On the first statistics exam, the coefficient of determination between the hours studied and the grade earned was 80%. the standard error of estimate was 12. there were 26 students in the class. develop an anova table for the regression analysis of hours studied as a predictor of the grade earned on the first statistics exam. (round your decimal answers to 2 places.)

Check the explanation below and the attached files.

Step-by-step explanation:

Hello!

With the given information you need to develop an ANOVA table for the regression analysis, with variables X: hours studied as predictor and Y: grade earned on the first statistic exam as the dependent variable.

The ANOVA table in the first attached file shows in a general way each element of it and how they are calculated.

The information you have is:

Sample n= 26

Se= 12 ⇒ Se²= 144

R²= 0.80

Number of independent variables k= 1

First: The easiest step is to calculate the degrees of freedom

Degrees of freedom of the treatment (regression) DfTr= k= 1

Degrees of freedom of the residual DfE= n - 2 = 26 - 2 = 24

Degrees of freedom of the totals DfT= n - 1 = 26 - 1 = 25

To calculate the Square Sums you have to use the values of Se and R²

The sample variance of the residues (Se²) (also the Mean Square of Residues) and you calculate it as the Sum Square of residues divided the degrees of freedom of the residues.

Se²= MSE = SSE/DfE

SSE= MSE * DfE = 144*24= 3456

The coefficient of determination is calculated as the division between the Square Sum of treatment and the Square Sum of the Totals.

R²= SStr/SST ⇒ SStr= R² * SST (1)

And the Sum of Squares of the Total is the sum of the Sum of Squares of the tratment and the Sum of Squares of the resudues.

(2) SST = SStr + SSE

Using a doble equation system, replacin (1) in (2):

SST = SStr + SSE ⇒ SST = (R² * SST) + SSE

SST - 0.8*SST= SSE

0.2*SST = SSE

SST = SSE/0.2 = 3456/0.2

SST = 17280

Then:

SStr= R² * SST = 0.8 * 17280

SStr= 13824

Now the only elements of the table that are left are the MSTr and the velue of the Statistic.

MSTr= SStr/DfTr= 13824/1 = 13824

F= MStr/MSE = 13824/144 = 96

Check the 2nd attachment, I've summarized all calculation into the table.

I hope it helps!

solution:

→c= a b , that is product of a and b is possible only when when a is m × n matrix and b is n × p matrix.

that is, number of columns of a= number of rows of b.

so, size of matrix , c= a b is , if a = (m × n)matrix

b= (n × p)matrix

then a b = [ (m × n)matrix] ×[   (n × p)matrix ]

→→c= a b= [m × p ] matrix

→→size of product a b will be equal to = number of rows of a × number of columns of b

answer: answer is c!

ok i got to organize the

3/15 and 12/55

3*

aha

3*4 equals 12 but not

a: no

b:

8/24 12/35

no

gautie

it's c

5/18 = 25/90

5*5=25

18*5 = 90

hope that you!

one more happy customer!

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step-by-step explanation: