Question

Part 2. In the following computational Problem 2.1 The accompanying datan (m/min) appeared in a recent study. shower in die density (Am) andyre of deposition t 0.24 40 1.20 60 1.71 80 222 2x -200. y. -5.37. x = 12000, X-92617. XX. - 331.4 1. (pts) Compute bo(A). simple linear regression model A) and the equation of the estimated regression line, assuming b. (2pts) What is the estimated change in the rate of deposition (am/min) when current density is decreased by 10 mA/em?? 4 of 7

Answer 1

	X	Y	X^2	Y^2	XY
	20	0.24	400	0.0576	4.8
	40	1.2	1600	1.44	48
	60	1.71	3600	2.9241	102.6
	80	2.22	6400	4.9284	177.6
SUM	200	5.37	12000	9.3501	333
n	4
Mean	50	1.3425
SSxx	2000	Sum(x^2) - ((Sum(x))^2 /n)	SSR	2.080125	slope * Ssxy	MSR	2.080125
Ssyy	2.140875	Sum(y^2) - ((Sum(y))^2 /n)	SSE	0.06075	SST-SSR	MSE	0.030375
Ssxy	64.5	Sum(xy) - (Sum(x)*Sum(y)/n)	SST	2.140875	Ssyy	F	68.48148
slope	0.03225	Ssxy/SSxx
intercept	-0.27	Mean Y - Mean X * Slope
Se	0.174284	SQRT(SSE/(n-2))
Sb1	0.003897	Se/SQRT(SSxx)
r	0.98571
r^2	0.971624

a)

Y intercept(b0) = -0.27

Slope(b1) = 0.03225

Y = -0.27 + 0.03225 * X

b)

If X = -10

Y = -0.27 + 0.03225 * (-10)

Y = -0.5925

c)

Hypothesis :

H0 : β1 = 0

Ha : β1 not = 0

Test :

t = b1 / Sb1

t = 0.03225/0.003897 = 8.2756

P value = 0.0143 ( Use t table or calculator), df = 2

P value > 0.01, It is not statistically significant

d)

90% prediction intervel Ô = MSE = V0.0304 a=0.1. df = n -2=4-2=2 te = 2.92 (Use t table) = 0.1743 PI = (Y - E, Y + E) Î = -0.27 +0.0323 x 45 E = ta/2,6- 1 in + - + (X-X) SSxx = 1.1813 = 2.92 x 1 0.03041 1 + + 4' (45-50) 2000 = 0.5684 PI = (1.1813 - E, 1.1813 - E) = (1.1813 -0.5684, 1.1813 +0.5684) = (0.6128, 1.7497)