Gonna need help with this worksheet...

ok here's the formula for simple regression: Y=B0+B1*X+E and here's the formula for multiple regression: Y=B0+B1*X1+B2*X2+.....Bk*Xk+E

Can someone start me off with #1 part a?

@amistre64 @phi @satellite73

@.Sam. @jhonyy9 @lalaly @LoveYou*69 @Luis_Rivera @Mertsj @tomo

i have no idea how to do a least squares, or any statistics really. sorry

:(

it's kinda like slope y=b+mx ...regression is: y=b0+b1x

AMISTRE!!!!!!!!!!!!!!!!!!!!!! :DDD

well, one thing i know off thebat is that the anchor point for the line will be the average x and average y values

do you recall all the Sxx Sxy stuff? if not i can look it up to refresh

ditto. im just having a hard time setting up the data so it fits the regression equation. for example, which regression equation should i use (simple or multiple)? and what do i put for B0?

what does Sxx and Sxy stand for again?

if i walk myself thru it ... they are something in the back of my head for standard errors i think

we want a line mx+b, such that it is the average of all the stated points, or rather that the distance squared, from the line to any given point is a minimum

i have SSE (sum of squares for erros) in my notes and SE (standard error of estimate)

those look useful

i also have the coefficient of determination (R^2)

http://stattrek.com/regression/linear-regression.aspx thats the site i was looking for

hold on, im gonna upload some notes i took

anchor point is: \((\bar x,\bar y)\) \[b_1=\frac{\sum[(x_i-\bar x)(y_i-\bar y)}{\sum(x_i-\bar x)}\] so b1 is the slope and the given anchor point can be inputed into the point slope format of a line

forgot a square on the denominator

ok, nevermind about posting notes. cant get them on my computer.

Hours studied 2 4 5 8 10 11; avg = 6' 2/3 Exam Score 50 67 75 90 95 99 ; avg = 79' 1/3

2-6-2/3 = -4 2/3 4-6-2/3 = -2 2/3 5-6-2/3 = -1 2/3 8-6-2/3 = 1 1/3 10-6-2/3 = 3 1/3 11-6-2/3 = 4 1/3 50-79-1/3 = -29 1/3 67-79-1/3 = -12 1/3 75-79-1/3 = -4 1/3 90-79-1/3 = 11 2/3 95-79-1/3 = 16 2/3 99-79-1/3 = 20 2/3 ((-4- 2/3)(-29- 1/3)+(-2- 2/3)(-12- 1/3)+(-1- 2/3)(-4- 1/3)+(1+ 1/3)(11+ 2/3)+(3+ 1/3)(16+ 2/3)+(4+1/3)(20+ 2/3))\((-4- 2/3)+(-2- 2/3)+(-1- 2/3)+(1+ 1/3)+(3+ 1/3)+(4+1/3)) the wolf hates me .....

wouldn't the 2 (xi-xbar)'s cancel out?

no, a sum over a sum does not cancel out like summands

\[\frac{a+b}{a+c}\ne\frac{b}{c}\]

umm, hours is "x" parts right?

yup

2 4 5 8 10 11; avg = 6' 2/3 6 12 15 24 30 33 -20 ------------------- -4 -8 -5 4 10 13 10/3 = sum (x-bar x) then

i have a question about the denominator. in the numerator we multiply to get the answer, but there's nothing to multiply by in the denominator, so do i just add up the answers i got for each x given subtracted by x bar?

o nvm i think you just answered my question ha ^^

-4/3 -8/3 -5/3 4/3 10/3 13/3 50 67 75 90 95 99 ; avg = 79' 1/3 150 201 225 270 285 351 -237 -------------------------- -87 -36 -12 33 48 114 -4/3( -87/3)-8/3(-36/3)-5/3( -12/3)+4/3( 33/3)+10/3( 48/3)+13/3(114/3) = 310 3(310)/10 = 3(31) = 93 as a slope?

6 12 15 24 30 33 -20 150 201 225 270 285 351 -237 where do these #s come from??

when adding or subtracting fractions, its best to multiply the whole number by the denominator, in this case that was 3 .... then you can do some normal subtracting 6 2/3 = 20/3

the slope should be about 5.2 .....

alright hold on a second. im gonna write all this out on paper and see how you got everything.

i forgot to square the demoninator

like always ;)

i went and let Excel do the work .... but i organized the data into what was needed

-4 -8 -5 4 10 13 ...so these are all the answers you get when multiplying the numerators together? -87 -36 -12 33 48 114 and ....these are all the answers when adding the denominators together?

4 things needed for the linear regression line average X average Y sum of the products sum of the xs squared \[Line:~y-avgY=\frac{sum~of~products}{sum~of~xs}(x-avgX)\]

by hand i got a value of 10/3 .. without squareing it :) so denominator is 100/9

pfft, still confused .....

2 4 5 8 10 11; avg = 6' 2/3 6 12 15 24 30 33 -20 ------------------- -14 -8 -5 4 10 13 , square these then add (196+64+25+16+100+169) / 9 ... since 3^2 = 9 196 169 164 25 16 ---- 570 570/9 = 63 1/3

im not even going to attempt a product of the xy parts by hand lol

lol hold on im working it out on paper...

ok, i must be doing something wrong... i got (300.8) / (4) when i multiplied everything in the numerator and added in the denominator

\[top:~(x_1-\bar x)(y_1-\bar y)+(x_2-\bar x)(y_2-\bar y)+(x_3-\bar x)(y_3-\bar y)\\~ \\~~~~~~~~~~~~~~+(x_4-\bar x)(y_4-\bar y)+(x_5-\bar x)(y_5-\bar y)+(x_6-\bar x)(y_6-\bar y)\] your top is worked out like this right?

ok, lemme type out what i wrote out...

( (-4&2/3 * -29&1/3) + (-2&2/3 * -12&1/3) + (-1&2/3 * -4&1/3) + (1&1/3 * 11&2/3) + ( 3&1/3 * 16&2/3) + (4&1/3 * 20&2/3) ) // ( (-4&2/3) + (-2&2/3) + (-1&2/3) + (1&1/3) + (3&1/3) + (4&1/3)

( ( -10/3) * -86/3) + (-4/3 * -35/3) + (-1/3 * -11/3) + (4/3 * 35/3) + (10/3 * 50/2) + (13/3 * 62/3) ) is what i did to the numerator first before multiplying and adding it

this worksheet says your allowed to use excel, but just not in configuring the linear regression equation ..... in other words, no letting excee process the final result. right?

im assuming he just wants us to show our work.

lol, there is a difference between showing work, and torturing yourself.

its a handwritten assignment that we turn in class.

then it torture ...

YES, that class IS torture! ha

ok so what am i doing wrong

i used a graphing calc to input everything to get the answers i got

\[top:~(\frac{6-20}{3})(\frac{150-238}{3})+(\frac{12-20}{3})(\frac{201-238}{3})+(\frac{15-20}{3})(\frac{225-238}{3})\\~ \\~~~~~~~~~~~~~~+(\frac{24-20}{3})(\frac{270-238}{3})+(\frac{30-20}{3})(\frac{285-238}{3})+(\frac{33-20}{3})(\frac{297-238}{3})\] \[bottom:~(\frac{6-20}{3})^2+(\frac{12-20}{3})^2+(\frac{15-20}{3})^2\\~ \\~~~~~~~~~~~~~~+(\frac{24-20}{3})^2+(\frac{30-20}{3})^2+(\frac{33-20}{3})^2\]

ok i didnt square each individual answer on the bottom... but for the top you dont get 300.8? want me to put it in my calc?

i get what i got on the excel picture i posted :)

for the top i get 328.67

thats lazy ;)

i know ... but im old

excuses excuses lol

you are close to it, so im sure there is just some input error in your calculator

i think i see it.....

ok i still got the wrong answer so ill just do it the way you did it... i dont know why im getting the wrong answer doing it my way

..cause my way is ALWAYS best ;)

\[top:~(\frac{-14(-88)}{9})+(\frac{-8(-37)}{9})+(\frac{-5(-13)}{9})\\~ \\~~~~~~~~~~~~~~+(\frac{4(32)}{9})+(\frac{10(47)}{9})+(\frac{13(59)}{9})\] \[top:~\frac{-14(-88)-8(-37)-5(-13)+4(32)+10(47)+13(59)}{9}\] \[top:~\frac{1232+296+65+128+470+767}{9}\] \[top:~\frac{2958}{9}=328.66666.....\]

this was fun, but i gotta get going home now. enjoy ;)

yeah, it was "fun" lol talk to you later.

For #1 Part B it asks what will happen if you studied 30 more minutes... so do I just plug in .5 or 1/2 to the line equation? y=5.1895x + 44.737

@phi

almost. all you care about is the slope you want the change in y (how many more points) with a change in x (hours studied)

so you do not care about the 44.737

so just plug in .5 for x and multiply by the slope?

Part B it asks what will happen if you studied 30 more minutes. It is like asking, "if you move 1/2 hour on the x-axis, how far up to you move on the y-axis?"

yes

ok sounds good. thanks. i'll let you know when i need more help.

fyi, I checked your numbers y=5.1895x + 44.737 and they look good

ok so i got 21.07921567 for the standard error of estimate for Part E but I dont know how to explain if the model fits well with the data.

i'm also needing help with Part F and for #2, do I basically do the same thing I did for #1 Part A first? i'm gonna logoff, but please do continue to help if you're online and i'm not. I will log on sooner than later to see what you say. thanks. :)

** i got 21.07921567 for the standard error of estimate for Part E** How did you get this number? According to wikipedia the standard error squared is \[\sigma^2= \sum\frac{\epsilon \epsilon }{n-p} \] n is the number of samples and p is 2 (# of parameters) the residual error is the difference between the data and the least-square estimate 50 67 75 90 95 99 55.1158 65.4947 70.6842 86.2526 96.6316 101.8211 just looking at the plot, you can see it is a good fit.

wait, how did you get the #s below the test scores again?

@amistre64

#2 is the same concepts as #1, just a new set of numbers

pages relate to X, and prices to Y

im still working on #1 haha (parts e and f) i was wondering where Phi got the #s underneath the test scores ^^^

im not sure what phis method was. youll have to consult the phi :)

can you help me finish parts e and f then? i have the error of estimate for e but i need to explain whether the model fits the data well. for part f i have an idea of what to do but im not 100% sure. do i draw a standard normal curve?

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