#$&* course Mth 163 1:25 pm10/3/2014 007.
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Given Solution: Your straight line should pass above the first and third points and beneath the second. If this is not the case the line can be modified so that it comes closer on the average to all three points. The best possible straight line passes through the y-axis near y = 2. The x = 2 point on the best possible line has a y coordinate of about 3, and the x = 7 point has a y coordinate of about 7. So the best possible straight line contains points with approximate coordinate (2,3) and (7,7). The slope between these two points is rise/run = (7 - 3)/(7 - 2) = 4/5 = .8. Note that the actual slope and y intercept of the true best-fit line, to 3 significant figures, are .763 and 1.79. So the equation of the line is .763 x + 1.79 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: OK ********************************************* Question: `q002. Plug coordinates of the x = 2 and x = 7 points into the form y = m x + b to obtain two simultaneous linear equations. Give your two equations. Then solve the equations for m and b and substitute these values into the form y = m x + b. What equation do you get? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: For the coordinates given above (2, 3) and (7, 7) substituted into y = mx + b is 3 = 2m + b And 7 = 7m + b And then use the elimination method 3 = 2m + b (* -1) -3 = -2m -b 7 = 7m + b 4 = 5m /5 M = 4/5 7 = 7*(4/5) + b 7 = 28/5 + b -28/5 35/5 - 28/5 = b B = 7/5 or 1.4 This results in the equation y = .8 x + 1.4 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: Plugging the coordinates (2,3) and (7, 7) into the form y = m x + b we obtain the equations 3 = 2 * m + b 7 = 7 * m + b. Subtracting the first equation from the second will eliminate b. We get 4 = 5 * m. Dividing by 5 we get m = 4/5 = .8. Plugging m = .8 into the first equation we get 3 = 2 * .8 + b, so 3 = 1.6 + b and b = 3 - 1.6 = 1.4. Now the equation y = m x + b becomes y = .8 x + 1.4. Note that the actual best-fit line is y = .763 x + 1.79, accurate to three significant figures. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Everything was correct and I did take not that the actual best fit line was as stated above. ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `q003. Using the equation y = .8 x + 1.4, find the coordinates of the x = 1, 3, and 6 points on the graph of the equation. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Finding the coordinates of y using x = 1, 3 and 6 into the equation y = .8(x) +1.4 is as follows: y = .8(1) +1.4 y= 2.2 y = .8(3) +1.4 y = 3.8 y = .8(6) +1.4 y = 6.2 Making the points (1, 2.2), (3, 3.8) and (6, 6.2) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: Evaluating y =.8 x + 1.4 at x = 1, 3, and 6 we obtain y values 2.2, 3.8, and 6.2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: OK ********************************************* Question: `q004. The equation y = .8 x + 1.4 gives you points (1, 2.2), (3, 3.8), and (6,6.2). How close, on the average, do these points come to the original data points (1,2), (3, 5), and (6, 6)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The first and third points are closest to the original points, being only .2 units off while the second point 1.2 units off from the original points. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: (1, 2.2) differs from (1, 2) by the .2 unit difference between y = 2 and y = 2.2. (3, 3.8) differs from (3, 5) by the 1.2 unit difference between y = 5 and y = 3.8. (6, 6.2) differs from (6, 6) by the .2 unit difference between y = 6 and y = 6.2. {}The average discrepancy is the average of the three discrepancies: ave discrepancy = ( .2 + 1.2 + .2 ) / 3 = .53. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I did not find the average discrepancy with my answer. ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `q005. Using the best-fit equation y =.76 x + 1.79, with the numbers accurate to the nearest .01, how close do the predicted points corresponding to x = 1, 3, and 6 come to the original data points (1,2), (3, 5), and (6, 6)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Y = .76 (1) + 1.79 Y= 2.55 This point is .55 from the original. Y = .76 (3) + 1.79 Y = 4.07 This point is .93 off from the original Y = .76 (6) + 1.79 Y = 6.35 This point is .35 units off from the original. The average of these numbers is: .55 +.93 +.35 / 3 = 0.61 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: Evaluating y =.76 x + 1.79 at x = 1, 3 and 6 we obtain y values 2.55, 4.07 and 6.35. This gives us the points (1,2.55), (3,4.07) and (6, 6.35). These points lie at distances of .55, .93, and .35 from the original data points. The average distance is (.55 + .93 + .35) / 3 = .61 from the points. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: OK ********************************************* Question: `q006. The average distance of the best-fit line to the data points appears to greater than the average distance of the line we obtain by an estimate. In fact, the best-fit line doesn't really minimize the average distance but rather the square of the average distance. The distances for the best-fit model are .55, .93 and .35, while the average distances for our first model are .2, 1.2 and .2. Verify that the average of the square distances is indeed less for the best-fit model. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: .55^2 = .3 .93^2 = .9 .35^2 = .1 This average is .3 + .9 + .1 / 3 = .43 .2^2 = 0.04 1.2^2 = 1.44 .2^2 = 0.04 This average is 0.04 + 1.44 + 0.04/ 3 = .50 The average of the best fit model does fit better. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 2
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Given Solution: The square distances for the best-fit model are .55^2 = .30, .93^2 = .87 and .35^2 = .12. The average of these square distances is (.30 + .87 + .12) / 3 = .43. The squared distances for the first model are .2^2 = .04, 1.2^2 = 1.44 and .2^2 = .04, so the average of the square distances for this model is (.04 + 1.44 + .04) / 3 = .51. Thus the best-fit model does give the better result. We won't go into the reasons here why it is desirable to minimize the square of the distance rather than the distance. When doing eyeball estimates, you don't really need to take this subtlety into account. You can simply try to get is close is possible, on the average, to the points. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: OK ********************************************* Question: `q007. If the original data points (1,2), (3, 5), and (6, 6) represent the selling price in dollars of a bag of widgets vs. the number of widgets in the bag, then how much is paid for a bag of 3 widgets? How much would you estimate a bag of 7 widgets would cost? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: A bag of 3 widgets would cost $5 A bag of 7 would be about $7. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: If we use the best-fit function y =.76 x + 1.79, noting that y represents the cost and x the number of widgets, then the cost of 3 widgets is y = .76 * 3 + 1.79 = 4.07, representing cost of $4.07. The cost of 7 widgets would be y = .76 * 7 + 1.79 = 7.11. The cost of 7 widgets would be $7.11. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I failed to use the best fit function. I think we were going on our original function. ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `q008. According to the function y = .8 x + 1.4, how much will a bag of 7 widgets cost? How many widgets would you expect to get for $10? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Using this function the bag of 7 would be 7$. For $10 we would do as follows: 10 = .8x +1.4 -1.4 8.6 = .8x /.8 X= 10.75 so we could only purchase 10 whole widgets. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: Using the model we obtained, y = .8 x + 1.4, we note that the cost is represented by y and the number of widgets by acts. Thus we can find cost of 7 widgets by letting x = 7: cost = y = .8 * 7 + 1.4 = 7. To find the number of widgets you can get for $10, let y = 10. Then the equation becomes 10 = .8 x + 1.4. We easily solve this equation by subtracting 1.4 from both sides than dividing by .8 to obtain x = 10.75. That is, we can buy 10.75 widgets with $10. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: OK If you understand the assignment and were able to solve the previously given problems from your worksheets, you should be able to complete most of the following problems quickly and easily. If you experience difficulty with some of these problems, you will be given notes and we will work to resolve difficulties. ********************************************* Question: `q009. Sketch a graph with the points (5, 7), (8, 8.5) (10, 9) and (12, 12). Sketch the straight line you think best fits the data points. Extend your line until it intercepts both the x and y axes. What is your best estimate of the slope of your line? What is your best estimate of the x intercept of your line? What is your best estimate of the y intercept of your line? If your graph represents the cost in dollars of a widget vs. the number of widgets sold, then what is the cost of 4 widgits, and how many widgets could you get for $20? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: My best estimate of the slope of this line was 1.03 It seems that the best x intercept is at (0.2, 0) Estimating the y intercept was (0, .3) The cost of 4 widgets would be about 3.78 For $20 we could get 20.71 widgets. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3 ********************************************* Question: `q010. Plot the three points (1, 2), (2, 3.5) and (3, 4) on a reasonably accurate hand-sketched graph. Sketch a straight line through the first and last points. What is the distance of each of the three points from the line? What is the sum of these distances? What is the sum of the squares of these distances? Sketch a line 1/4 of a unit higher than the line you drew. What is the distance of each of the three points from the new line? What is the sum of these distances? What is the sum of the squares of these distances? Which line is closer to the points, on the average? For which line is the sum of the squares of the distances less? How far from the line y = x + 7/6 is each of the three points? What is the sum of these distances? What is the sum of the squares of these distances? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Since we sketched the line through the first and last points there is no distance between them and the line by the 2nd point is the y value is .5 off The sum is .5 and the sum square is .25 The new line has the points (1, 2.25), (2, 3.25) and (3, 4.25). and the distances are .25 on each point. This is a total of .75 The sum of the squares is: .0625 + .0625 + .0625 = 0.1875 (which of course averages to .0625) The second line is closer to the original points by average. For the function y = x + 7/6 For (1, 2), (2, 3.5) and (3, 4) is as follows: y = 1 + 7/6 y = 2.16 y = 2 + 7/6 y = 3.16 y = 3 + 7/6 y = 4.16 (1, 2.16), (2, 3.16) and (3, 4.16) make the first and 2nd point .16 away from the originals and .34 from the middle point. The sum of these distances is: .66 and these distances squared were: 0.0256, 0.0256 and 0.1156 This averages out to be 0.0556 making this line closer to the points than the first 2. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3 ------------------------------------------------ Self-critique rating: OK " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: OK " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!