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Plot (iii)What type Pattem exists the data?Trend and Seasonal DamernUse Excel Solver to find the coefficlents of = multipl regression mode Krith dummy variables as ...

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Plot (iii)What type Pattem exists the data?Trend and Seasonal DamernUse Excel Solver to find the coefficlents of = multipl regression mode Krith dummy variables as follows develop Qtr2 Quarter otherwise= Qtr3 Quarter othenmse Round vour answers decima places:equation account for seasonal effects the data. Qtr]Quarterothenmse;Qir1Qtr2Qtr3Let Penod reler Ine observation Quarter of year Period I0 refer [0 Lhe Observation Quanter Year 1; and Period rerer the observaticn Quarter of Year Using the dum

Plot (iii) What type Pattem exists the data? Trend and Seasonal Damern Use Excel Solver to find the coefficlents of = multipl regression mode Krith dummy variables as follows develop Qtr2 Quarter otherwise= Qtr3 Quarter othenmse Round vour answers decima places: equation account for seasonal effects the data. Qtr] Quarter othenmse; Qir1 Qtr2 Qtr3 Let Penod reler Ine observation Quarter of year Period I0 refer [0 Lhe Observation Quanter Year 1; and Period rerer the observaticn Quarter of Year Using the dummy variables defined in part (b) and Period, develcp a equation account for seascnal effects and any linear trend the time series using Excel Solver: Round your answers to two decimal places. If Your answier negative value enter minus sign. Qtr] Qtr2 Qtr3 Period Based upon the seasonal effects the data and linear trend, compute estimates quartery sales for year Round Your ansriers cne decimal place. Quarter fcrecast Quarter icrecast Quarter fcrecast Quarter fcrecast 196.1



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Weeds among the corn Lamb’s-quarter is a common weed that interferes with the growth of corn. An agriculture researcher planted corn at the same rate in 16 small plots of ground and then weeded the plots by hand to allow a fixed number of lamb’s-quarter plants to grow in each meter of corn row. The decision of how many of these plants to leave in each plot was made at random. No other weeds were allowed to grow. Here are the yields of corn (bushels per acre) in each of the plots:$^{8}$
(a) A scatterplot of the data with the least-squares line added is shown below. Describe what this graph tells you about the relationship between these two variables. Minitab output from a linear regression on these data is shown below.
(b) What is the equation of the least-squares regression line for predicting corn yield from the number of lamb’s quarter plants per meter? Define any variables you use.
(c) Interpret the slope and y intercept of the regression line in context.
(d) Do these data provide convincing evidence that more weeds reduce corn yield? Carry out an
appropriate test at the A
0.05 level to help answer this question.

Hello, everyone, Uh, this is C seven off computer excites and check the tree. So the person we wanted to use to estimate the model that max 10 TV cool to beta zero plus beta one. A lot of expenditure plus beta to lunch program. Plus you. Here. Matt Kenney is the percentage of students passing the media Matt can like Spanish is a lot of expenditure and explained that trees in it's for per student in terms of dollars and lunch program is the percentage of students in the school lunch program. So first, we estimate the model that last one is equal to constant plus bait along like expanded. A possibility, too. This is the code progress Matt Salmon Longest finishing lunch program. So the costume tomatoes you had is minus 20.36 based on one hat, which is the car coefficient on. Well, I expect she's 6.23 and made a two hand, which is the coefficient on lunch program is negative point Trio five and our squad from dis regression ease points 18 are the signs of stop coefficients what they expected. So let's see, um, Beta one had, which is a coefficient of long extended. She's positive, which means that spending more per student will increase the percentage of students passing the test. This is something would expect. Beta two hat has a negative, Uh, is the negative side. So this is the coefficient of lunch program betweens. As a percentage of students in school lunch program increases the percentage of students passing that test decreases. That's, um Well, that could be because, uh it could mean that in some schools where the percentage of students intervention program, it might be that they're spending that there could be a lot of factors, but this is a little bit weird. Maybe unexpected. Uh, OK. Second question is, what do you make with the intercept you estimated in part one in particular. Does it make sense to set it to explain it? Three variables to zero. So because your hat in the first part we find is minus 20.36 which means one large expenditure at lunch program is he called zero. Then the percentage of students passing that test is gonna be negative. 20 which, of course, doesn't make sense, because it should be in between 0 100 but ah, So let's see if that if it makes any sense explanatory variables to equal to zero. And when we actually look at the date that we see that the minimum that the large expense you get this it points something. So it doesn't make sense to set the vehicle to zero. And, uh, also in the daytime, a minimum. Uh, Thea Lunch expenditure lunch program percentages 1.4. So he could make sense to set the lunch from that vehicle to zero. That's like nobody is in the school lunch program, but from later we see that doesn't make sense to set of a stag nature to be equal to zero. Okay, okay. The question is, now run the SIM progression of maths final, just large expenditure, and we'll make lunch program and we see that now the constant is minus 69 Betas. You Matilda point beautiful on basil until days 11 point of 16. And within a compared the slope coefficient to estimate we find in the part one and we are asked to, uh is the estimated spending effecting a larger or smaller than the part one. So in part one, the crawfish most 6.23 now in part to impart tree. It is 11.16. So the effect of expenditures on students passing the math test it's bigger. 11 is bigger than 6.23. And, uh, okay, and part four is asking us to find the correlation between large expenditure and lunch program. And instead I just fruits coronation might be spending much program. It gives me coalition here, give me the coalition table. Some people expenditure and lunch programs is gonna be minus 0.19 is the coronation and the park five is asking this to use this correlation to explain the difference between these two qualifications from two different Rick rations. So in ah, regression in turn part, we have only one extra natural available large expenditure, and in part one, we have to log expended on lunch program. So when we don't include lunch program in this regression, this knocked expenditure is also gonna capture the effect of one from him on that test. There's gonna be only too available by us to lunch. For them has a negative effect. All mad at 10 and two expended Chandler for the negative correlation. Overall, we're gonna have negative negative. We're gonna have a positive by us. So which means that log, the effect of flogged expenditure on math 10 is overestimated compared to the 1st 1 So in the first part we have 6.20 to know you have 11 point of 16 which is higher, just overestimated compared to the first part. And that is because lunch program even negatively affecting that and and not explain the children's program has a negative correlation. So overall, these two negatives makes it positive, and we have a positive bias. All right. I hope that help. Thank you for watching.

All right, We're dealing with the temperatures that they gave us in table 4.3. And if you do a scatter plot of this, we end up with their the low being January or months number one on the something that would be January. And then we have months number 12 December there. And then if we just follow this pattern, we're going to continue with a Sinus soil type graph here. So this 1 to 12 that's going to give us our period for the grass. And so that means we're looking at, in part a finding the value of B in our equation. Well, the period equals to buy divided by be in our general equation. So multiply both sides by B 12 B was too high. Divide by the 12 and we end up with B equaling high over six. No. Uh, the high and low temperatures in the table are going to give us the range of our function here. So are low temperature is at 48 degrees and our high temperature being at 82 degrees. So her part B, we're gonna find our amplitude by doing the maximum value minus the minimum value divided by two. So 82 minus 48 divided by two gives us an amplitude off 17. And then we have a midline here in our graph, and that location of the mid line is the value of K. That's the amount of our vertical shift. And so that's simply going to be our amplitude below the maximum or above the minimum. So 82 minus 17 or 48 plus 2 17 I'm gonna do the 82 minus 17. Gives us a value of K uh 65 greets and then for part C, find a value of H that will put the minimum at T equals one and the maximum at T equal seven. Okay, well, we want our minimum here at one. And normally, when you're dealing with a sign, which is what they want us trying to use, normally we start at the origin and go up from there. So that means that the origin would be in the middle of our maximum and minimum. So our value of H is gonna be the midpoint of seven had one. So seven plus one over to gives us an age value of four. So we now have the equation of lie equals or tea for temperature equals whatever your variables. Gandhi. 17 time society of by over six times equality T minus four. Close My parents, Susan, my brackets plus 65 and in part D. You were supposed to take this equation and graft that on top of your scatter plot, and you should see that it is a very good hitting. Most of the points now for party. I'm supposed to use this model to predict when the temperature is going to be 70 degrees and we're assuming that our time t equals zero is January 1st. So let's open up a new window here. And so we would have the temperature the y value of 70 equally her function 17 times by over six time quality T minus four plus 65. So now we have to solve this and figure out what forgot sign in there. Let me squeeze that in. So we should have 70 equals 17. Sign by over six times. T minus four. Plus is 65. That's much better. All right. So we would subtract 65 from both sides. Toe work on getting he signed by itself, and then we divide by 17. So we end up with five. Seventeenth's equals a sign. Uh, I over six times a quantity T minus four. So now we've got to do a little bit more algebra. First, we have to get rid of our drink function. So we do the inverse sine on both sides that gets rid of our drink function. And then once we get this decimal, we have to solve that for tea. So the inverse sine of 5/17 is going to give us point 2985 rounded off to form a decimal places that illegal pi over six times the T minus four. And then if we multiply both sides by six over pi and then at four to both sides, we end up with one t value of four point 57 But remember, when you're dealing with trig functions, you're supposed to get two different values. When you do your inverse sine, the Catholic is only going to give you one. So here's our first value, and then we would have to find our second value. Basically, our Quadrant four answer that would relate to that. Instead of the 40.2985 We would subtract that from two pi to get the value in Quadrant four and go through the same answer to get tea by itself. And that T value is approximately nine point for three. So what? Over four and 1/2 months, almost nine and 1/2 months and depending on how you calculate fractional part of the months, I'm assuming four months is going to put us into May with our fraction and doing 0.57 times 31 days for May I get May 18th for this and the same thing. This is past September, so we're in the October 0.43 times. 31 would give us October 13th by calculating it that way. The answer in the back of the book actually is off a day. They have May 19th October 14th. So they were calculating. There's probably based on number days and not using the actual months. February being a short month kind of messes. It's up to you. The calculations. So where was in a day off their answer. I think that will be all right.

Part one. The coefficient on their attempt trend is minus point 0139 and it's standard error is 1390.12 So monthly unemployment claims is a dependent variable and it enters Theis equation in lock. So the change of this kind of variable should be interpreted as percent a percentage change. The size of the Tan Trinh implies that monthly unemployment claims falls 1.4% almost 1.4% per month on average. Yeah, and given the result for the coefficient of the temp, Trine, we can calculate the T statistic. The T strut is with a head divided by its standard error. And we have a very small number of send it Errol here. So we should have a fairly large T statistic. And given the value of the T statistic, you can conclude that the trend is significant. Okay, so that is the trend of unemployment claims about their seasonality. We will look at the estimation results for the monthly dummies. We will evaluate their individual significance and their joint significance. So you will look at the t the individual T statistic. You may find that six out of 11 monthly dummy. Dummy variables have a high T statistic. Yeah. Yeah. Okay. So I can write Six out of 11 dummies are highly significant for joint significance. You in run an F test and find a p value of the F statistic. You may get the P value of the F statistic as mhm going. Oh 01 So these monthly dummies are also jointly highly significant. You can conclude that there is a very strong seasonality in unemployment claims. Okay, in part to doing ad variable Easy to the regression. It's estimated coefficient is minus 0.508 The standard error is 0.146 We can interpret this result as unemployment claims are estimated to fall. Okay, because the Beta head has a minus sign, it is negative. We we need to convert. We we need to calculate very quickly. Thio, find the change in unemployment claims, so you will need to take one minus the exponential of beta hat minus 0.5 08 so e to the power of minus 0.58 and to get their percent change, you would multiply the whole bracket with 100 and what you get is unemployment claims are estimated to fall by almost 40% after Enterprises Zone designation. For part three, we must assume that around the time of enterprise zone designation, there were no other external factors that can cause a shift down in the trend of the lark of unemployment claims. Yeah, so no other factors influence unemployment claims, unemployment claims trend around the time of easy designation. Okay, we already controlled for a time, trend and seasonality, but this may not be enough.


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