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# I need help in statistic. Can’t figure it out.?

4. An experiment was conducted involving 12 subjects to study the effect of the percentage of a certain drug in the bloodstream on the length of time (in seconds) it takes to react to a stimulus. The data are as follows.

a. Construct a scatterplot of the data (label axes appropriately).b. Find the value of the linear correlation coefficient, r.* (Extra Credit) Perform a hypothesis test using α = 0.05 to determine whether there is sufficient evidence to support a claim of linear correlation between the two variables. Use the test indicated (t-test or PPMC table).c. If there is a significant linear correlation, find the linear regression equation, letting the first variable be the predictor x variable.d. Find the predicted reaction time for a subject with 5.2% of the drug in his bloodstream.

### 1 Answer

- AlanLv 78 months agoFavorite Answer
a. (hopefully, you can do a scatter plot)

my plot does look to linears

To do a linear regression,

you need

Σ(x) = 1 + 1.5 + 2 + 2.5 + 3 + 3.5 + 4 + 4.5 + 5 + 5.5

+ 6 + 6.5

Σ(x) = 45

x_bar = 45 /12 = 3.75 (will also be needed letter)

Σ (y) = 1 + 0.8 + 1.8 + 1.4 + 2.1 + 1.8 + 2.2

+ 3 + 2.8 + 3 + 4.1 + 4.9

Σ (y) = 28.9

Σ (xy)= 1*1 + 1.5*0.8 + 2*1.8 + 2.5*1.4 + 3*2.1 + 3.5*1.8

+ 4*2.2 + 4.5*3 + 5*2.8 + 5.5*3 + 6*4.1 + 6.5*4.9

Σ (xy)= 131.15

Σ (x^2) = 1^2 + 1.5^2 + 2^2 + 2.5^2 + 3^2 + 3.5^2 + 4^2

+ 4.5^2 + 5^2 + 5.5^2 + 6^2 + 6.5^2

Σ (x^2) = 204.5

Σ (y^2) = 1^2 + 0.8^2 + 1.8^2 + 1.4^2 + 2.1^2 + 1.8^2 + 2.2^2

+ 3^2 + 2.8^2 + 3^2 + 4.1^2 + 4.9^2

Σ (y^2) = 85.99

R = [ n*Σ (xy) - Σ(x)Σ (y) ]/

[sqrt( n*Σ (x^2) - ( Σ(x))^2 ) * sqrt( n*Σ (y^2) - ( Σ(y))^2)

R = (12*131.15 - 45*28.9 )/ ((sqrt(12*204.5 - 45^2)*

(sqrt(12*85.99 - 28.9^2)))

b. (answer )

R = 0.94089666

(much better R than in your other question)

r^2 = 0.885286524

these are a good sign of linear correlation

*Extra credit

This is backward to

do a t-test first to find out whether

slope is good unless you find the slope first

You have to actual do the linear regression

prediction first.

m = (n* Σ (xy) - Σ(x)Σ (y) ) / (n Σ (x^2) - ( Σ(x))^2)

m = ( 12*131.15 - 45*28.9) / ( 12*204.5 - 45^2)

m = 0.637062937

b = (Σ (y)- m Σ (x))/n = (28.9-0.637062937*45) / 7

b = 0.019347319

so our NULL Hypothesis

slope = 0

measured slope = 0.637062937

now we need

SE_slope = sqrt [ Σ(yi - ŷi)^2 / (n - 2) ] / sqrt [ Σ(xi - x_bar)^2 ]

x_bar =3.75

ŷi = m*xi + b = 0.637062937*xi + 0.019347319

plugging the values

ŷi= 0.656410256

0.974941725

1.293473193

1.612004662

1.930536131

2.249067599

2.567599068

2.886130536

3.204662005

3.523193473

3.841724942

4.16025641

SE_slope = sqrt( (1-0.656410256)^2 + (0.8-0.974941725)^2

( (1.8-1.293473193)^2+ (1.4-1.612004662)^2

+ (2.1-1.930536131)^2 + (1.8 - 2.249067599)^2 + (2.2- 2.567599068)^2

(3-2.886130536)^2 + (2.8-3.204662005)^2+ (3-3.523193473)^2

(4.1-3.841724942)^2+ (4.9-4.16025641)^2) / 11 ) /

sqrt( (1-3.75)^2 + (1.5-3.75)^2 + (2-3.75)^2 + (2.5-3.75)^2

(3-3.75)^2 + (3.5-3.75)^2 + (4.0-3.75)^2 +

(4.5-3.75)^2 + (5.0-3.75)^2 + (5.5-3.75)^2 + (6.0-3.75)^2

(6.5-3.75)^2 )

SE_Slope = sqrt ( 26.30994172 /11) / sqrt( 35.75 )

SE_Slope = 0.258657799

t = measured_slope- 0 / SE_Slope = 0.637062937/ 0.258657799

t = 2.462956617

t_test = 2.463 DF = 12

gives:

so looking up

DF = 11 , alpha = 0.05 (two-tailed) t_critical = 2.201

so

so t_test> t_critical

2.463> 2.201

so it is significant

so now you have to do the rest of the question .

c.

we already did this , because I don't know

how measure the significance without doing

it first

m = (n* Σ (xy) - Σ(x)Σ (y) ) / (n Σ (x^2) - ( Σ(x))^2)

m = ( 12*131.15 - 45*28.9) / ( 12*204.5 - 45^2)

m = 0.637062937

b = (Σ (y)- m Σ (x))/n = (28.9-0.637062937*45) / 7

b = 0.019347319

but your question

says let x = d in the question

y = x ( a guess just to confuse you)

x = 0.637062937d + 0.019347319

d.

If d = 5.2 %

x = 0.637062937*5.2 + 0.019347319

x = 3.332074592 seconds

x = 3.332 seconds (or however , you want to round it)

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