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Albrely [Beoales ecele wfndol| HapKollWATH TO2 ALL (797 unrcad)Dersrll GraphingcultarLoying the Groundworkccalcourses/62194 sionments/225757nethlix twitter reddm Tt...

Question

Albrely [Beoales ecele wfndol| HapKollWATH TO2 ALL (797 unrcad)Dersrll GraphingcultarLoying the Groundworkccalcourses/62194 sionments/225757nethlix twitter reddm Ttub Imparted From Sa. roCtonline cumn- psych lext drive measured kz is a little off: patients with high k2 or low k2? Use the graph LYOU generated to explain your reasoning:Tcp HatLog Floating section of the Frasier River has J bend: Logs, which we model as straight line segments; float down the river: The logs must stay in the river

albrely [Beoales ecele wfndol| Hap Koll WATH TO2 ALL (797 unrcad) Dersrll Graphing cultar Loying the Groundwork ccalcourses/62194 sionments/225757 nethlix twitter reddm Ttub Imparted From Sa. roCtonline cumn- psych lext drive measured kz is a little off: patients with high k2 or low k2? Use the graph LYOU generated to explain your reasoning: Tcp Hat Log Floating section of the Frasier River has J bend: Logs, which we model as straight line segments; float down the river: The logs must stay in the river (not go on the bank} and we assume they float flat on the surface Of the water: The south bank of the river has the shape 91 and the north bank nas the shape !2 Sr" + 2 for some interval around I = 0. OSHARivecpdf R 3a, What is the longest log that can float around the bend? (You shouid justify vour answer fully rather tnan just giving numben; You may uSe numerical approximation instead of finding the exact answer ) the questions above_you found the minimum value of some function f (a). What is the physical interpretanon (in terms of logs and Tvets and 50 on) of f (a)? what is the physical interpretation of its variable: &? OSHA Criteria Ratings 1.0 pts



Answers

Use the data in AIRFARE for this exercise. We are interested in estimating the model
$\begin{aligned} \log \left(\text { fare }_{i t}\right)=& \eta_{t}+\beta_{1} \text { concen }_{i t}+\beta_{2} \log \left(d i s t_{i}\right)+\beta_{3}\left[\log \left(d i s t_{i}\right)\right]^{2} \\ &+a_{i}+u_{i t}, t=1, \ldots, 4 \end{aligned}$
where $\eta_{t}$ means that we allow for different year intercepts.
(i) Estimate the above equation by pooled OLS, being sure to include year dummies. If
\Deltaconcen $=.10,$ what is the estimated percentage increase in fare?
(ii) What is the usual OLS 95 $\%$ confidence interval for $\beta_{1} ?$ Why is it probably not reliable? If you have access to a statistical package that computes fully robust standard errors, find the fully
robust 95$\%$ CI for $\beta_{1} .$ Compare it to the usual CI and comment.
(iii) Describe what is happening with the quadratic in log(dist). In particular, for what value of dist does the relationship between log(fare) and dist become positive? [Hint: Figure out the turning point
value for log(dist), and then exponentiate. Is the turning point outside the range of the data?
(iv) Now estimate the equation using random effects. How does the estimate of $\beta_{1}$ change?
(v) Now estimate the equation using fixed effects. What is the FE estimate of $\beta_{1} ?$ . Why is it fairly similar to the RE estimate? (Hint: What is $\hat{\theta}$ for RE estimation?)
(vi) Name two characteristics of a route (other than distance between stops) that are captured by $a_{i}$ Might these be correlated with concen $_{i t}$ ?
(vii) Are you convinced that higher concentration on a route increases airfares? What is your best estimate?

What we want to conduct A p D. T. A pair differences tests at the alpha equals 1% confidence level. Testing the claim that the population means A bar next pr are not equal. We have the data for A and B. Given below assuming amount shapes, not your distribution on the right. I've already calculated D. Bar noted that an equal seven and calculated SD as 70.47 So we proceeded to do the five steps listed below to solve this first. We evaluate the requirements and hypotheses. So the requirements to use the students distribution have been met because the distribution shape we have degree of freedom and minus one equals six. Are null hypothesis is mute equals zero. Or alternative is beauty does not equal zero. And we're testing at alpha equals 00.1 confidence Next will compute the test statistic and P value. So the statistic is T equals D. Bar divided by SD over. Uh huh. 2.083 from a tea table. This puts p between .1.05. So we can conclude that P is greater than alpha, which means we fail to reject the null hypothesis, which means that we lack evidence that beauty does not equal to about.

All right. Hello, one. So I've written down the table. You're given Andi about a two lines s. So the first thing to do is to take the difference between the number of cases in one period on bond that of the previous period. Eso, for instance, here, uh, 220 to minus 95 gives you 100 27 on that basically gives you the number of cases. Total number of cases between the two intervals. Eso again here 417 minus 2022 gives you two and 48 and so on. But now the question asked you the average number of cases in each interval on. But then, to do that, you have to divide by the number of days in each interval and the first interval has five days and all the other ones have seven days. So you then divide here 127 by five to obtain 25.4 on That basically tells you that on average and you have 25.4 new cases in the first five days and then the following seventies eso between the fifth day and the 12 day you have 35.4 Sorry. 35.4 cases on average a zoo divide two and 48 by seven. And then you continue for all the different values. Eso Exactly. It's the same thing. You divide this number. So, for instance, this number here during eight by seven, 2, 15, 44. so I've just written down values. Um, I stopped a bit before because it basically is. It's the same thing for later values, Onda. That gives you three answer for question A on now, regarding the next question. Um, so when did the PMO just have evidence to locate the rate of new cases began to slow? Well, either you then simply have to take the average rate on DSI. When that starts to decline on approximately, it's more or less here. Okay? Because so at T equals 19. So, unfortunately, we don't have a day by day data. Uh, we only have every weekly data set. Um, but we can say that approximately after 20 days theme, the rate of inflection started to slow down as the average number of cases per day source to decline. Um, Okay, Right. So why would an expansion exponential model be inappropriate. So the two ways of seeing this firstly well, as you can see, it slows down. Whereas an exponential model, uh, simply grows forever on grows out of increasingly fast rate eso. In a way, the exponential model works quite well. In the early phases of this, um would fit quite well the data in the first few days on board. That's generally the case. And that means that is the case for logistic models, logistic growth models. Um, but as you can see, the number of cases slowly starts to decline, and then at the end, it becomes quite stable, which is definitely not the case for an exponential growth model. On. Additionally, there is some form of absurdity which arises through most exponential models is it's because exponential models grow to infinity on bond. Um, it would imply that at some point P would take the value, Let's say 10 billion and there are no there. There aren't 10 billion people on this earth. So, um, there is some form of of impossibility through expansion holds. Yeah. Um, Okay. So, um, getting so then we're told that a logistic model fits the data quite well. Um, And now regarding the inflection point, we know that the inflection point is the point at which the rate of growth starts to decline on. But that is basically equivalent to the question be, that is, as you can see again, growth here starts to decline. AT T equals 19 eso. That is our inflection point. And when it comes to limiting value of P, um, you know that the limiting value off P, um, which is the carrying capacity also of this logistic growth model, eyes twice the value taking at the inflection point. Now again, we don't have a precise, precise data set. A, we can say is that it is approximately a T equals 19. So what you do is you double the value taken by P. AT T equals 19, which is 800. So the model would predict more or less a total number infection off 1600. Now again, if we have a date today data, we would maybe see that this increase carries on the YouTube 20 40th or 21st day, and we'd have a slightly higher number on when you look at the the limiting value it seems to converge towards 1750 something. Okay, so we're not far off. Um, okay. And so were given a precise, uh, precise logistic function which fits this data in some sense. Andi, you asked for limiting value of team. So let me write down P, which is a function of time. Okay, 50 is equal to 1600 divided by one plus 17.53. Sorry. I eat to the power of minus zero point 14. 0, 80. Okay, Now, if you already familiar with realistic group logistic girls, you know that this year is the carrying capacity. Andi, that, uh, dysfunction p converges towards this point or another way of seeing this is to calculate this limit explicitly on. Do you know that e to the power of minus something which goes to infinity converges to zero. So this everything here converse to zero, and you're again left to left with 7 1760 divided by one. So 1760 which again fits the data quite well. We know that after 87 days, we're at 1000 and 55 the number of cases really seems to die out at this point. Thank Andi. Yes, that's that's it.

Were given the set of data points listed at the top of this whiteboard X fly. And we want to use these data points to answer the following questions. A through F. Starting off with part A on the left, we want to produce a scatter plot of these data points. I've already included the scatter plot. As you can see where the data points X. Y are demarcated by the black crosses or exits next to the right and part B. We want to compute the sum is relevant to the state to as well as the Pearson correlation coefficient. R The sums are given by following the forms exactly. So some access to some of the X values. Some why is some of the individual Y values and so on. To compute are we use the following formula which takes us input, our sample size and and the Sun is just computed. This gives our equals .9126. Next below. In parts you want to find the equation of the line of best fit which requires finding these parameters first are simple mean X bar and a sample mean Y bar are given by the sum of our X values about it by n 6.25 And some of our Y values over M 32.8. Yeah, we can find the parameters for our best fit. Line being a. As follows. The slope B is given by the equation here, which takes us input and the sample size and the sums we found above Plugging In. We get the equal 22 and then plugging in. Ry bar be an X bar to our A equation on the right gives us intercept negative 104.7. This means we have equation for the line of best fit why hat equals negative 104.7 plus 22 X. Next part Do we want to return to the scatter plot on the left and graph ry hat. Doing so we want to make sure we include our X. Men and women, which looks like this next in the bottom right part. You want to calculate? The coefficient of determination are square and interpret its meaning. This is simply the square of the correlation coefficient 0.83 to eight. We interpret this to mean that roughly 83 of the variation of the data can be explained by the corresponding variation and excellently squares line 17 of the data accordingly cannot be explained by this. Finally, in part, after the bottom, we predict y where x equals 6.5 Plugging into our white hat, we obtain 38.3.

Were given the set of data points listed at the top of this whiteboard X fly. And we want to use these data points to answer the following questions. A through F. Starting off with part A on the left, we want to produce a scatter plot of these data points. I've already included the scatter plot. As you can see where the data points X. Y are demarcated by the black crosses or exits next to the right and part B. We want to compute the sum is relevant to the state to as well as the Pearson correlation coefficient. R The sums are given by following the forms exactly. So some access to some of the X values. Some why is some of the individual Y values and so on. To compute are we use the following formula which takes us input, our sample size and and the Sun is just computed. This gives our equals .9126. Next below. In parts you want to find the equation of the line of best fit which requires finding these parameters first are simple mean X bar and a sample mean Y bar are given by the sum of our X values about it by n 6.25 And some of our Y values over M 32.8. Yeah, we can find the parameters for our best fit. Line being a. As follows. The slope B is given by the equation here, which takes us input and the sample size and the sums we found above Plugging In. We get the equal 22 and then plugging in. Ry bar be an X bar to our A equation on the right gives us intercept negative 104.7. This means we have equation for the line of best fit why hat equals negative 104.7 plus 22 X. Next part Do we want to return to the scatter plot on the left and graph ry hat. Doing so we want to make sure we include our X. Men and women, which looks like this next in the bottom right part. You want to calculate? The coefficient of determination are square and interpret its meaning. This is simply the square of the correlation coefficient 0.83 to eight. We interpret this to mean that roughly 83 of the variation of the data can be explained by the corresponding variation and excellently squares line 17 of the data accordingly cannot be explained by this. Finally, in part, after the bottom, we predict y where x equals 6.5 Plugging into our white hat, we obtain 38.3.


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