Assignment 20

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course Mth158

7/20 4pmAlso, my Chapter R test grades are not on blackboard. I have a notice on my feedback page that says it should be.

020. `* 20

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Question: * 4.2.8 / 2.6.8 (was 2.5.6). graph like basic stretched cubic centered around (20,20)

How well does the graph appear to indicate a linear relation?

Describe any significant deviation of the data from its best-fit linear approximation.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

The line is fairly linear, but does deviate quite dramatically from this line. At the lower end, the

points lie above the best fit line, and towards the upper side, they are lower than the line.

confidence rating #$&*2

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Given Solution:

* * The graph is curved and in fact changes its concavity. The data points will lie first above the best

-fit straight line, then as the straight line passes through the data set the data points will lie below

this line. **

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Self-critique (if necessary):OK

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Self-critique rating #$&*OK

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Question: * 4.2.22 / 3.1.90. Sales S vs. advertising expenditures A. 335 339 337 343 341 350 351 vs. 20,

22, 22.5, 24, 24, 27, 28.3 in thousands of dollars.Does the given table describe a function? Why or why

not?What two points on your straight line did you pick and what is the resulting equation?What is the

meaning of the slope of this line?Give your equation as a function and give the domain of the

function.What is the predicted sales if the expenditures is $25,000?

Not a function, 24 has two values.

????? I am unsure how to find the line of best fit without using the calculator. I continued the problem

using the calculator. Is this acceptable?????

The slope is 2.06

y= 2.06x + 292.88

f(x) = 2.06(25) + 292.88

f(25)=51.5 + 292.88

f(25)= 344.85

The table does not describe a function because ordered pairs that have the same first element and a

different second element. Specifically 24,000 is paired with both 343,000 and 341,000.

Student Solution:

I picked the points (20000,335000) (27000,350000).

INSTRUCTOR COMMENT: These are data points, not points on the best-fit straight line graph. You should have

sketched your line then picked two points on the line, and the line will almost never pass through data

points.

STUDENT SOLUTION CONTINUED:

The slope between these points is

slope = (350000-335000)/(27000-20000) = 15000/7000 = 15/7 = 2.143 approximately.

Our equation, using this slope and the first chosen point, is therefore

y-335000=2.143(x-20000)

y- 335 = 2.143x-42857.143

y= 2.143x+29214.857 equation of the line

Expressed as a function we have

f(x) = 2.143x+292142.857.

Predicted sales for expenditure $25000 will be

f(25000) = 2.143(25000) + 292142.857

= 53575 + 292142.857

= 345717.857

We therefore have predicted sales

f(25000)= $345,717.86

INSTRUCTOR COMMENT: Excellent solution, except for the fact that you used data points and not points on

the best-fit line.

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Self-critique (if necessary):See question above.

You should be able to plot a graph by hand and sketch a reasonable approximation to the best-fit line, then using two points on that line you should be able to determine its equation using the two-point form of the equation of a straight line.

The exercise at

http://vhcc2.vhcc.edu/dsmith/forms/ph1_ph2_fitting_a_straight_line.htm

takes you through this step-by-step. You aren't required to do the exercise or submit it, or if want to submit the work for only one or two of the graphs this is also acceptable. Of course you're welcome to submit the whole thing, but you can probably get the idea from just one or two of the graphs.

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Self-critique rating #$&*1

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Question: * extra problem (was 2.5.12). Sketch a graph of y vs. x for x = 5, 10, 15, 20, 25; y = 2, 4,

7, 11, 18. Fit a good straight line to the data and pick two points on this line. Use these points to find

an estimated equation for your line. **** What two points did you select on the line you graphed, and what

is the equation of the line through those points? **** What is the equation of the best-fit line and how

well does the line fit the data? Describe any systematic deviation of the line from the best-fit line.

****

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Your solution:,

I chose points (0,0) and (25,15), which means the slope is 3/5.

y - 15 = 3/5(x - 25)

y = 3/5x

The line fits fairly well. The first 4 points are below the line, with the last being above it.

confidence rating #$&*2

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Given Solution:

* * STUDENT SOLUTION WITH INSTRUCTOR COMMENT:

I chose the points (5,2) and (10,4)

The slope between these points is

slope = rise / run = (4-2)/(10-5) = 2/5

so the equation is

y-4 = 2/5(x - 10), which we solve for y to get

y = 2/5 x.

INSTRUCTOR COMMENT:

This fits the first two data points, but these are not appropriate points to select. The data set curves,

with increasing slope as we move to the right.

You need to sketch the best-fit line, as best you can see it, and pick two points on that line. The best-

fit line is not likely to pass through any of the data points, and you should never use data points to

determine the equation of the best-fit line.

Make an accurate sketch of the data points. Sketch your best-fit straight line, the straight line that

comes as close as possible on the average to the points. Extend the line slightly beyond the data set.

Estimate the y coordinates of the x = 1 and x = 20 points of this line. Find the equation of the straight

line through these points.

The coordinates of your points should be reasonably close to (1, 5.5) and (20, 30), though because it's a

little difficult to judge exactly where the line should be you are unlikely to obtain these exact results.

The equation will be reasonably close to y = .8 x - 3. **

STUDENT QUESTION

I chose points (5,2) and (20,11), and got model y = .6 x - 1.

INSTRUCTOR RESPONSE (why you don't want to choose data points)

Your chosen points (5, 2) and (20, 11) are data points.

You should sketch the graph, estimate the best-fit line, and use two points on the line, not two data

points, as a basis for your calculations.

If you choose two data points, you are accepting whatever error there was in measuring those points. If

you fit the trend of the data with a straight line, you the presumably random errors in the data points

tend to 'cancel out', leaving you with a much better model.

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Self-critique (if necessary):OK

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Self-critique rating #$&*OK

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Question: * extra problem (was 2.5.18). Individuals with incomes 15k, 20k, 25k, 30, 35, 40, 45k, 50k,

55k, 60k, 65k, 70k (meaning $15,000, $20,000, etc; 'k' means 'thousand') have respective loan amounts

40.6, 54.1, 67.7, 81.2, 94.8, 108.3, 121.9, 135.4, 149, 162.5, 176.1, 189.6 k. Sketch a graph of loan

amount vs. income, fit a good straight line to the data and use two points on your line to estimate the

equation of the line.

Using the points (30,12) and (160,60), you find the slope 24/65

y - 60 = 24/65(x-160)

y = (24/65)x

That's a pretty good model. You would get y = 24/65 x - 24/65 * 160 + 60

- 24/65 * 160 + 60 isn't very large, but it's not zero.

However I believe you graphed income vs. loan amount rather than loan amount vs. income. The points would have been (12, 30) and (60, 160), meaning (12 000, 30 000) and (60 000, 160 000).

ERRONEOUS STUDENT SOLUTION WITH INSTRUCTOR COMMENT:

Using the points (15,000, 40,600) and (20,000 , 67,700) we obtain

slope = rise / run = (67,700 - 40,600) / (20,000 - 15,000) = 271/50

This gives us the equation

y - 40,600= (271/50) * (x - 15,000), which we solve for y to obtain

y = (271/50) x - 40,700.

INSTRUCTOR COMMENT: You followed most of the correct steps to get the equation of the line from your two

chosen points. However I think the x = 20,000 value is y = 54,100, not 67,700; the latter corresponds to x

= 25,000. So your equation won't be likely to fit the data very well.

Another reason that your equation is not likely to be a very good fit is that you used two data points,

which is inappropriate; and in addition you used two data points near the beginning of the data list. If

you were going to use two data points you would need to use two typical points much further apart.

{]In any case to solve this problem you need to sketch the best-fit line, as best you can see it, and pick

two points on that line. The best-fit line is not likely to pass through any of the data points, and you

should never use data points to determine the equation of the best-fit line.

Make an accurate sketch of the data points. Sketch your best-fit straight line, the straight line that

comes as close as possible on the average to the points. Extend the line slightly beyond the data set.

Estimate the y coordinates of the x = 10,000 and x = 75,000 points of this line. Find the equation of the

straight line through these points.

The coordinates of your points should be reasonably close to (5000, 19000) and (75000, 277,000). It's

fairly easy to locate this line, which does closely follow the data points, though due to errors in

estimating you are unlikely to obtain these exact results. The equation will be reasonably close to y =

2.7 x - 700 .

If we let y = 42,000 we can solve for x:

42,000 = 2.7 x - 700 so

2.7 x = 42,700 and

x = 42,700 / 2.7 = 15,800 approx..

Your solution will differ slightly due to differences in your estimates of the line and the two points on

the line. **

**** What is the equation of the line of best fit? **** How well does the line fit the scatter diagram of

the data? Describe any systematic deviation of the line from the best-fit line. **** What is your

interpretation of the slope of this line? **** What loan amount would correspond to annual income of

$42,000?

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Self-critique (if necessary):OK?

??????I do not understand what happened, but I do not think I received a correct answer. Any help would be

appreciated?????

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Self-critique rating #$&*1"

&#Good responses. See my notes and let me know if you have questions. &#

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