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naaar 491 Finding " mlerpreting the Concopts equaltun Equation = of the drdu regression Regression ' line for Ihe data Line Jcrictian Eacuc' Then USe Ihe linte: (Fach Tken COnstruct @ scaftes plot 0} Uhe ifmeaninaful regression Elutiiluri variablc Ifthe _ predu Vuldua significant correlalion ) If < Om Vulie chieue, meaningful to predict thaoaluc for cach du Icchnalogy. of % explain whv Height und = Number ' ol Ihe nine tallest or Stories bullding F The heights (in lect) andthe Heurton_ erat uMucT5 uxores Height; (Sexwees Exeporis Corpovareo} I02 Slutic , Sl price. Square footuge; 490 122 IZ 145.0 81.9 119.9 4,9 289.0 229.0 189.9 950 (eet 850 feel MX) fcet 70 feet Square Fooldge Home Sule Price The s4uarc (in thousands funlagcs md sal- plccs dollars) 4vcn nomcs AkrOM Ohio, #e shown table at the left. Lnela Ho Maritu 1450 square feet 2720 square lect 2175 sqquare feet 8'0 squure fect Houn Studyin Test Scores The numhet he Wuleue spent studying for test and Iheir scores O1 (hat test 1509 248 1786 TABLE FOR EXERCISE 18 4| " ' 4"|' ounn eucanur, Uealecurie n hwn 13 houts Ion 4| ' Goalytud Wins The number ol Boals scored and the numbet Wins for the top I0 teams in the 2016-2IZ Enelish Premier League cion LN' D Loguc) 4( | ' ( 4 |e P; [ Wins, Eleetrocurdivgrum hour Gunlt gonls goaks goals goals



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Please do the following. (a) Draw a scatter diagram displaying the data. (b) Verify the given sums $\Sigma x, \Sigma y, \Sigma x^{2}, \Sigma y^{2},$ and $\Sigma x y$ and the value of the sample correlation coefficient $r$. (c) Find $\bar{x}, \bar{y}, a,$ and $b .$ Then find the equation of the least-squares line $\hat{y}=a+b x$. (d) Graph the least-squares line on your scatter diagram. Be sure to use the point $(\bar{x}, \bar{y})$ as one of the points on the line. (e) Interpretation Find the value of the coefficient of determination $r^{2} .$ What percentage of the variation in $y$ can be explained by the corresponding variation in $x$ and the least-squares line? What percentage is unexplained? Answers may vary slightly due to rounding. Data for this problem are based on information taken from Prehistoric New Mexico: Background for Survey (by D. E. Stuart and R. P. Gauthier, University of New Mexico Press). It is thought that prehistoric Indians did not take their best tools, pottery, and household items when they visited higher elevations for their summer camps. It is hypothesized that archaeological sites tend to lose their cultural identity and specific cultural affiliation as the elevation of the site increases. Let $x$ be the elevation (in thousands of feet) of an archaeological site in the southwestern United States. Let $y$ be the percentage of unidentified artifacts (no specific cultural affiliation) at a given elevation. The following data were obtained for a collection of archaeological sites in New Mexico: $$\begin{array}{c|ccccc}\hline x & 5.25 & 5.75 & 6.25 & 6.75 & 7.25 \\\hline y & 19 & 13 & 33 & 37 & 62 \\\hline\end{array}$$ Complete parts (a) through (e), given $\Sigma x=31.25, \Sigma y=164, \Sigma x^{2}=197.813, \Sigma y^{2}=6832, \Sigma x y=1080$ and $r \approx 0.913$. (f) At an archaeological site with elevation 6.5 (thousand feet), what does the least-squares equation forecast for $y=$ percentage of culturally unidentified artifacts?

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.

So we find that when we enter the data into the calculator warranty. Your computer, we find at 50.3 is your y intercept and the 0.12 to roughly. 7697 is the slope. And this entering value is the square footage of the home and then the output is the value or the cost of the home in thousands. And so when you grab that we'll find it has a positive slope and the data does look to be quite linear. And so let's do some predicting. So when we plug in the 14 50 square feet, that value is a legitimate value. And it would predict the house cost of $127,900. When we put in B the 2700 20 square feet, that ends up is not meaningful because that's outside that's extrapolating. So the model we have is not including those values. Part C the 22,175 square feet, that's within the domain that comes out to be 216,000, basically $800 would be the prediction, and for part D 1000, 890 square feet that predicted value would be 100 and 81,000 $900.

This is through the estimation results. You have points as a function of experience. Experience. Square age and college are simple. Has almost 270 observations. And our square of this regression is 0.1 for one. So here I have their estimated coefficient written in blue and their standard errors in green in the bracket. In part two, you will define a turn around point of experience. You will find the partial derivative of points. Wow, with respect, Thio experience. Yeah, and you will get the coefficient of experience 2.364 miners to times the coefficient of Experience Square, which is 0.7 37. The turning point of experience is the level of experience that makes this partial derivative equal to zero. So you in set, uh, forgot times experience here. You will set this equation. Yeah, equals zero, and you can find the value of turning point experience that is 2.364 divided by two times 0.77 and you wouldn't get roughly 15.4. So the increase from 15 to 16 years of experience would actually reduce salary. This is a very high level of experience, and we can essentially ignore this prediction. Only two players in the sample of 269 have more than 15 years of experience in the next part, part three. You may see that college has a negative and statistically significant coefficient. And based on the hint of the problem written in the problem, okay, you may know that many of the most promising players leave college early or, in some cases, forgo college altogether to play in the n. B. A. These top players command the highest salaries. It is not more college that hurts salary, but less college is indicated of superstar potential. In Part four, you will add Age Square to the regression from Part one. It's coefficient is is it 10536 with a standard error of 0.492 It's T statistic. It's very above one, and recall that he start is the ratio of beta hat over the standard error of beta hat. You have 0.5 divided by 0.4 That is slightly above one so we can drop this variable. It is not significant in the same regression, the coefficient on age is minus 3.984 with a standard error of 2.6 89 Together, these estimates imply a negative increasing return to age. The turning point is roughly at 74 74 years old. And again you find that turning point by finding the value of age that makes these partial derivative of points. We respect you. Age equals zero in any case, the linear function of age since sufficient Part five This is the L s results where you regret lock of which on points experience, age and the quadratic of experience and age intercept is 6.78 coefficient of points is 0.78 of experience coefficient is 0.218 For Experience Square, it is minus 0.71 for H is 0.48 And for college. Oh, I should say there is no quadratic form of Asia. We consider we conclude we should drop it from part four. So the last variable is college. We have 0.4 as it's coefficient. The our square is point 49 Let me filled out their standard error really quick. We have 0.85 for the intercept point. 0074 points 0.5 for Experience 0.28 for Experience Square, 0.35 for age and 0.53 for college now last part part six in part six you in test whether age and college are jointly significant in the regression from Part five. Yeah, the non hypotheses is beta of the two variables, age and college both equals zero you will get you will do an F test. The F statistic you get with two and 263 degrees of freedom is about 1.19 and the P value associate it with this F test is roughly 0.31 So we are unable to reject the null hypotheses. Age and college are statistically insignificant. This means on scoring and years played are controlled, for there is no evidence for which differentials, depending on age or years, played in college.

In this particular example were asked to look at a residual plot. Now in my book, the plot is already printed for you. But in case in your textbook it is not. Let's quickly go through and do a plot. I like to use days most dot com to do this. So I'm going I selected the plus sign table and now you see I have an X. One Y one. So I'm going to type in the X values and then I'll type in the Y values. I'll pause the video and then when I come back it'll be all done. So I've entered the information into my chart and then I also went ahead and did the least squares regression. Again, that's provided for you. And we're really just looking at the residual plot. So I won't go through how to do the least squares. But you see here here's the residuals. This is what I want to see. Now notice I can see my line of fit but I can't see my information. So you can just press the minus sign and change your viewing screen a little bit screen and then we can see our dots. So remember the residual plot is plotting the distance between the green dots and our line a regression model. So with this particular program I can click on the button plot and I'm going to see all these red dots around the X axis. These are my residuals. So I'm gonna take this and I'll move this to my other screen so we can answer the question. So here is the residual plot and you could have done that yourself. You could have graft your various points using the information from the chart that they give us. But we're told that it's uh least squares. Regression model is usually good model. If our points seem to be randomly distributed around the X axis, no real pattern, a little symmetry. And this seems to be the case here. There's some randomness here, pretty much the same number as on the top is on the bottom. So in this case the answer is yes, it is a good fit and then let her be is asking about an outlier. So if a point on the plot seems far outside, it could represent an outlier and that could influence our least squares regression model. And looking here, there does not seem to be any dots that are too far away from the other ones. I mean this could be close, but not enough to really matter.


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