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Reserve Problems Chapter 12 Section ProblemDuring research the amount of Internet users was measured_ Each time three random groups of 10,000 people of the average ...

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Reserve Problems Chapter 12 Section ProblemDuring research the amount of Internet users was measured_ Each time three random groups of 10,000 people of the average age of 20 , were considered_ The data are as follows (11 the number of years since the beginning of the research, X2 age_ number of users):and 601700 1450 220 3300 2800 570 4750 4410 1110 6490 5930 1520204040(a) Fit multiple linear regression model using Xj and Xz as the regressors_ Round your answers to one decimal place (e.9 98.7)_(

Reserve Problems Chapter 12 Section Problem During research the amount of Internet users was measured_ Each time three random groups of 10,000 people of the average age of 20 , were considered_ The data are as follows (11 the number of years since the beginning of the research, X2 age_ number of users): and 60 1700 1450 220 3300 2800 570 4750 4410 1110 6490 5930 1520 20 40 40 (a) Fit multiple linear regression model using Xj and Xz as the regressors_ Round your answers to one decimal place (e.9 98.7)_ (b) Estimate Round your answer to the nearest integer (e.g. 9876)_ (c) Find the standard errors of the regression coefficients_ Round your answers to one decimal place (e.9. 98.7). Bo "'8z (d) Use the model to predict the number of users in random group of 40 years after years of research. Round your answer to the nearest integer (e.g. 9876)_ Numbcr of uscrs Activate Windows Go t0 Settings to activate Windows.



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Use the results from Problems $25-30$ in Section $4.1 .$ (Refer to Problem 28, Section 4.1.) The American black bear (Ursus americanus) is one of eight bear species in the world. It is the smallest North American bear and the most common bear species on the planet. In $1969,$ Dr. Michael R. Pelton of the University of Tennessee initiated a long-term study of the population in Great Smoky Mountains National Park. One aspect of the study was to develop a model that could be used to predict a bear's weight (since it is not practical to weigh bears in the field). One variable thought to be related to weight is the length of the bear. The following data represent the lengths of 12 American black bears. (a) Find the least-squares regression line, treating total length as the explanatory variable and weight as the response variable. (b) Interpret the slope and $y$ -intercept, if appropriate. (c) Suppose a $149.0-\mathrm{cm}$ bear is captured in the field. Use the least-squares regression line to predict the weight of the bear. (d) What is the residual of the 149.0 -cm bear? Is this bear's weight above or below average for a bear of this length? (TABLE CAN'T COPY)

Okay, So what do we have in this question? Now? Ben Ford's law states that the first non zero digits off numbers Ronald random from a large, complex data file have a certain type of a probability distribution, and it has been given to us. Oh, kit. So that is his draw. The stable. This is very important. So this is going to be first non zero digit first, non zero did it All right. Now, the probability according to Ben Ford's law. So this is going to be the probability. As for Ben Fords law, Okay. And then we're going to have the sample frequency. Let us call these the observed values. The observed values. Okay, so, non zero digits, these rains all the way from 1 to 9. So this is going to be a long table. So this is 1234 567 eat and nine. Okay. Now, what are the probabilities that are given to us According to this law, zero point 301 0.301 Then this is 0.176 Then this is 0.125 Then this is 0.97 This is 0.7 nine. 0.670 point 058 0.51 and then we have 0.46 Yes. Okay. This thing is disturbing as well. Again, It's not pay attention to that. Because if I refresh the page, it'll go. I left to right. All of this again. So what were the observed values? The argument was 83 49 30 to 22 83 49 32 20 to 25 18, 13, 25 18, 13. Then we have 17 and 16, 17 and 16. Okay, these are the observed values. Now, this is all the data that is given to us. And this total sample size is 2 75. Mr. Oral, sample size is 2 75. All right, now, what is the question they're saying? Use a 1% level of significance to test the claim that the distribution, our first non zero visits in this accounting file, follows the band Ford's Law. Okay, So the first point is that our Alfa 0.1 our Alfa is 0.1 What is our null hypothesis are null hypothesis is going to be that the distribution, the distribution off first, non zero desserts off first, non zero. Okay, digit in the accounting I am file follows the band. Ford's law doesn't mean that the distributions are basically the same. Follows the Ben Fords a lot. What was what is going to be the alternative hypothesis? The alternative hypothesis will be that the distribution off first non zero digits in the accounting file in the accounting file doesn't follow the band. Ford's law. Okay, now, in order to as this plane, we're going to perform ah Chi Square test and one of the first step to performing a chi Square test, it is finding the expected values finding the expected values for all the categories in the formula. For that is the formula to find this between values is sample size, the sample size that we have multiplied by the probability of the proportion off each category multiplied by the probability or the proportion off each category off each category. So Okay, so let me just put this formula interaction over you. This table will give us the expected values. This table is going to give us the expected values. Okay. All right. What is our sample size? It is to so many faith. So what is going to be the expected value for one? The expected value in this case will be to 75 multiplied by 0.301 82.775 82.775 Then for the next one is to 75 multiplied by 0.176 This is 48.4 48.4. Then it is to 75 multiplied by 0.125 34 point 375 Then we have 2. 75 multiplied by 0.97 This is 26.675 26.67 Faith. Then this is to 75 multiple of a 0.79 This is 21.725 21.72 Faith. Okay, then it is to 75 multiplied by 0.67 This is 18.4 to 5. 18.4 to five. Then there's 2 75 multiplied by 0.58 This is 15.95 15.95 Yeah. Then this is 0.51 multiplied by who's 35 that has 14.25 14.25 And then we have 0.46 multiplied by 2. 75. This is 12 0.65 These are the expected values. What is the next step? We're going to calculate the individual chi square values for all of these categories. Okay, so how is this given? Well, the formula to find the individual keister values is find the difference between the observed and the expected values square it divided by the expected value. Now, once you find all the individual values, you will some them all up, and it will give you the overall Caires questions. Stick for the problem. Let us look at this formula in action. So we're here for the first category. The difference is going to be between 83 on 82.775 We're going to square this and divide this by the expected value that is 82.775 So this is 0.6 So let me just write this zero. Then we have difference between 49 48.4. We square this and divide this by 48.4. This is 0.7 0.7 Then the difference between 34.375 and 32. We square this and divide this by 34 point 375 This is 0.16 0.16 This is the difference between 26.675 and 22. We square this 4.675 and divide this by 26 point six. Certain faith. So this is almost 0.82 Then it is the difference between 25 21 point 7 to 5. We square this and divide this by 21.75 21 0.7 to fight. This is 0.493 0.493 Okay, then we have a difference between 18 point 4 to 5 and 18. So this is 180.4 25 We square there's and divided by 18.4 to 5. So this is 0.90 point 0098 So, Yeah, this is 0.98 Okay, then the difference between 15.95 and 13, we square this 2.95 and divide this by 15.95 This is 0.0 point 545 Then we have the difference between 17 minus 14.25 We square this and divide this by the expected well do 14 0.0 to fight 0.63 0.63 Then we have the difference. Between 16 on 12.65 We square this and divided by 12.65 and we get over here 0.89 Okay, Now we're going to add all of these up. So this is zero plus 0.7 plus 0.16 plus 0.82 plus 0.493 plus 0.98 plus 0.5 for five. 0.63 plus 0.89 This is the total edition comes out to 3.553 point 55 So I can say that my chi square for this entire question is 3.55 Now what else do I need in order to find the P value? Well, I need the degrees of freedom which is given by the formula number off categories, number off categories minus one. How many categories do I have? 123456789 So this is going to be nine minus one, or I can write this as eight. All right, So I have my chi square statistic, and I also have my degrees of freedom. Now what I'm going to use is I'm going to use an online too, to get my p value. So my chi square statistic is 3.55 My visa freedom is eight. My level of significant if I look at the question is 1%. So this is 0.1 I have calculated and I find that my P value is 0.89 My people live is 0.895 Okay, it is 0.895 What was my Alfa? My Alfa was 0.1 So I can see that my P value is much greater than my Alfa hence, I will say that I fail to reject my null hypothesis. H not. This means What do I say? I say that I do not have enough statistical evidence. Studies tickle evidence to suggest that to suggest that what was the wording? To suggest that the distribution off first, non zero digits, that the distribution off first non zero digits in this accounting file in this accounting fine in this accounting file does not follow does not follow Ben Ford's lock When Ford's No right, what was the alternative hypothesis? It waas That doesn't follow the went for lawyer. So I do not have enough statistical evidence to suggest that the distribution of first non zero deserts in this accounting file does not follow Ben Folds Law, and that's how we go about doing this question.

For the match pairs given on the right. We want to conduct a sign tested matched pairs testing P does not equal 0.5 at alpha equals 0.5 significance. This question is testing our understanding of non parametric tests in particular how to conduct a scientist of matched pairs. We proceed there steps A through D below to solve. So first, in a we stayed alpha hypotheses, this gives alpha equals 0.5 H. And r P does not equal five. H. A P does not equal five. So the no hypothesis P is equal to have, the alternative is not A and B. We compute the test out. So are signs for these matched pairs are as follows and equals 12 is the total number of matched pairs. So X equals the number of plus or the total number plus and minus which is 4/12. Thus we have seen it equals x minus 5/45 over and equals negative 1.155 Thus the p value is for a normal distribution, P equals two PZ greater than zero equals 248 Thus we conclude that we fail to reject asian off because he is greater than alpha, which means that we lack evidence to support the claim P does not equal five.

Part one. This is the L S estimate We've been used. The regression results in part one for a white test. The white test is for hetero scad elasticity. For the white test, we would run another regression. The dependent variable in this regression is the square of the residual from the previous regression. Let's say it's you square and we will regress You square on the fitted value and square a fitted values from the previous equation. So we have college G p A at and College GP a head square. This is a regression with an intercept and the white test has the non hypotheses of no hetero scholastic city in mathematic form. This now hypotheses imposes doubt. Taiwan equals Delta two equals zero and data one and data to are the coefficients. I was fitted value terms. In this equation, we have to restrictions, so we will calculate the F statistic. The related degrees of freedom are two and 158 respectively. You may find that the F statistic is 3.58 and it's P value is 0.31 Given this p value Bush, we are able to reject the null hypotheses at the 5% level. Okay? Meaning there is significance evidence. What happened of hetero Scholastic city in the errors of the college G p A equation. Yeah. Okay. To fix this problem, we will run weighted least square. First, we need to examine the fitted values from the regression. In the first part, the minimum value of the fitted value is polling 027 and the maximum fitted values. The maximum fitted value is 0.165 So we can confirm that fitted values are within zero and one. Yeah, we are able to use thes fitted values as the weights in a weighted least square regression. And this is the result for after we run weighted least square regarding PC variable, there is a very small difference in its estimated coefficient. The LST statistic and W L s T statistic are very close. The R square in this regression is somehow larger than that from the regression in part one. But the old ls result and WLS result are not quite comparable. In part four. We win estimate the equation again with all l s, but with Robert standard error. Okay, you should get exactly the same estimate s in part three now. It should be the same result as in part three. Uh, not part one. So if you get a different any different estimated coefficient, you must do something wrong. The only difference is is the standard error. You wouldn't get a larger standard, Errol for almost all estimated coefficients.


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