#$&*
course Phys201
6/18@6:42pm
fitting a straight line to data#$&*
Phys201
Your 'fitting a straight line to data' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** Fitting a Straight Line to Data_labelMessages **
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2 hrs
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Copy this document, from this point to the end, into a word processor or text editor.
Follow the instructions, fill in your data and the results of your analysis in the given format.
Any answer you given should be accompanied by a concise explanation of how it was obtained.
To avoid losing your work, regularly save your document to your computer.
When you have completed your work:
Copy the document into a text editor (e.g., Notepad; but NOT into a word processor or html editor, e.g., NOT into Word or FrontPage).
Highlight the contents of the text editor, and copy and paste those contents into the indicated box at the end of this form.
Click the Submit button and save your form confirmation.
Note that the data program is in a continual state of revision and should be downloaded with every lab.
Most student report spending between 1 and 2 hours on this lab exercise. A few report as little as 40 minutes. Some report significantly longer times. If you are adept with graphs you will probably tend toward the shorter times; if the exercise requires longer then there's a good chance you need the practice. This topic is important, and its importance goes well beyond your physics class.
The figure below depicts a series of six data points and a straight line segment.
The line segment is part of the 'best-fit line' for the data.
You may think of this as the line that comes as close as possible, on the average, to the data points. This definition can be improved upon--it doesn't specify exactly how 'closeness' is defined or how 'closeness' is averaged--but it gives you the general idea, and it is sufficient for 'eyeball' estimates of the best-fit line.
The specific definition, which you don't really need right now and might not need in this course is as follows:
Each data point lies either above or below the line, at some specific vertical distance from the line. This vertical distance is the deviation of that point from the line. The best-fit line is the straight line that minimizes the sum of the squared deviations--there is no other line with a smaller sum of squared deviations.
The exercise here will be to obtain eyeball estimates of the best-fit line.
Any graphing calculator or computer spreadsheet will calculate the best-fit line for a given set of data, but neither calculator nor spreadsheet should be used in this exercise.
Take a piece of string and stretch it over the line (you may do this on the screen or you may use a printout). Then raise and lower the string, and see how the distances to the various points change. Vary the slope of the string.
Can you see that if the line in the figure below is raise or lowered a significant amount, and/or if its slope is changed significantly, the average distance of the points from the line will change?
Answer this question below and indicate also by how much the slope of the string needs to change before the increased average distance between the string and the points is apparent.
The slope would need to decrease by about .01 before the increased average distance between the string and the points is apparent.
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Give in the first line below the horizontal and vertical coordinates, in comma-delimited form and in that order, of the rightmost point on the line, as best you can estimate them.
In the second line give the same information for the leftmost point on the line.
Starting in the next line give a brief statement telling in your own words what the numbers you have entered mean.
(.53,17.7)
(18.7,.13)
&&&
(.53,17.7)
(.13,18.7)
&&&&&
@&
I don't believe you provided the coordinates in the order x then y in both of your answers. In one, you did. In the other, you appear to have reversed them.
*@
The numbers entered are the coordinates of the leftmost and right most points on the graph. This would be (x,y). You could potentially calculate slope from these points. &&&& These points indicate direction or how far you go on the x axis and y axis from the origin&&&&&
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At you move from the leftmost point to the rightmost, by how much does your horizontal coordinate change, and by how much does the vertical coordinate change? Indicate in the first line using comma-delimited format. Use + or - with each result to indicate whether the change in the coordinate is positive or negative.
Starting in the next line give a brief statement telling in your own words what these numbers mean and how they were obtained.
As you move from the leftmost point to the rightmost, your horizontal coordinates increase. The vertical coordinates decrease.
(.53,17.7)-
(.5,18.3)-
(.4,17.8)-
(.25,18.4)+
(.13,18.7)+
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&&&&
+.4, -1
+.4 means that you are moving to the right horizontally .4 units, -1 means that you are moving down vertically 1 units. They were obtained by finding the differences between the x coordinates and the y coordinates.&&&&&
The two numbers in the first line are the coordinates of your first point.
I can't tell what the numbers in the next four lines represent.
However, only two numbers were requested, and I don't see anything here that appears to answer the question.
Can you provide the two numbers that answer the question, and indicate in detail how you obtained them?
*@
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The change in horizontal coordinate is called the 'run', and the change in vertical coordinate is called the 'rise', between the two points.
The slope between the two points is the rise divided by the run: slope = rise / run.
In the first line give the run and the rise, delimited by commas.
In the second line give the slope.
Starting in the next line give a brief statement telling in your own words what these numbers mean and how they were obtained.
Runs: 17.7,18.3,17.8,18.4,18.8 Rise:.53,.5,.4,.25,.12
Slopes:33.4,36.6,44.5,73.6,156.7
The Rise is going to your y coordinates. The Run is going to be your x coordinates. The slope is your rise divided by your run and is calculated between the two points on a graph.
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&&&&&
.4,-1
Slope=-2.5
As you travel from the leftmost point to the rightmost point, how much you go in the y direction is the rise and how much you go in the x direction is the run. The slope is your rise divided by your run or the ratio between the rise and the run.&&&&&&
Only one rise and one run were requested.
There is only one rise between the points, and one run.
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The vertical axis of a graph occurs at horizontal coordinate 0. If a graph doesn't show an axis through horizontal coordinate zero then it doesn't show the vertical axis. Not all graphs will show the vertical axis. For example of the data are clustered closely together at a great distance from the vertical axis, a graph would have to be extended through a lot of 'blank space' to get to that axis, and the detailed behavior of the data would be obscured.
Does the vertical axis show on the above graph?
If the straight line is extended until it intercepts the vertical axis then the point where this occurs is called the vertical intercept of the graph.
If this straight line is extended, does it intercept the vertical axis?
If so, at about what vertical coordinate does this occur?
Answer these questions in the below. Answer each in a concise yet complete sentence that includes the reasons for your answer.
Yes the vertical axis shows on the above graph. If the straight line was extended, it would intercept the vertical axis. The coordinate at which this would occur would be approximately (0,19).
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&&&&&& The vertical axis does not show on the above graph. If you extend the straight line until it intercepts the vertical axis at (0,19).&&&&&
The equation of a straight line in the x-y plane can be expressed in the common form y = m x + b, where m is the slope of the line and b the y-intercept.
Substituting your estimates of the slope and vertical intercept for m and b in the form y = m x + b, you obtain an estimate of the equation of the best-fit line.
Give your estimated equation for this line below. Add a brief statement about the meaning of your answer and how it is related to previous answers.
y=mx+b
y=(0)(0)+19
y=19
&&&&&&
y=-2.5x+19 -2.5 is your slope (which was found previously) so that would go in the m position. 19 is where the line crosses the y axis.&&&&&&&&&&
The y coordinate would be 19. It would make my hypothesis true that if the straight line was extended, it would intercept the vertical axis at (0,19).
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In may cases it is not appropriate to include x = 0 in the scale of the graph, in which case it is not possible to extend the line to the actual y axis in order to estimate the y intercept. If this is the case we can still estimate the equation, based on the coordinates of our two points.
You have already estimated the slope of the line. Leaving x, y and b in symbolic form, substitute m into the form y = m x + b. What equation do you get?
y=mx+b
y=0x+b
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&&&&y=-2.5x+b&&&&&
Now substitute the x and y coordinates of either of your two points. It doesn't matter which point you choose. Choose a point, substitute its x coordinate for x and its y coordinate for y. What is your equation? What symbol(s) remain in the equation?
y=mx+b
17.7=(m)(.53)+b
the symbols b and m remain in the equation
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&&&&17.7=-2.5(.53)+b
17.7=-1.325+b&&&&&&&
You can solve this equation for the one remaining symbol b. Do so, and indicate your solution in the first line below. Starting in the second line show and explain the steps in your solution, in detail.
17.7=(m)(.53)+b
17.7=(33.4)(.53)+b
17.7=17.702b
divide each side by 17.702
b=.99
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&&&&&
17.7=-2.5(.53)+b
17.7=-1.325+b
b=19.025&&&&&&
How does the value of b compare with the y-intercept you estimated earlier? How far in actual distance on the y axis is the b from the vertical intercept you estimated earlier? Why can we not expect the two values you obtained to match exactly?
&&&& Earlier I estimated that the y intercept would be located at (0,19). When I calculated b it equaled 19.025. This answer is almost identical. The two answers will not match exactly because the first y intercept calculation was an estimate. The second was an actual calculation. Therefore, there is a .025 difference between the two answers.&&&&&&
Answer these questions in complete statements that indicate your answers as well as the meaning and basic of your answers.
The value of b is far from the y-intercept I estimated earlier. It is approximately 18 spots away. The two values are not going to match exactly because of human error and the numbers earlier were an estimation.
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The actual best-fit equation for the above data set is y = -2.2032x + 18.982.
By what percent of the actual value -2.2032 of m in the above equation did your estimate differ from that value?
By what percent of the actual value 18.982 of b in the above equation did your estimate differ from that value?
Give your two results, delimited by commas, in the first line below.
Starting in the next line give a brief statement telling in your own words what these numbers mean and how they were obtained.
I guessed it would be 0 so at about 2%.
My value of 19 differed less than 1% of b.
(0,19)(-2.2032,18.982)
The first numbers were obtained through my estimation. The second set of numbers were given.
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&&&&&13.47%,.09%
The two numbers differ by percentage. The actual slope and the slope we estimated differs by 13.47% and the actual y intercept and the y intercept we estimated differs by .09%. -2.2032—2.5/-2.2032=13.47% and 19-18.982/19=.09 &&&&&&
@&
I can't locate your specific answers to the above questions, and cannot connect your subsequent answers to those answers.
Can you clarify your work on this graph?
*@
Now stretch your string out to form the line you think best fits the data in the figure below.
Note that the line that best fit the data in the first graph did not pass through any of the data points.
A best-fit line is very unlikely to actually pass through and data points. When fitting a best-fit line you should do your best to avoid allowing it to pass through any data point, and should allow this only if it is very clear that this is the only way to get the best fit to the entire set of data.
Your goal is to come as close as possible, on the average, to the data points. Unless you have a very good idea what that means, don't worry about the squared-deviation terminology. Just try to get as close a possible, on the average.
Estimate and record the coordinates of two widely-separated points on your line, but do not include the y-intercept among these two points.
This means two points on your line, not two data points.
Estimate and record the y-intercept of your line, if the graph does indeed pass through the true y axis. If this is not the case, indicate it.
Give the x and y coordinates of the first point on your line, separated by a comma, in the first line below.
Give the x and y coordinates of the second point on your line, separated by a comma, in the second line below.
Give the y coordinate of the y intercept in the third line.
&&&&(.1,2)
(.3,4)
(0,0.8)&&&&&
Starting in the next line give a brief statement telling in your own words what these numbers mean and how they were obtained.
( .2)(0)
(.1,1.98)
(.3,3.98)
Rise- your point at y
Run-your point at x
Slope-calculated between the 2 points on the graph.
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&&&&& The two x and y coordinates (.1,2)(.3,4) are random points located on the line of best fit drawing. The y intercept would be the point in which the line of best fit intersects the y axis. This point is located at (0,0.8)&&&&&&
Give the rise, run and slope between the two points on your line. Give these results in three lines, one number in each line. Starting in the fourth line explain how you obtained the rise, run and slope.
Rise: (3.98-1.98)=2
Run: (.3-.1)=.2
Slope=2/.2=10
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&&&&
Rise =2
Run=.2
Slope=10
I obtained the rise by doing 4-2=2. I obtained the run by doing .3-.1=.2. The slope was calculated by dividing the rise by the run 2/.2 which gives you a slope of 10.
Give in the first line the equation of your line, based on the slope and the y intercept. Starting in the second line give a complete explanation of how you obtained your equation.
y=mx+b
.2=10(2)+b
.2=20b
divide each side by 20
b=.01
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&&&&&y=10x+.8
I obtained this by using the equation y=mx+b. I then plugged the slope of 10 into the m spot and .8 into the y intercept position.&&&&&
Give the equation of your line, based on your two points (as before, substitute the slope into the y = m x + b form, substitute the x and y coordinates of one of the two points, and solve for b). Give the equation in the first line, then starting in the second line explain how you obtained your result for b.
y=mx+b
y=(10)(.3)+3.98
y=6.98
b=y coordinate
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&&&&&2=10(.1)+b
2=1+b
subtract 1 from each side
b=1&&&&&&&
Repeat this exercise for the graph below:
Give the x and y coordinates of the first point on your line, separated by a comma, in the first line below.
Give the x and y coordinates of the second point on your line, separated by a comma, in the second line below.
Give the y coordinates of the y intercept in the third line.
(1,1.4)
(2.1,2.1)
(0,1)
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&&&&& (1,1.5)
(3.8,2.5)
(0,1.3)&&&&&
Give the rise, run and slope between the two points on your line. Give these results in three lines, one number in each line.
Rise=(2.1-1.4)=.7
Run= (2.1-1)=1.1
Slope=.7/1.1=.64
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&&&&
Rise=2.5-1.5=1
Run=3.8-1=2.8
Slope=1/2.8=.36
&&&&&&&&
Give the equation of your line, based on the slope and the y intercept.
1=.64x+b
&&&&& y=(.36)x+1.3&&&&
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Give the equation of your line, based on your two points (as before, substitute the slope into the y = m x + b form, substitute the x and y coordinates of one of the two points, and solve for b). Give the equation in the first line. A subsequent explanation is optional, provided you feel you understand the process.
y=.64(1)+1.4
y=2.04
&&&& y=(.36)x+1.3
1.5=(.36)(1)+b
subtract .36 from each side
b=1.14&&&&&
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The figure below shows the best-fit line for the second of the three graphs shown so far.
How did your estimates compare with this result? Summarize briefly below:
My estimate of the y intercept was exact with coordinates of (0,1).
&&&& My estimates were very close. I had guessed a y intercept of (0,1.3). I obtained through calculations a y intercept of (0,1.14). The graph showed a y intercept at about (0,1). Therefore, I was very close.&&&&&
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The figure below shows the best-fit line for the third of the graphs shown so previously.
How did your estimates compare with this result? Summarize briefly below:
My estimates were a little off but not by much.
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Stretch your string over your estimated best-fit line for the graph below. You don't need to write down any data.
How close do you think you can come to the actual best-fit line?
Are you reasonably confident that you could come within 1 unit of the y intercept?
Are you reasonably confident that your line would lie no more than 1 vertical unit away from the actual best-fit line?
Are you reasonably confident that you could come within .1 unit of the y intercept?
Are you reasonably confident that your line would lie no more than .1 vertical unit away from the actual best-fit line?
In the space below, indicate how close you think you could come to the correct vertical intercept, and how close you could stay (in the vertical direction) to the actual best-fit line.
I think that I could come within 1 unit of the y intercept and my line would lie no more than 1 vertical unit away from the actual best fit line. I am not certain if I could come within .1 unit of the y intercept or that my line would lie no more than .1 vertical unit away from the actual best fit line because its such a small unit.
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The two figures below show the actual best-fit line. The second figure also includes the equation of that line.
Look at these figures and in the box below either give revised estimates to your previous answers, or state why you think that no revision is necessary:
I dont think a revision is necessary because I still think there is room for human error once you get to the .1 units.&&&& My answers were so close, once rounded, they will be exact answers if those given&&&&
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Repeat the above exercise for the graph below.
How closely do you think you can come to the actual best-fit line?
Are you reasonably confident that you could come within .2 unit of the y intercept?
Are you reasonably confident that your line would lie no more than .02 vertical unit away from the actual best-fit line?
Are you reasonably confident that you could come within .02 unit (which on this graph would be one-tenth of a verticial division) of the y intercept?
Are you reasonably confident that your line would lie no more than .02 vertical unit away from the actual best-fit line?
In the space below, indicate how close you think you could come to the correct vertical intercept, and how close you could stay (in the vertical direction) to the actual best-fit line.
I am confident that I could come within .2 unit of the y Intercept. I am not confident that my line would lie no more than .02 vertical units away from the actual best fit line, that I could come within .02units of the y intercept, or that my line would lie no more than .02 vertical units away from the actual best fit line.
&&&& This is because the points are more spread out on this graph. I could most likely come within .5 units of the correct vertical intercept and could stay within 1 unit of the actual best fit line.&&&&&&.
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The points of this graph are clearly more scattered from any straight line than in any of the preceding graphs. It is clearly harder to tell exactly where the best fit line should be. That is, there is more uncertainty than before when attempting eyeball a best fit to this data.
Using your string, estimate how much uncertainty there might be in the y intercept of your straight line:
Among all straight lines that seem to fit the data about equally well (i.e., the lines which on the average appear to be about as close to the points as any other), what is the maximum y intercept, and what is the minimum?
Give these two numbers in the first line, separated by a comma.
(.1,1.1)
(1.8,.9)
&&&&.8,.6&&&&&
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The lines x = 6, at the right of the graph region, and y = 1.8, at the top of the graph region, for the top and the right-hand sides of what we will call the graph rectangle.
It is likely that some of your estimated best-fit straight lines will pass through the top of this rectangle, and some through the right-hand side.
What is the x coordinate of the point furthest to the left at which a reasonable best-fit line could pass through the top of the graph rectangle (i.e., the smallest x coordinate at which a reasonable best-fit line could pass through the line y = 1.8)? Give this number in the first line of the space below. If you don't think any reasonable attempt at a best-fit line will pass through the top of the graph rectangle, enter 'none'.
What is the y coordinate of the lowest point at which reasonable best-fit line could pass through the right-hand side of the graph rectangle (i.e., the smallest y coordinate at which a reasonable best-fit line could pass through the line x = 6)? Give this number in the second line of the space below. If you don't think any reasonable attempt at a best-fit line will pass through the right-hand side of the graph rectangle, enter 'none'.
If your first answer was 'none', then what is the y coordinate of the highest point at which reasonable best-fit line could pass through the right-hand side of the graph rectangle? If your answer before was 'none' then enter your numerical answer in the third line; otherwise enter 'none' in the third line.
If your second answer was 'none', then what is the x coordinate of the point furthest to the right at which reasonable best-fit line could pass through the top of the graph rectangle? If your answer before was 'none' then enter your numerical answer in the fourth line; otherwise enter 'none' in the fourth line.
Two of your answers should be numbers, and two should be 'none'.
.9
.9
none
none
&&&&&3.5
1.6
none
none
&&&&&&&&
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The actual best-fit line is depicted in the figure below. My best guess had the line a little steeper and a little lower on the y axis. How did yours compare?
Mine was a little closer, still needs to be loser on the y axis though.
&&&& Mine was fairly close.&&&&&
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The figure below includes the best-fit line and its equation.
See where you think the best-fit line for each of the two graphs below intercepts the y axis, and the y coordinate of the point at which here it passes through the right-hand side of the graph rectangle. Jot down your answers. Don't look at the pictures below these. Just see what you think and see how you do. There are no 'wrong' answers and when I myself do this my own estimates aren't that great, so there's certainly margin for error, but try to be as accurate as you can.
Give your results in the space below, with the y-intercept and the y coordinate of the right-hand point in the first line separated by commas, and the same information for the second graph in the second line of the space below:
(.1,1)
(3,5.8)
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&&&&& 0,0
.5,0&&&&&&&
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I can't follow your work related to the first graph.
I haven't yet looked at your work on the subsequent graphs. I need to first see the process you are using for the first one.
&#Please see my notes and submit a copy of this document with revisions, comments and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).
Be sure to include the entire document, including my notes
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&#Good work on this lab exercise. Let me know if you have questions. &#