Question

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.

Answer 1

Part 3: Find the residuals, and decide if your model is a good fit. Explain your method.

Answer 2

well, what does your scatter plot look like? can you work with excel?

Answer 3

I already did a scatter plot I just need help with part 2 and part 3

Answer 4

a few things you will need to determine the line of best fit

sum of the y parts
sum of the x parts

sum of the xy parts
sum of the xx parts

and the number of points used

you will also need the average X and Y values.

Answer 5

I dont understand, do you mind explaining? Like I know the line of best fit is when you draw a line through the points.

Answer 6

One second, I will try to attach my scatter plot here.

Answer 7

the line of 'best' fit is the one that has the least amount of total error ... as such, we could either develop the formulas needed for a slope and a point; or, we could just memorize the formulas already stated in the textbooks.

Answer 8

either way, we need to know those 7 values to play with since:\[y=mx+b\]
\[m=\frac{n\sum xy-\sum x\sum y}{n\sum xx-\sum x\sum x}\]
\[b=\bar y-m\bar x\]

Answer 9

i spose they generally name the equation as y hat: \(\hat y\), to distinguish it from the observed y values

Answer 10

Ok here is the scatter plot

Answer 11

something that will make life easier if we have to do this by hand is to simply subtract the smallest value from all the points to reduce the size of the numbers involved. its the same line, just moved. but if this is done by a computer program, then thats irrelevant

Answer 12

(52 37 31 9 26 40 22 10 45 34 19 60) - 40

(12 -3 -9 31 -14 0 -18 -30 5 -6 -21 20)

12-3 -9+31-14+0-18-30+5-6-21+20

-14-18-1 = -33, average is -33/12
......................................................................



95 84 72 58 77 86 72 43 90 81 62 98

(95 84 72 58 77 86 72 43 90 81 62 98)-70

25 14 2 -12 7 16 2 -27 20 11 -8 28

25+14+2-12+7+16+2-27+20+11-8+28

2+76 = 78, average is 78/12
........................................................................

these values will give us a shifted line by (-40,-70) that we will have to shift back

sum of xx, and sum of xy of course would be more time consuming by hand

Answer 13

do you have excel or some equivalent program?

Answer 14

I don't have excel, but I still don't understand what your doing. Like what the lesson tells me on how to find the line of best fit is to use two of the points on the line and the point-slope formula, y−y1=m(x−x1), to write the equation of the line then convert the equation of the line to slope-intercept form so that I can make predictions or answer questions about the data.

Answer 15

So the two points i used was (52, 58) and (60,98) i found the slope and got 3/8 or 0.375 then the equation for the line i got was y=0.375x + 75.5

Answer 16

so you did not create the regression equation, you used a 2 point guesstimation

Answer 17

those two points were from the data

Answer 18

\[\hat y=\frac{n\sum xy - \sum x \sum y}{n\sum xx - \sum x \sum x}(x-avgX)+avgY\]
\[\hat y=\frac{12(32120) - 385(918)}{12(15117) - 385(385)}(x-32.08333)+76.5\]

which simplifes to something like:

m = 0.9647
b = 45.5471

Answer 19

when we guess a line of best fit by using 2 data points, its not generally not going to be a good fit.

Answer 20

Oh okay

Answer 21

the slope between the points that you choose is more like 5 ...

60, 98
-52, -58
--------
8, 40 ... 40/8 = 5

Answer 22

Wont it be 52,95 and 60,98

Answer 23

yes, i was just checking to see if you had used valid points to start with :)

Answer 24

id used 9,58 since its relatively the lowest/leftmost point instead of 52,95

Answer 25

oh ok

Answer 26

60, 98
-9,-58
--------
51, 40 that gives us a more realistic slope closer to the best fit line

Answer 27

40/51 = 0.7843

Answer 28

oh because it is closer to 0

Answer 29

not that is closer to 0, its just a better fit to me, visually that is.

Answer 30

ok

Answer 31

now i have to find the equation of the line

Answer 32

notice your line in red, versus my line in black.

Answer 33

Right, so yours is more accurate.

Answer 34

the library im at 'upgraded' to the newer version of excel .... i like the older one better.

Answer 35

visually, mine if more accurate :)

Answer 36

So what do i do now? I find the equation, correct?

Answer 37

I got y=0.7843x +50.9413

Answer 38

with my points? seems fiar, let me dbl chk

Answer 39

ok

Answer 40

y = 0.7843x +50.9412

but yeah, thats good

Answer 41

Okay, so after i got that, I dont know what to do next

Answer 42

lets call the line; yh instead of just y

y represents the actual data values that have been recorded, yh is just a model that we think will predict the values.

Answer 43

Ok

Answer 44

well, use your yh equation for each time value so determine how well it predicts it. you are going to create a whole new set of data points which we should call yh.

Answer 45

we know when x=9, y=58 since we used that to create the line with to start with, and the last point will be exact as well. the others are going to differ to some degree

Answer 46

so we have to make up an x value?

Answer 47

no, the x values stay the same, we are trying to predict the outcome of a given time value; not predict the time value itself

Answer 48

like this

Answer 49

Oh

Answer 50

the residuals are just the difference between yh and y

Answer 51

so i would write 57.9999?

Answer 52

well, since we used an approximation for the slope fraction, yes ... there is going to be some slight differences when it comes to the points we used.

Answer 53

how accurate your yh is, is strictly up to you.

Answer 54

we are "Finding the predicted score for each time listed in the table". by using our yh equation with the given time values

Answer 55

I get it, but like i dont know what to write on the paper, so far i just have my equation work.

Answer 56

um, i used excel to find the yh values .... and posted a screenshot of it.

Answer 57

create a new table, and define your x and yh values.

Answer 58

the yh is the y with the ^ on top?

Answer 59

yes :)

Answer 60

y hat

Answer 61

so the values in yh is the predicted scores?

Answer 62

heres is what has happened so far, so that you can see the big picture that you might be missing.

a data set we given, this formed a scattered set of points on the graph

we used 2 of those points to construct an equation that we can use to predict those points, and others.

we are now finding how our equation plays with the time values, so yes yh is the predicted scores since we are using our equation to try to determine a given score with it.

Answer 63

realistically we should have our points stated in terms of (time,grade) or simply g(t)

we dont have an equation for g(t), we have a bunch of scattered about time,grade points and so we construct a mathematical model that we hope we can use to some effect.

so we develop the grade prediction model as: g' = 0.7843(t) +50.9412

we use g' to compare with the known time,grade points to determine how good or bad a fit our model is

Answer 64

Im sorry its still confusing i get part of it but not fully and i dont know what to write on the paper, this assignment is holding me back from doing the others

Answer 65

i dont know what you need to write either, but my guess is that you need to write up something like the first 3 rows of this.

http://assets.openstudy.com/updates/attachments/53c010f6e4b00f624a91300f-amistre64-1405100458010-untitled.jpg

Answer 66

ok

Answer 67

the 4th row is the residuals, the difference between our observed and predicted values:

yh - y

Answer 68

the last row, is just taking the differences and squaring them .... not sure if thats the process you are expected to determine the goodness of fit with tho