Question

Help please :)

Answer 1

Question 1

The table represents data collected on the time spent studying (in minutes) and the resulting test grade.

Time Spent Studying (min) 52 37 31 9 26 40 22 10 45 34 19 60
Grade on Test 95 84 72 58 77 86 72 43 90 81 62 98
Part 1: Create a scatter plot with the predicted line of best fit drawn on it. Determine the type of correlation (if any) and predict the model that will be used.

Part 2: Find the line of best fit for the data either by hand or using technology. Explain your method. Find the predicted score for each time listed in the table.

Part 3: Find the residuals and decide if your model is a good fit. Explain your method. (If your model is not a good fit, complete Part 2 again with a different set of points or choose a different model.)

Answer 2

I don't understand what part 2 is asking.

Answer 3

im not suree @Luigi0210 help her

Answer 4

We compute the predictions for each of these tools for all possible single amino acid substitutions in the Ensembl human proteome.

Answer 5

The SIFT and PolyPhen predictions are precomputed and stored in the variation databases and predictions are accessible in the variation API by using the sift_prediction, sift_score, polyphen_prediction and polyphen_score methods on a Bio::EnsEMBL::Variation::TranscriptVariationAllele object. For users wanting to access the complete set of predictions from the MySQL database or API, an explanation of the format used is provided here.

Answer 6

The predictions and associated scores are stored as a matrix, with a column for each possible alternate amino acid and a row for each position in the translation. Each prediction for a position and amino acid is stored as a 2-byte value which encodes both the qualitative prediction and score encoded.

Answer 7

Part 2 requires that you find the "regression line." You can do this easily on a TI-83 or -84 calculator, or you can go through the laborious process of calculating the "regression line" by hand. Let me know if you want further help with this.

Answer 8

Yea, but HOW do I do that?

Answer 9

@mathmale

Answer 10

Do you have a TI-83 or -84 calculator?

Answer 11

I don't even know what that is, no.

Answer 12

Makayla: Are you taking this course through classroom instruction, or online, or both?
Do you have a textbook? any kind of calculator? In other words, what resources are you able to access?

Finding regression lines becomes easy once you've done it a few times, but there's a lot of basic math operations involved, making it a bit hard to explain through typing. That's why I'm asking whether you have any resources with you or online that would explain at least some of the concept and practice necessary to find regression lines.

Answer 13

I am taking it online, I don't have a textbook, and the calculator is normal, not high tech. Ive already made the line of fit, Im just not sure how to make the y=mx+b formula or what to put into it.

Answer 14

Here's an example of one reasonably appropriate resource available online:

http://www.statisticshowto.com/how-to-find-a-linear-regression-equation/

I assume you then have online learning materials. What do those materials have to say about "regression lines" or "linear regression?"

Answer 15

That link helped, thank you. Sorry for being so complicated xD

Answer 16

I have developed a regression line using my TI-83 Plus calculator. The line is

y = 0.965x + 45.547. Of course you yourself have to be able to obtain such lines through your own efforts. Now I'd like to demo how you'd use this line.

Answer 17

Looking at your data, I see that the first person spent 52 minutes preparing for the test and obtained a test score of 95.

Please take that 52 and substitute it into the regression equation y = 0.965x + 45.547.

What do you get?

Answer 18

Where do I plug the 52 in, the x?

Answer 19

Yes. 52 becomes your x value.

On your calculator, enter 52. Then multiply that by 0.965. Result?

Answer 20

50.18

Answer 21

Good. Now please add 45.547 to that. (That 45.547 comes from the regression line constant."

Answer 22

Then when I add it to 45.547, I got 95.727

Answer 23

OK. You got 95.727 by using the regression line, right? You plugged in x=52 and got y=95.727 as your PREDICTED value. Now look at your table of values. For x=52, the ACTUAL value is 95. Pretty close, eh? You'll need to do the same thing with all of the other given x-values: Calculate the predicted y value and compare it, in each case, to the actual y value.

Answer 24

What you are doing here is testing the accuracy of the regression line as a predictor of y values.

I've calculated the regression line for you because the arithmetic involved is pretty long and tedious, though not necessarily difficult.

Answer 25

I actually redid it and got 2285.54846

Answer 26

never mind, I was multiplying, sorry the 95 is correct.

Answer 27

I would prefer not to have to explain the whole process of obtaining regression lines through actual calculations from scratch. However, if you were willing to read the web page whose address I've given you, and to find similar reference material in your online materials, then we could make an appointment to meet again to discuss how to do this. But it'd probably take about 30 minutes.

Answer 28

Okay, I greatly appreciate your help :)

Answer 29

Again: If the student studies 52 minutes and obtains an actual score of 95%, that is an OBSERVED value of y.
The corresponding calculation with the regression line involves PREDICTING y using that regression line. Substituting 52 for x yields y=95.725, which is quite close to 95.

And so on.

Send me a message if you do find reference materials, read them and want to go through the process of finding regression lines manually.

Answer 30

Nice working with you! Appreciate your positive attitude and willingness to stick with this work.

Answer 31

So will I always use that equation while doing this problem and just keep plugging in the x value?

Answer 32

y = 0.965x + 45.547.

Answer 33

that's what you're supposed to do in Part B, yes.
For every GIVEN x value, you must calculate the predicted y value, and then you must compare the two different y values.

For example, for x=52, your actual y value was 95, and your predicted y value was 95.727, and so the "residue," the difference between the two y values, is 0.727. (or -0.727).

Lot of work, but these calculations are quite easy.

Again, let me know, later, if you 've found material explaining how to find regression lines manually and want to discuss this further.

Answer 34

I need help with part c !