#$&*
course Mth 158
3/12/13 Around 1:30PM
020. `* 20
*********************************************
Question: * 4.2.8 / 2.6.8 (was 2.5.6). graph like basic stretched cubic centered around (20,20)
How well does the graph appear to indicate a linear relation?
Describe any significant deviation of the data from its best-fit linear approximation.
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
Linear. All of the plotted points are clustered near each other and form a uphill line with very moderate slope. I cannot find the best-fit linear approximation because I don't own a graphing calculator and the points aren't specifically numbered as they are just general scatter graphs for practice determining between linear and nonlinear relations.
@&
You can always sketch the graph and sketch an approximate best-fit line. Estimate the slope and the y-intercept of that line, and you'll have the equation.
This is what you will be expected to do on tests, where graphing calculator output is in any case not a legitimate way to solve a problem.
If you're unsure how to do this, let me know and I'll send you an exercise I use at the beginning of my physics classes. The exercise seems to work very well.
*@
confidence rating #$&*: 3
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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. **
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary):
I don't believe the graph I have for 4.2.8 is curved.
------------------------------------------------
Self-critique Rating: 3
*********************************************
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?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
I cannot say whether or not this table describes a function because this assignment skipped the chapter that covered functions.
@&
The basic definition of a function should have been covered in prerequisite courses. However this course does cover it soon, and the definition won't give you any trouble.
*@
I used points (20 , 335) and (27 , 350) to find the slope and equation.
(350 - 335) / (27 - 20) = 15/7
@&
Were you to graph these points and sketch a good best-fit line, it wouldn't go through any of the data points.
Basing your line on two arbitrarily chosen data points ignores all the rest of the data, which should be taken into account when sketching the best-fit line.
The general procedure is to sketch the graph, sketch your best approximation to the best-fit line, then use two points on the line (not two data points) as a basis for finding the equation of the line.
Once more, you could easily do this, and once you have the points the procedure is identical to the one you've used here.
*@
The slope of this line represents the relative increase of sales when advertising expenditures are increased.
I cannot give my equation as a function because I haven't covered those chapters yet. Here is the equation:
y - 335 = 15/7(x - 20)
y - 335 = 15/7x - 42.857 (rounded)
y = 15/7x + 292.14
@&
This equation expresses y as a function of x.
*@
If expenditures = 25
y = 15/7 * 25 + 292.14
y = 53.57 + 292.14 (rounded)
y = 345.71
Sales would be $345,710.
confidence rating #$&*: 2
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution:
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.
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.
STUDENT QUESTION
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?
INSTRUCTOR RESPONSE
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.
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary):
I used the same data points as in the incorrect solution. I am going to look at the exercise of finding the best-fit line, but I didn't know you could find the line without using a graphing calculator (which I don't have) because the chapter section didn't explain any other way to find it except by using a graphing calculator or using two reasonable data points.
Also, I am doing the chapters and sections in the order that they are done in the assignments. By doing so, when we skip ahead, I can't perform part of the questions because I haven't done those sections.
Was the work I did do acceptable????
@&
Your work was fine. The parts you hadn't covered yet are pretty easy and once you get to those topics you'll have no trouble with them.
*@
@&
That link is the exercise I mentioned earlier.
*@
------------------------------------------------
Self-critique Rating: 1
*********************************************
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. ****
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
The points on the graph I picked (after VERY briefly glancing at the best-fit line exercise to figure out how to do so without a graphing calculator) are (10 , 5) and (23, 15).
y - 5 = 10/3(x - 10)
y - 5 = 10/13x - 7.69
y = 10/13x - 2.69, approximately
I don't know the actual best-fit line because I don't have a graphing calculator.
confidence rating #$&*: 1
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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.
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary):
Since 10/13 approximately = 0.77, is my guess at a best-fit line and equation okay???? I just eyeballed it.
@&
Unless the data are widely scattered, it's not difficult to get close to the best-fit line by eyeballing.
At this level I much prefer eyeballing, which carries a lot more meaning, to the calculator, which does it for you.
Of course if you need to do an accurate analysis you'll use the calculator, but once you understand by estimating some best-fit lines, use of the calculator is very easily learned.
*@
------------------------------------------------
Self-critique Rating: 2
*********************************************
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.
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?
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary):
I chose the points (20 , 55) and (60 , 163) for my line of best-fit.
Slope:
(163 - 55) / (60 - 20) = 108/40 = 2.7
Equation:
y - 163 = 2.7(x - 60)
y - 163 = 2.7x - 162
y = 2.7x + 1
If income = 42
y = 2.7(42) + 1
y = 113.4 + 1
y = 114.4
$114,400
I am confused. I have the income on the x-axis and the loans on the y-axis. Is this wrong???? When you say ""Sketch a graph of loan amount vs. income"", I thought that equates to ""y vs. x"" and the loans should be on the y-axis. Also, +1 is a lot different from -700 which is what you said the equation should be relatively close to. Is this because I messed up the y and x axis????
If I solve with y = 42
42 = 2.7x - 1
41 = 2.7x
x = 15.185; $15,185 approximate.
------------------------------------------------
Self-critique Rating: 2
&#This looks good. See my notes. Let me know if you have any questions. &#