#$&* course Mth 163 9/30 4 007.*********************************************
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Given Solution: Your straight line should pass above the first and third points and beneath the second. If this is not the case the line can be modified so that it comes closer on the average to all three points. The best possible straight line passes through the y-axis near y = 2. The x = 2 point on the best possible line has a y coordinate of about 3, and the x = 7 point has a y coordinate of about 7. So the best possible straight line contains points with approximate coordinate (2,3) and (7,7). The slope between these two points is rise/run = (7 - 3)/(7 - 2) = 4/5 = .8. Note that the actual slope and y intercept of the true best-fit line, to 3 significant figures, are .763 and 1.79. So the equation of the line is .763 x + 1.79 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: ok ********************************************* Question: `q002. Plug coordinates of the x = 2 and x = 7 points into the form y = m x + b to obtain two simultaneous linear equations. Give your two equations. Then solve the equations for m and b and substitute these values into the form y = m x + b. What equation do you get? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (2, 3) and (7, 7) 2m + b = 3 7m + b = 7 To solve 7m + b = 7 2m + b = 3 = 5m = 4 = m = 4/5 = .8 Then you get 2(.8) + b = 3 1.6 + b = 3 b = 1.4 So the equation would be y = .8x + 1.4 confidence rating #$&*: 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): ok ------------------------------------------------ Self-critique rating: ok ********************************************* 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: y = .8(1) + 1.4 = 2.2 y = .8(3) + 1.4 = 3.8 y = .8(6) + 1.4 = 6.2 (1, 2.2) (3, 3.8) ( 6, 6.2) confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Evaluating y =.8 x + 1.4 at x = 1, 3, and 6 we obtain y values 2.2, 3.8, and 6.2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: ok ********************************************* Question: `q004. The equation y = .8 x + 1.4 gives you points (1, 2.2), (3, 3.8), and (6,6.2). How close, on the average, do these points come to the original data points (1,2), (3, 5), and (6, 6)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: These on average come pretty close. For (1, 2.2) it is a .2 difference on the y coordinate than the original (1,2) For (3, 3.8) it is a 1.2 difference on the y coordinate than the original (3, 5) For (6, 6.2) it is a .2 difference on the y coordinate than the original (6, 6). confidence rating #$&*: 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 do the average discrepancy, which is adding all the unit differences and dividing by 3 to get .53. ------------------------------------------------ Self-critique rating: ok ********************************************* 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: When plugged in you get the new y values 2.55, 4.07, and 6.35. This is a difference of .55, .93, and .35. There is an average difference of .61. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Evaluating y =.76 x + 1.79 at x = 1, 3 and 6 we obtain y values 2.55, 4.07 and 6.35. This gives us the points (1,2.55), (3,4.07) and (6, 6.35). These points lie at distances of .55, .93, and .35 from the original data points. The average distance is (.55 + .93 + .35) / 3 = .61 from the points. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: ok ********************************************* Question: `q006. The average distance of the best-fit line to the data points appears to greater than the average distance of the line we obtain by an estimate. In fact, the best-fit line doesn't really minimize the average distance but rather the square of the average distance. The distances for the best-fit model are .55, .93 and .35, while the average distances for our first model are .2, 1.2 and .2. Verify that the average of the square distances is indeed less for the best-fit model. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: .2^2 = .04 1.2^2 = 1.44 .2^2 = .04 .04 + 1.44 + .04/3 = .506… .55^2 = .3025 .93^2 = .8649 .35^2 = .1225 .3025 + .8649 + .1225 / 3 = .4299 So the best-fit model is less. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: To get the more accurate price, use the best-fit model. y = .76(x) + 1.79. y = .76(3) + 1.79 = $4.07 would be the cost of 3 widgets. y = .76(7) + 1.79 = $7.11 would be the cost of 7 widgets. confidence rating #$&*: 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): ok ------------------------------------------------ Self-critique rating: ok ********************************************* 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: y = .8x + 1.4 y = .8 (7) + 1.4 = $7.00 for a bag of 7 widgets To get $10 worth 10 = .8x + 1.4 8.6 = .8x x = 10.75 You could get 10.75 bags of widgets for $10 confidence rating #$&*: 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: 3 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 for intercepts at x and y axes is (0, 2) and (-2.5, 0). The slope would be .8 y = mx+b b = 2 and m = .8 y = .8x + 2 Cost of 4 widgets y = .8(4) + 2 y = $5.20 Could get with $20 20 = .8x + 2 18 = .8x x = 22.5 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* 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: The distance from the first point to the line is 0, the second point is about .25 and the third is 0. The sum of these distances is .25 with an average of .833… The sum of the squares of those distances is .25^2 = .0625 When sketching a line ¼ unit higher than the original line, it runs through the second point, and above the first and third point. The distance from the line of the first point is .25, the second is 0, and the third is .25. The sum of those distances is .5 with an average of .166… The sum of the squares of the distances is .25^2 = .0625 and .25^2 = .0625. So .0625+.0625 = .125 The second line is closer to the points on average because it has an average of .166… For the line with the sum of the squares of the distances the first line is closer because it is .125. With y = x + 7/6 for points (1,2) (2, 3.5) and (3, 4). y = x + 1.16666 y = 1 + 1.16666 = 2.16666 y = 2 + 1.16666 = 3.16666 y = 3 + 1.16666 = 4.16666 Sum of the differences is .16666+.033334+.16666 = .366654 with an average of .122218 Sum of the squares is .16666^2 = .02777. .16666^2 = .02777. and .033334^2 = .00111. So .02777+.02777+.00111 = .05665 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-critique rating: ok " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ok " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!