Question

9. An experiment was performed on a certain metal to determine if the strength is a function of heating time (hours). Results based on 20 metal sheets are given below. Use the simple linear regression model.

∑X = 40

∑X2 = 200

∑Y = 80

∑Y2 = 1120

∑XY = 388

Find the estimated y intercept and slope. Write the equation of the least squares regression line and explain the coefficients. Estimate Y when X is equal to 5 hours. Also determine the standard error, the Mean Square Error, the coefficient of determination and the coefficient of correlation. Check the relation between correlation coefficient and Coefficient of Determination. Test the significance of the slope.

Answer 1

Sample size, n =	20
Ʃ x =	40
Ʃ y =	80
Ʃ xy =	388
Ʃ x² =	200
Ʃ y² =	1120
x̅ = Ʃx/n =	2
y̅ = Ʃy/n =	4
SSxx = Ʃx² - (Ʃx)²/n =	120
SSyy = Ʃy² - (Ʃy)²/n =	800
SSxy = Ʃxy - (Ʃx)(Ʃy)/n =	228

Slope, b = SSxy/SSxx =    1.9

y-intercept, a = y̅ -b* x̅ =    0.2

Regression equation :

ŷ = 0.2 + 1.9 x  

If x is increased by 1 unit then there is an increase of 1.9 units in y.

Predicted value of y at X = 5
ŷ = 0.2 + 1.9 * 5 = 9.7  

Sum of Square error, SSE = SSyy -SSxy²/SSxx = 366.8
Standard error, se = √(SSE/(n-2)) = 4.51418

Mean square error = se² = 20.37778

Correlation coefficient, r = SSxy/√(SSxx*SSyy) = 0.7359

Coefficient of determination, r² = (SSxy)²/(SSxx*SSyy) = 0.5415

Slope significance test:
Null and alternative hypothesis:                  
Ho: β₁ = 0 ; H1: β₁ ≠ 0          

Test statistic:

t = b /(se/√SSxx) = 1.9/(4.51418/√120) = 4.611

df = n-2 =18     

p-value = T.DIST.2T(4.611, 18) = 0.0002              

Decision :   

p-value < 0.05,  Reject the null hypothesis.