Read through the list of goals at the end of the chapter frequently.
In this chapter you will work with bivariate quantitative data and relationships between two categorical variables. For the quantitative part, you will learn to "straighten" x-y data, that is to use a transformation function to create a new relationship between f(x) and g(y) that is approximately linear. Find the least-squares relationship between the transformed data, then find the inverse of the original transformation function to transform the model into a curve which passes through your original x-y data. It's pretty cool to accomplish this and magnificently powerful math.
The second part, the categorical part, covers conditional and marginal probabilities. For instance, break the class into m/f and soph/jun/sen identifiers. Each person falls into exactly one of the gender groups and exactly one of the class year groups. Overall, what is the likelihood that a randomly-selected person is in a particular class? What is the probability that they are a particular gender? If they are a girl, then what is the probability that they are a senior? If they are a senior, what is the probability that they are a guy? Also, if guys do better than girls in 1st period and guys do better than girls in 5th period, how could the combination of the two classes indicate that girls are doing better than boys?
Dress appropriately for the weather and for doing activities that involve sitting on the floor this week. See you 10/8 at CiCi's???
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Yes! Read Chapter 2 of Freakonomics, but skip the bad words. You can check for daily updates using the homework link on the main blog page.
See you tomorrow!
Welcome. There's always next time.
Go to the classhomework site to get specifics about the assignment for tonight. (4.1)
Use your best judgement. What tool could you use to predict when the number of members would reach 100,000?
Did you look at the positing that said PASSWORD?
lassitermath
Post your questions!
Sorry, thought we had fixed this.
Re: the Facebook problem
Why are you taking e^L2? In our experience we exponentiated both sides of the LSRL formula, not the data in a list. Is this how you are trying to transform the data? Well, e is approx 2.7, so to take e^L2 you are asking for numbers like 2.7^100,000. Thats a huge number and won't likely help straighten the data.
To save data, go to PROG and create a NEW Program. Name is something you can remember. Hit enter to get a colon prompt. Use "2nd RCL L1 ENTER STO L1 ENTER" to save the existing data and have it erase and refill L1 when you run the program. Use a different list name for each set of data.
To get 0 when you use e^something, you must be raising e to a very negative number. This is the opposite of the overflow error problem, but is solved the same way. Instead of breaking the original LSRL into two parts a and b^x or x^b, just leave it as e^(a+bx) to evaluate. The calculator will evaluate the exponent, which will be reasonable, and exponentiate.
Now, if you are using silly inputs you will get bad results anyway, so make sure that you are making logical moves.
Good luck, kids!
Carmel HS and Broken Arrow, two perennially top-notch bands, brought their A-games. It must have been a marvelous night of marching bands for all who attended. Rumor had it that THIS was the must-see BOA event for the year--more than Grand Nationals!
I can't wait to see the video of the LHS performances. Our band makes me so proud. :)
y=c10^kx
That's what you get if you had an exponential to start with, then you took the common log of the y values to straighten the data.
After you ran the LSRL through the straightened data you could express that equation as
log y-hat = a + bx.
To re-transform the "linear" equation to a curve so it will pass through your original data, you have to exponentiate both sides of the equation:
10^(log y-hat) = 10^(a + bx)
which is the same as
y-hat = 10^a * 10^bx.
Now define 10^a = c and b = k
and you have the equation you asked about.
The funny looking stuff is just numbers.
Exponential: log or ln of the y variables
Power: log or ln of BOTH x and y
...usually.
Then you will use one function of x (x or log of x) and one function of y (y or log of y) to generate your regression equation.
Re: problem 4.1
The reason you may not have found the same answer the book gets is that you probably rounded off.
Let y2 =
10^(-1094.507083676+.55577057368x).
Then find Y2(1982).
If you use 10^(-1094.51+.5558x), you will get something closer to 12,178,673.85.
Of course, if you try to use
10^(-1094.51) etc. you will get an overflow error.
The decimal places matter when you are talking about millions of acres.
Good luck, kids. I have to go to class.
Mrs. L
OK, class hasn't started yet.
To get the Predicted: If you have stored your regression formula in Y1, use Y1(L1), because you are finding f(x), where the function is whwat you stored in Y1 and the x is what you used in L1.
To get the residuals, subtract those values from the actuals (in your case, the values you have stored in L3).
Break time.
Yes, you can always do that, but it doesn't give you any insight into the way the graph grows--for instance, if you leave it in "tentothe" form, you don't know what the coefficient and common ratio are in an exponential model. That's a huge sacrifice!
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