Question

Please help with AP Stats!!! I will fan and medal!!!

A. What’s the equation of the regression line you’d use to predict stopping distance based on speed?

B. Interpret the slope of the regression line.

C. Find a 95% confidence interval for the slope of the regression line.

D. Test the hypothesis H0: β1 = 0 and Ha: β1 ≠ 0.

E. Suppose, instead of testing H0: β1 = 0, you decided to test H0: β1 = 5 against Ha: β1 ≠ 5. What P-value would be associated with this test?

Answer 1

The following image goes with the question.

Answer 2

help me an I'll help you? Deal

Answer 3

yeah i already learned it

Answer 4

the equation is simple enough, it just takes a lot of time by hand.

using point slope formula: y-Yavg = m(x-Xavg)

finding the average x and y is simple enough .... the slope can be harrowing tho but theres a simple formula to gather the steps:
\[m=\frac{n\sum xy-\sum x\sum y}{n\sum xx-\sum x\sum x}\]

Answer 5

we need to calculate the sums ...

sum of x^2, sum of xy, sum of x, sum of y and use them in the formula

Answer 6

So it that just for A?

Answer 7

if the data could be copied and pasted, excel makes short work of it

Answer 8

Hmm... It could be attached as a file

Answer 9

or if you can type them in here directly, i can copy paste them into an excel sheet

Answer 10

You just want me to copy and paste the picture I had attached?

Answer 11

i cant say that im all that familiar with the CDE parts ... the data for speed and stop distance. i cant copy paste from the pictures, and its not worth the effort for me to transcribe it by hand :)

Answer 12

Data Display

Row Speed StopDist
1 25 63
2 25 56
3 30 84
4 35 107
5 45 153
6 45 164
7 55 204
8 55 220
9 65 285
10 65 303

Descriptive Statistics

Variable N Mean Median TrMean StDev SEMean
Speed 10 44.50 45.00 44.38 15.36 4.86
StopDist 10 163.9 158.5 160.0 88.4 28.0

Variable Min Max Q1 Q3
Speed 25.00 65.00 28.75 57.50
StopDist 56.0 303.0 78.7 236.2

Regression Analysis

The regression equation is
StopDist = __________________________________

Predictor Coef Stdev t-ratio p
Constant .89.99 12.68 −7.10 0.000
Speed 5.7053 0.2707 ___________

s = 12.47 R-sq = 98.2% R-sq(adj) = 98.0%


Does this help at all?

Answer 13

http://stattrek.com/regression/slope-confidence-interval.aspx im reviewing this to see how to do parts C ... part B should be straight from your course material.

Answer 14

This is probably easier to see and copy and paste.

Answer 15

speed^2 speed stop sp*st
625 25 63 1575
625 25 56 1400
900 30 84 2520
1225 35 107 3745
2025 45 153 6885
2025 45 164 7380
3025 55 204 11220
3025 55 220 12100
4225 65 285 18525
4225 65 303 19695
21925 445 1639 85045

10 85045 445 1639 121095 5.705300353
10 21925 445 445 21225 slope

Answer 16

Thank you! What letter does that refer to?

Answer 17

the columns are simply the parts needed, the at the bottom of the columns are the sums of the column....

the rest is getting it into the formula and dividing it out to get a slope of 121095/21225

or simply put as 5.71 for an approximation ... how accurate it gets is up to your material a spose

Answer 18

Xavg is given in the data as 44.50

Yavg is given in the data as 163.9

this creates the formula as: y = 5.71(x-44.5) - 163.9 again youll have to determine the decimal accuracy for the slope based on your materials suggestions

Answer 19

Thanks for the help, so that final answer refers to part A?

Answer 20

yep, and excel has a regression line function that does it all for us if need be, but i always like to do the work to keep in shape :)

Answer 21

i didnt calculate the average xy parts, i assumed the data was correct as given ... might need to dbl chk

Answer 22

yeah, they are good

Answer 23

Thank you!

Answer 24

gotta typo in the equation ...

y = 5.71(x-44.5) + 163.9 , thats better, i had -163

Answer 25

yep

the stat link gives us a standard error formula to work out the confidence interval with ...

SE = sb1 = sqrt [ Σ(yi - ŷi)2 / (n - 2) ] / sqrt [ Σ(xi - x)2 ]

Answer 26

10180.81 380.25
11642.41 380.25
6384.01 210.25
3237.61 90.25
118.81 0.25
0.01 0.25
1608.01 110.25
3147.21 110.25
14665.21 420.25
19348.81 420.25
8791.6125 2122.5

2.035215584 is what im getting for the standard error, we use this as a standard deviation

Answer 27

got that a little off, the y parts done use the average y, they use the regression line

Answer 28

\[\hat y=mx+b\]is what is used to determine\[\sum(y-\hat y)^2\]
not \(\bar y\)

Answer 29

that seems to be a better value of the standard error, the deviation of the actual y values from the predicted y values that the line equation gives us

Answer 30

the stat link says that we want the t score related the the test statistic (our equations slope), based on n-2 degrees of freedom

Answer 31

any of this ringing any bells? im having to teach myself as we go for CDE

Answer 32

the confidence interval is similar to previous CI calculations:\[m\pm t_{n-1}*(se)/\sqrt{(\sum(x-\bar x)^2}\]

Answer 33

t_(n-2) that is

Answer 34

It is making more sense. Is this referring to part C?

Answer 35

hopefully :)

Answer 36

yeah, finding a confidence interval

Answer 37

i see the 0.2707 in the data

Answer 38

i need to revise my interval notation: one website gave us the whole SE, and another im viewing split it into parts :)

\[m\pm t_{n-2}*SE\]

Answer 39

Okay! I don't mean to ask for too much, but before you leave, is there any way you could consolidate the final answers into A) ___, B) ____, C)____ ... etc, because I am having a hard time keeping track. Thank you so much, you have already gone above and beyond.

Answer 40

B will have to come from your course material, its just the definition of what the slope tells us about the information -- something to do with correlation.

A, im just now gathering from the data ... 'm' value is sitting right there near the bottom next to speed. sneaky critters, so we just use that and the averages to determine the equation with

Answer 41

A, y = 5.7053x - 5.7053(44.5) + 163.9, simplify as needed

C, the SE is right next to the place the slope is at .... so we need to determine the t score for a 99% CI with 10-2=8 degrees of freedom

Answer 42

Oh right, I will look for B in the textbook.

Answer 43

http://onlinestatbook.com/2/calculators/inverse_t_dist.html this looks useful, even tho they cant spell confidence .... 8 degrees at 99% gives us a t score of 3.355 so out confidence interval is simply\[5.7053\pm3.355(.2707)\]

Answer 44

But it is asking for 95% isn't it?

Answer 45

these tired old eyes of mine .... yeah, so just replace it with the appropriate t value

10-2 = 8 degrees of freedom at 95% :)

Answer 46

hypothesis testing is simple enough, youll just have to see if there are any special deals when playing with the slope of the line i got no idea what we would use off hand for determing the parameters of the distribution :/ good luck with the rest of it tho.

Answer 47

it looks like the test statistic for the hypothesis of Ho=0, is simply the slope divided by the SE

Answer 48

\[test stat=\frac{5.7053-H_o}{.2707}\]

Answer 49

the test statistic has a student t distribution, not a normal distribution .... so the P-value is the area of the 2tails associated with it.

Answer 50

..and thats all i got for this :) again, good luck

Answer 51

Thank you very much! I appreciate all the help.

Answer 52

youre welcome