Assignment 7

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course Mth163

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:

My straight line lies above the first and third points but underneath the middle point. My two points lie at (2,3) and (7,8).

The y axis points for x=2 is 3 so we get (2,3). The y axis points for x=7 is 8 so (7,8).

The slope is rise / run so we take 8-3 / 7-2 = 5/5 = 1.

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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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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:

Plugging the numbers x=2 and x=7 into the equation y=mx+b we get

3= 2m+b

7=7m+b

To solve for m and b we first have to subtract the first equation from the second to eliminate b.

(7m+b = 7)-(2m+b = 3) =

5m = 4;divide from both sides

m=4/5 or .8

Now we plug m=1 into the first equation and solve for b.

2*.8 + b = 3;multiply

1.6+b=3; subtract 1.6 from both sides

b=1.4

So now the equation is y = .8x + 1.4

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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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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:

X=1

Y=.8(1)+1.4

Y=2.2

(1, 2.2)

X=3

Y=.8(3)+1.4

Y=3.8

(3, 3.8)

X=6

Y=.8(6)+1.4

Y=6.2

(6, 6.2)

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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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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) and (1, 2) only changed .2 in the y values.

(3, 3.8) and (3, 5) only changed 1.2 in the y values

(6, 6.2) and (6, 6) only changed .2 in the y values.

To find the average of the 3 we have to add .2 + 1.2 + .2 together and divide by 3.

.2 + 1.2 + .2 / 3 = .53

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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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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

Y=2.55

(1, 2.55)

X=3

Y=.76*3+1.79

Y=4.07

(3, 4.07)

X=6

Y=.76*6+1.79

Y=6.35

(6, 6.35)

The y points lie at the differences of .55, .93, .35. So the average is .55+.93+.35 / 3 = .61

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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.

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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:

The square distances for the best-fit model .55, .93, and .35 are

.55^2 = .30

.93^2 = .86

.35^2 = .12

The average of these is .30+.86+.12 / 3 = .43

The square distances for first model .2, 1.2, .2 are

.2^2 = .04

1.2^2 = 1.44

.2^2 = .04

The average of these is .04 + 1.44 + .04 / 3 =.51

So the best-fit model is less.

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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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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:

Using the best-fit model y=.76x + 1.79 we get

For 3 widgets y=.76*3 + 1.79 = 4.07 or $4.07

For 7 widgets y= .76 *7 + 1.79 = 7.11 or $7.11

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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.

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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:

For 7 widgets y= .8*7 + 1.4 = 7 or $7.00

For 10 dollars worth of widget we get

10 = .8x + 1.4; subtract 1.4 from both sides

8.6 = .8x ; divide by .8

10.75 = x

So we can buy 10.75 widgets for $10.00

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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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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?

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Your solution:

To find the slope of my line with points (4,5) (15, 13) I used the formula y^2 - y^1 / x^2 - x^1 = 13-5 / 15-4 = 8 / 11 = .73.

The best estimate of my line crossing the x axis is x=-2.

The best estimate of my line crossing the y axis is y=1.8

For 4 widgets it would cost $5.

For $20 you could get 22.5 widgets.

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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?

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Your solution:

Using the points (0, 1.5) and (3.5, 5) for my line I obtained the equation y= 1x + 1.5 so after plugging in the numbers x=1, 2, 3, the distance between the 2 lines was .5, 0, .5.

The sum of the distances is .5 + 0 + .5 = 1, the average distance is .33

The sum of the squares is .5^2 = .25, 0^2 = 0, .5^2 = .25 so .25 + 0 + .25 = .5, the average distance is .17

Using my new points ( .25, 1.75) and (3.75, 5.25) I obtained the equation y= 1x + 1.5 so after plugging in the numbers x=1,2,3 the distance between the lines now is .5 + 0 + .5 = 1 with the average being .33.

The sum of squares is the same as earlier .5 with the average being .17

The line that is closest is the sum of squares line because it is .17 which is less than .33.

With the line y=x+(7/6) x=1,2,3 the distance is .17, .33, .17.

The sum of these numbers is .67, with an average of .22.

The sum of squares is .17^2 = .03, .33^2=.12, .17^2 = .03 so .03+.12+.03=.18

The average of these squared numbers is .06

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