course Mth 163
My solution doesn't match the given solutions in any of these problems.
Assignment 7:007.
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Question: `q001. Note that this assignment has 8 questions
Sketch a graph of the following (x, y) points: (1,2), (3, 5), (6, 6). Then sketch the straight line which appears to come as close as possible, on the average, to the three points. Your straight line should not actually pass through any of the given points.
Describe how your straight line lies in relation to the points.
Give the coordinates of the point at which your straight line passes through the y axes, and give the coordinates of the x = 2 and x = 7 points on your straight line.
Determine the slope of the straight line between the last two points you gave.
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Your solution:
The straight line can lie from zero up It is closest to (1,2) then closer to (3, 5) than it is at (6, 6).
Mine passes right thru the origin. (2,3) and (7, 8)
My line lies below the first and second point and above the third.
The slope is 5/5 = 1
confidence rating:
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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
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Self-critique (if necessary): I have some of the solution right but other parts of it doesn't match the given solution.
You did OK, but there is no reason at all to expect the line to go through the origin. The given points determine where the line lies, and unless the origin is one of your given points, it is very unlikely that the line would go through it.
Had you not diverted your line through the origin, you would almost certainly have been closer to the ideal answer. However your answer was still reasonably close.
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Self-critique Rating: Am I doing this wrong or is it the given answer?
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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?
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Your solution:
y = mx + b
y = m(2) + b
y = 2m + b
3 = 2m +b
y = mx +b
y = m(7) + b
y = 7m + b
8 = 7m + b
5 = 5m
Divide this by 5 and 1 = m
3 = 2(1) + b
3 = 2 +b
subtract 2 from each side
1=b
y = 1x +1
confidence rating:
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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.
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Self-critique (if necessary): My solution doesn't match given solution.
Your solution can't be expected to match the given solution. It should be reasonably close to the given solution, and it is.
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Self-critique Rating: Solutions don't match.
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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.
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Your solution:
y = .8x + 1.4
x = 1------ y = .8(1) + 1.4 = 2.2
x = 3------ y = .8(3) + 1.4 = 3.8
x = 6------ y = .8(6) + 1.4 = 6.2
My equation:
y = 1x + 1
x = 1----y = 1(1) +1 = 2
x = 3----y = 1(3) +1 = 4
x =6----y = 1(6) +1 = 7
confidence rating:
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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.
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Self-critique (if necessary): Solution doesn't match.
You were supposed to use the given equation for this one, as you did. Those solutions matched.
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Self-critique Rating: My solution and the given solution doesn't match.
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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)?
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Your solution:
(1, 2.2) is .20 difference and (6, 6.2) is . 20 difference. They are close to the original point. However (3, 3.8) is a little further away. It is 1.2 difference. The average discrepancy is . 53.
My equation:
(1,2) is exact and (3, 4) and (6, 7) are close to the original points. 1.0 difference for both. My average discrepancy is 1.0.
confidence rating:
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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.
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Self-critique (if necessary): Solution doesn't match
Your result matches the given solution exactly.
There was nothing in this problem about your model; however you did correctly find the discrepancy for your model.
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Self-critique Rating: My solution and the given solution doesn't match.
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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)?
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Your solution:
x = 1----- y = .76(1) + 1.79 = 2.55
x = 3------ y = .76(3) + 1.79 = 4.07
x = 6------ y = .76(6) + 1.79 = 6.35
the average discrepancy is . 61
confidence rating:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 = .58 from the points.
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Self-critique (if necessary): My solution and the given solution don't match up.
(.55 + .93 + .35) / 3 = .61; there was an arithmetic error in this given solution.
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Self-critique Rating: Not sure
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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.
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Your solution:
My best fit line is larger than the estimate. I did everything the solution says to and rechecked my calculations but it still turned out the same answer. It was always the same answer never the given solution.
confidence rating:
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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.
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Self-critique (if necessary): N/A
Self-critique Rating: N/A
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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?
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Your solution:
3 widgets --- .76(3) + 1.79 = $4.07
7 widgets---- .76(7) + 1.79 = $ 7.11
y = .76x +1.79
confidence rating:
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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 = 4.05, representing cost of $4.05.
The cost of 7 widgets would be
y = .76 * 7 + 1.79 = 7.11. The cost of 7 widgets would be $7.11.
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Self-critique (if necessary): The given solution calculations are wrong in numerous problems. I have used a calculator and double checked my work and the given solution still says different every time. I don't know what I am doing wrong or if I am at all.
There are a couple of errors in the given solutions. There were a number of solutions based on your estimated best-fit line where the goal was to be reasonably close, but could not expect to be exact. Be sure you see my notes.
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Self-critique Rating: not sure
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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?
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Your solution:
y = .8x + 1.4
7 widgets .8(7) + 1.4 =$7
10 widgets 10 = .8x +1.4
8.6 = .8x
Divide b .8 and the answer is 10.75 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.
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Self-critique (if necessary): OK
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Self-critique Rating: OK
You're doing everything well. See my notes.