozygous Dominant 0 Heicrozykous 574 0.12 {0ds Reressvc HGIort Jorovun and reduce anxiety when penetics temu to clarity student thinking TABLE 3.2 Still comparing (populatlon5 lequallon; uslng Ihe Hardy-Weinberg Frequencies Alcie H-W Equivalent Biological Tcrm 2001 2012 0.65 Dominant Allelc Recesslvc Allele conducted determine whetherthe genotype frequencies of Chi-square analysis was different from 2001 to 2012.Note: The same analysis this population were significantly for thisandthe next question we canbe used compare the allele frequencies; however; results: genotype frequency The geneticists found the followingz will focus only genotype frequencies between 2001 and 2012 signifi- 20.0 and 0.001.Were the cantly different? Explain how you know: 12 Microevolution defined as 'the change in the allele or genotype frequency of a popu- lation over time (Brooker et al,, 2012) Did the white-footed mouse population evolve? Explain how you know:



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? $$\begin{array}{|c|c|c|c|c|} \hline {}{} {\text { Questions Arranged from Easy to Difficult }} \\ \hline 24.64 & 39.29 & 16.32 & 32.83 & 28.02 \\ \hline 33.31 & 20.60 & 21.13 & 26.69 & 28.90 \\ \hline 26.43 & 24.23 & 7.10 & 32.86 & 21.06 \\ \hline 28.89 & 28.71 & 31.73 & 30.02 & 21.96 \\ \hline 25.49 & 38.81 & 27.85 & 30.29 & 30.72 \\ \hline\end{array}$$ $$\begin{array}{|c|c|c|c|} \hline {}{} {\text { Questions Arranged from Difficult to Easy }} \\ \hline 33.62 & 34.02 & 26.63 & 30.26 \\ \hline 35.91 & 26.68 & 29.49 & 35.32 \\ \hline 27.24 & 32.34 & 29.34 & 33.53 \\ \hline 27.62 & 42.91 & 30.20 & 32.54 \\ \hline\end{array}$$

So the variable under consideration is hours slept. The two populations we are looking at is some a long word. Virtual. Just abbreviate to LH And the second one there is one C. Is going to be just be the differences between these populations. Okay. And can we use these differences? Yes we can. Since they are differences. Yeah. So our alternative hypothesis is going to be the thing that one of them is greater than the other. Actually it's gonna be, the first one is greater than the other. So um you want greater than youtube making that a right tailed test. Now they give us that D. bar is 2.33. So put that there & SD. is 2.02. Mhm. Yeah. And we have how many samples? 345 10. So two divided by Square root of 10. This is going to be approximately equal to let me bust out my calculator here. 2.02 Divided by Square Root of 10. 2.33 divided by 6.66 there. Yeah .63. This is about equal to 3.680. This is going to get us a p value of Oh, I gotta go all the way back the book here. It wasn't at the right page. And by analysis of critical values 3.688 lands as well, past the 1.833 for the 5% critical value And for the 2.4-1 critical value at the 1% significance level. So in both cases, we will reject the null hypothesis.

What do we have this time? Well, the types off browse favored by deer are given to us in the table. Using binoculars, volunteers observed the feeding habits of a random sample of 320 years. Because our sample size in his 320 what we have to do is we have to use a 5% level of significance, which means that our Alfa our Alfa is 0.5 okay, 5% level of significance to test the claim that the natural distribution off the brows fits the deer feeling pattern. Okay, so it's going to be our null hypothesis are null hypothesis for this question is going to be that the natural distribution, the natural distribution off browse fits the dear feeding fat. Um, okay. And what is going to be the alternative hypothesis? The alternative hypothesis will be that the natural distribution off brows does not fit the deer feeding pattern, or I can say, and the deer feeding pattern and the dear feeding Batam are not the same or similar are not similar. All right, now we know our alphas. 0105 Now we have been given a table So that is Just look at the table. Now. First we have the type of brows, the type off. Gross. All right, this is going to be the first column, or here we have sage brush, rabbit brush. We're here. We have sage brush. Then we have Excuse me. Uh, how does this building? Okay, rabbit brush, Rabbit brush. Then we have sold brush, sold brush, then we have serviceberry. Then we have serviceberry and then the other category. Okay, so now what is the plan? Composition in steady area that is also given to US land composition in study area in study area. Now, this is also given to us. This is 32% 32%. Then this is 30 8.7%. Then this is 12%. Then this is 9.3% and then it is 8%. Okay, now the observed number of deer feeding on this plant. Which means let me just write this as the observed value, the observed values What are we observed? Values that we have? It is 102 Then it is 1 25. Then it is 43 27 in 23. Now What exactly is the sum of all of this? We need this because that is going to be our sample size. And this is 3 20. Okay, so this is our sample size. The next thing that we do in order to calculate the Chi Square statistic is to find the expected values, the expected values, the expected values for all the categories. And they're given by the sample size, the sample size, which happens to be 3. 20 in this case which is also been ordered by in multiplied by the probability the probability off each category off each category Okay, the probability or the proportion off his category that is given to us. So let us see how this is calculated. We're here. We're going to have the expected value, the expect dead values. All right, now to the calculator, I go. I use my calculator to find all the respected values. Now, for the first category is going to be 32% off 3 20 against 0.32 multiplied by 3 20 which is 102.4102 point four. Then I have 38.7% or 0.387 multiplied by 3. 20. This is 1. 23.841 23.84 Then we have 12% 0.12 multiplied by 3. 20. This is 38.4, 38.4. Then we have 9.3%. 0.93 multiplied by 3. 20 it was 29.76 29.76 and then we have 8% off 3. 20 0.8 multiplied by 3. 20. So this is 25.6 25.6. These are the expected values. Now I need to find the individual chi square values. Okay, so administrate this as individual as individual chi square values. Okay, now how exactly? What? Calculate these. Okay? The formula to calculate the individual rascal values is the difference between observed minus expected whole square divided by the expected value. And once I do this for all the categories, I will sum them all up. And this is going to give me the total high square value for my problem. So let us just look at this. This is going to be the difference between observed and expected values. So in this case, it is going to be 0.4. I square this right, I am going to square this. So the 0.16 this is divided by the expected value which is 10 2.4, which is 0.15 Or let me just write this as 0.2 Similarly, over here there is a difference of 1 25 and 1 23.84 I square this square 1.16 and I divide this by 1. 23.84 This is 0.108 Or I can just write this 0.1 Okay, then there is a difference. Off 43 38.4. I square this and I divide this by 38.4, which is 0.550 point five fight. Then there is a difference off 29.76 and 27. I square this 2.76 and I divide this by the expected value off 29.76 which is 2.0 point 255 Now I can just write this as 0.26 All right, then there is a difference off 25.6 and 23 2.6. I square this and this is divided by 25.6. So this time I have 0.2640 point 264 And then what I do is I some all of these individual chi square values up. So this is going to be 0.2 plus 0.1 plus 0.55 plus 0.26 plus 0.264 This is going to be 1.86 This is 1.86 This is the value off my chi square statistic. This is the value of Mike Rice. Question District 1.86 All right now I have my guy square statistic. Now, I wish to find out the p value right for p value. I need to find the degrees of freedom that is DF and it is given by the formula number off categories number off categories minus one. Now, how many number of categories do I have? Well, I have this away. A sage brush, rabbit brush, salt brush. So it's very and the other. So there are five categories. Right, So this is going to be fine. Minus one. Okay, this is five minus one. This is right. This a little properly. Okay, so this is five minus one. Okay, which is four now? In order to find the P value, you can either use the chi square table, which will give you an approximate value. Arrange off evil. You will not give you the exact value. Or you can use a statistical software like SPS s or R python or a chi square calculator. As I'm doing over here, I'm going to use an online calculator. I have my chi square statistic as 1.86 Okay, 1.86 animal You The freedom is four. My significant level is 5%. Right. So this is going to be at 50.5 and I hit. Calculate, this is gonna be a me the exact banding and a P value of 0.8964 and I can see that my P value is 0.8964 Now I can see that my P value is much greater than Alfa. What was my Alfa? My Alfa was 0.5 right? This is my Alfa, so I can see that my P value is much greater than my Alfa. And hence I will say that I same to reject my null hypothesis statement. Okay, let us go away and look at the alternative hypothesis. And in l a hypothesis Manal hypothesis waas that the natural distribution off browse fits the deer feeding patterns. So I will say that I do not have enough statistical evidence to say I will say that I do not have enough statistical evidence to say that the distribution this line is going to be an answer. That the distribution Okay, just a moment. Let's just look at what that line is. The distribution, the natural distribution of browse fits the distribution off browse fates and the deer feeding pattern and the deer feeding pattern are different. Are different are I can also say that whatever it in over here Yeah, I will say that I do not have enough statistical evidence to say that the natural distribution of rows fates that the natural distribution off off brows does not fit the deer feeling pattern. Okay, at 5% level of significance. So this is how we go about doing this question on this is my answer.

Okay, Let us look at this question. Now, the fish and the game department stopped Lake Lulu with fish in the proportions that were given to us. Okay, so what we have here are different kinds of fish. So let me just make a column over here. So this is going to be kind off fish. Okay, so first we have catfish, we have catfish, then we have bus. Then we have blue kin. Then we have bluegill, and then we have bike. So this is going to be 1 20. This is these are the observed values that we have. 85. This is 2. 20. And this is 75. Okay, now they stopped them in different proportions. 13 30%. 15%. 40%. 15%. Again five years later. So this was the observed value. These are the observed values, the observed values. Okay, now, over here, we'll read the original proportions. Original proportions, original proportions. Now, the original proportions, who are 30% for catfish in this was 15. This was 40%. And the last one is 15 again. All right. Now it was found that 500 fish in the sample were distributed the way it has given the observed values. In the five year interval, did the distribution of the fish changed at 0.5 level? So for this question, we can say that our Alfa is 0.5 Okay, Now, what is going to be another hypothesis? Arnel hypothesis will be that the distribution, the distribution did not change did not change in the past five years in the past five years. All right. And what will be the alternative hypothesis? The alternative hypothesis will be that the distribution changed in the past five years in the past five years. All right, so now the tests that were conducting is the Chi Square test and what is the first step in order to calculate the value for the chi square statistic. First, we need to find the expected values, the expected values, the expected values for all the categories. Now it is given by sample size. The sample size for a question is 500 right sample size multiplied by the probability or the proportion probability Oh, are the proportion off each category off each off each category. Okay, off each category. All right, So what is my sample say's my sample size is 500. If I just add all of these up, I'm going to get 500 rate. So what will be my expected values? My expected values. My expected values are going to be if I just use my calculator over you. Just a moment. Yes. They are going to be 30% off. 500 now, which is going to be 1 50 Then we're going to have 15% which means 75. This is also going to be 75. And this one is 40%. So 40% is going to be 1 60 right? Really? This is going to be 1 60? Nope, This will not be 1 16. So this is going to be 200. All right. 1 1503 100. 200. 500. Okay, these are expected values. Now, what is the next step? The next step is finding the guy sequester distinct. How do we do that for all the categories we're going to perform? The operation observed value minus the expected value. Whole square. That is we are squaring the difference between the observed and the expected values we divide this by the expected value and in the end, with some all of the individual values up. So this will give me the chi Square statistic for the entire problem. So if I go up, these are going to be my individual chi square values for all the categories. Right? Individual high school values. Now, what exactly did we do now? We're going toe. Do the observed minus the expected and then hold square. So this is the difference between 1. 15 and 1. 20. Now we're going to square this and we'll to divide this by the expected value. Okay, so this is 1 50 minus 1 20. We square this. So this becomes 900 on DWI. Divide this by 1. 50. So this is going to be six. Okay, then. 85 minus 75. We square this and direct this by 75. This is 1.331 point 33 All right, then we have the difference between 202. 20. So the square is 400 divide this by 200. So this is to And then I think over here the difference would be Cedo, right? Oh, minus zero. Yeah, Now what I'm going to do is I'm with some all of these individual values. So this is six plus 1.33 plus two and this becomes 9.33 So I can say that my chi square statistic turns out to be 9.33 Now, what else do I need In order to find the p value? I need the degrees of freedom. Degrees of freedom is what DF it is given by the formula number off categories number off care degrees minus one. All right. How many categories do we have if we look away? There are four categories catfish based, Google and pipe. So this is going to be four minus one or I can say that this is equal to three. My degrees of freedom is equal to three. Now, my chi square value is 9.33 and my degrees off freedom is three. Now, we can either use a chi squared even for this or what we can do is we can use a statistical package or a software which will give us an exact P value. What is our Alfa for this question? Are Alfa for this question is 0.5 Okay, so I put in all the values and I hit Calculate. I get my P value as 0.25 to 1. So we're here. I will say that my p value is 0.25 to 1. All right, Our Alfa was 0.5 and we can clearly see that RPI value is less than Alfa hands. We can say that we will reject that we reject are null hypothesis. H not know what was on a hypothesis. The null hypothesis waas that the distribution is still the same after five years. So I can say now this is the most important line. I have enough statistical. I have enough statistical evidence Enough statistical evidence to suggest do suggest that the distribution, the distribution, the distribution off fish off fish off fish in the league in the league has changed. Yeah, in the league has changed over the past five years over the past five years and this is how we go about doing this question

So using the data in the G p a two on 4107 college students, we have an equation, which is Yeah. So this is a question on the cold G p a is measured on ah force Cape four point skill. The H S P R C is the present out in the high school. What do you think class on the SNC is? The combined mats of a boss cause on the achievement. Now the first question says, Why does it make sense for the coefficients on the h S e r c to be negative? So why does it make sense for the girl officials to be negative? It makes sense for the commission to be negative, because if the commission was positive, the students, because their corruption is negative, right, it was positive the students or it's called less potential, would have been getting higher GPU in the college. So So it makes sense. That's the coefficient is negative, because if it was positive students Ohad scored less percent. House would have been getting higher GP. So that's that's now. The second question said, what is the predicted collar GP when the yeah, is it? Quote to 20 on the s etc. Is it called 1000 and 50? Why is pretty tight collar G p A. So if this is the cost 20 and this is cost 1000 and 50. The call g p a. We quotes one point three nights two point minus 0.13 five multiplied by 20 which is this plus zero points 001 for eight multiplied by 1000 and 50 on when you solve this the Kogi P o. B 2.6 76 So this is the predicted Uh huh g ph She now the third question safe. Suppose that so high school graduate and be glad to take any state percentile from high school but a since students eight sec school was was 1 40 points. Iowa. What is the predictable difference in the color G P A. For these two students on is the difference lodge. So we're going to solve the so since the HSB are is the same for is the same for am be so it's is the same. Yeah, for A and B therefore, the predictor difference between them with A and B scholar GPS is just a difference of their s a t score. He predicted Big difference between am Bs G p A is just It's just a difference off there s it is cause multiplied by the coefficient or variable s etc. In the question come so therefore the critical difference ngp a be equal to the 1 ft see points IRA multiplied by 0.14 AIDS and that's a 0.72 So the difference does not seem to be lodge now. The final question holding hs PRC fixed What difference SS is called lead through a predicted co g p a difference in 0.50 or one off off a great point. Our comments on your answer. So the difference entities caused list three predicted, called g p a difference off 0.5. So the formula will be 0.5 divided by 0.70148 on this is 33 seven points eight 37 eggs on this is proved so therefore it is proved on. This is the predicted This is the difference as it is caused. Oliseh predicted co g p. A difference off 0.50


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