Using the results of the problem below A Determine the coeff

Using the results of the problem below: A. Determine the coefficient of the determination, r², and interpret it\'s meaning. B. Determine the standard error of the estimate. C. How useful is this model for predicting the tear rating based on the plate gap in the bag sealing equiptment. (Previous problem dealt with a machine that bags coffee beans, and the tear rate vs plate gap on the machine.)

Tear	Viscosity	Pressure	Plate Gap
0.00	350.00	180.00	0.00
0.00	350.00	170.00	0.00
0.45	319.00	186.00	1.80
0.85	380.00	174.00	1.80
0.35	350.00	180.00	0.00
0.30	300.00	180.00	0.00
0.70	400.00	180.00	0.00
1.90	350.00	190.00	0.00
0.25	350.00	180.00	0.00
0.10	319.00	186.00	-1.80
0.15	380.00	186.00	-1.80
3.90	350.00	180.00	3.00
0.00	380.00	174.00	-1.80
0.55	350.00	180.00	0.00
0.00	350.00	180.00	-3.00
0.05	319.00	174.00	-1.80
0.40	319.00	174.00	1.80
4.30	380.00	186.00	1.80
0.00	350.00	180.00	0.00


b. Least squares regression equation: =b tb
b= nxy - ( x) ( y) over n x²- ( x)²
Equals 19(21.96)-(0)(14.25) over 19(43.92)-(0)²
Equals 417.24-0 over 834.48-0 = 417.24 over 834.48 = 0.5

Slope term, b = 0.5
b = - b , xbar=y over n - b x over n
Equals 14.25 over 19 - (0.5)0 over 19 = 0.75

Intercept term, b = 0.75
Least squares regression equation: = 0.75 + (0.5)x

Slope of (0.5) implies that for every one unit increase in plate gap(x), we expect an increase of 0.5 in tear(y)
Intercept of (0.75) is the expected value of tear(y) when plate gap(x) is zero.

At x=0 = b + b x
= (0.75) + (0.500)(6) = 0.75
x-0 = 0.750

$Using the results of the problem below: A. Determine the coefficient of the determination, r2, and interpret it\'s meaning. B. Determine the standard error of t$

Using the results of the problem below A Determine the coeff

Solution

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