HW due 11/16:
4.2, 4.3, 4.4, 4.5, 4.6, 4.10, 4.12, 4.13, 4.14, 4.41-4.48 (no calculations), 4.50, 4.51, 4.54, 4.57
Please note that there are a wide variety of school activities that will be taking students out of class this week. If you miss class, please make sure that you keep up by reviewing what we have on the blog and completing the homework.
The test will be on November 17th.
Friday, November 11, 2011
Today we learned the steps for transforming exponential data into linear form and then (ta da!!!) used the linear regression formula to model the curve of the original data.
First, we collected and entered the data. If you want to play along, please use these (x,y) pairs:
(0,50), (1,42), (3,29), (5,20), (7,14), (9,10)
Graph them and observe that they look like they follow an exponential decay pattern.
We determine that if the y values have an exponential relationship with the x values, then the ln y values will have a linear relationship with the x values.
Put the ln y values in L3. Graph L3 against L1 and confirm that the data look approximately linear.
Run the LSRL through L1 and L3. Confirm that the LSRL goes through the data.
Now, write down the equation the way the calculator wrote it.
y = 3.90099 - .1802 x
We used the REAL x values to create this equation, so leave the x alone.
We used the LN of y to create this equation, so change y to ln y. And add a hat to make it a predictor.
(ln y)hat = 3.9099 - .1802 x
Unfortunately, we can't enter this equation in this form to graph on our calculators. We have to solve for y.
y hat = e3.9099 (e-.1802)x, which is roughly equivalent to
y hat = 49.89 (.8351)x
This equation makes sense because the theoretical model would be y=50(5/6)x. This is pretty close!
Now, for the big finish. Graph this curvy equation through the original curvy data (L1, L3).
If you've done it according to this scheme, then your curve should go through the data pretty well.
You would follow up with an analysis of the residuals. . .
For power function models, follow the same procedure, but use the natural logs of BOTH the x and y values. You'll see this in practice on Monday.
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An explanation break with a story.
My birthday was Saturday. My brother Rob, whose family is Chinese, wrote his birthday greeting for me on my Facebook wall in Chinese. I wanted to respond in Chinese, but, having no skill in this area, had to use Google translater to compose my comment. I wrote my phrase in English, ran it through the translator, and then checked the answer by running the Chinese version back into English. It took a few tries before my response in Chinese resembled my English intentions. I posted my response, and my brother has not acknowledged my excellent command of another language!
This is similar to what we're doing with these non-linear data sets. We're recognizing that they are a different language (shape) from what we can work with (linear). We translate the data into our familiar form and make sure that the fit is good. We get our phrase (formula) and translate it back into the unfamiliar (non-linear) form.
谢谢你，我的第二个弟弟
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Monday, November 14, 2011
Today we collected data again, this time related to a power function. Consider the following pairs of (circumference, volume) data for balls:
(20.1, 5/8), (31, 2), (22.6, 13/16), (13.6, 1/6), (11.1, 1/10), (21, 2/3)
Circumferences were measured in cm. Volumes were measured in cups.
Note that a ball of 0 circumference would have no volume. Then, take the natural log of both the circumference and the volume to straighten the data. Work with the transformed data: find the LSRL, check the residuals. Transform the LSRL back into a curve of the form y-hat = a * x^b.
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Quick hint for transformations: if the data would hit an axis if the curve were extended, then take the natural log of that variable. Exponentials-ln of y only. Power-ln of both x and y.
Be sure to get a start on the HW problems. There are too many to work in just one night.
Wikipedia's article for Archimedes has a nice explanation of the displacement concept described in class, as well as an animated graphic that shows a more likely method for answering the king's question.
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Wednesday, November 16, 2011
The test is tomorrow. We checked HW today. If you did not have the homework (which was assigned Friday of last week!!!), then you have a lot of work to prepare for tomorrow's test.
Remember to read through section 4.3 about causation, common response, and confounding. Know the difference between these concepts. Realize that a lurking variable can influence both the presumed explanatory variable and the response variable or it can influence just the response variable. Once it is identified, the "lurking" nature is revealed and it becomes an explanatory variable!
You will not be given credit on the test for using expreg or powerreg functions from the calculator. Your curriculum requires you to straighten the data, so the test will be set up for you to demonstrate your skills.
And, speaking of skills, here's copy of the worksheet from today. You need to be able to read the computer output to create a least squares regression line for the straightened data, then convert the line into the curved form that would match the original data.
