X | Y | XY | X² | Y² |
6 | 15 | 90 | 36 | 225 |
20 | 31 | 620 | 400 | 961 |
0 | 10 | 0 | 0 | 100 |
14 | 16 | 224 | 196 | 256 |
25 | 28 | 700 | 625 | 784 |
16 | 20 | 320 | 256 | 400 |
28 | 40 | 1120 | 784 | 1600 |
18 | 25 | 450 | 324 | 625 |
10 | 12 | 120 | 100 | 144 |
8 | 15 | 120 | 64 | 225 |
Ʃx = | Ʃy = | Ʃxy = | Ʃx² = | Ʃy² = |
145 | 212 | 3764 | 2785 | 5320 |
Sample size, n = | 10 |
x̅ = Ʃx/n = 145/10 = | 14.5 |
y̅ = Ʃy/n = 212/10 = | 21.2 |
SSxx = Ʃx² - (Ʃx)²/n = 2785 - (145)²/10 = | 682.5 |
SSyy = Ʃy² - (Ʃy)²/n = 5320 - (212)²/10 = | 825.6 |
SSxy = Ʃxy - (Ʃx)(Ʃy)/n = 3764 - (145)(212)/10 = | 690 |
e)
Sum of Square error, SSE = SSyy -SSxy²/SSxx = 825.6 - (690)²/682.5 = 128.0175824
Standard error, se = √(SSE/(n-2)) = √(128.01758/(10-2)) = 4.00027
f)
Slope, b = SSxy/SSxx = 690/682.5 = 1.010989
y-intercept, a = y̅ -b* x̅ = 21.2 - (1.01099)*14.5 = 6.540659
Regression equation :
ŷ = 6.5407 + (1.011) x
Predicted value of y at x = 12
ŷ = 6.5407 + (1.011) * 12 = 18.6725
Critical value, t_c = T.INV.2T(0.1, 8) = 1.8595
90% Prediction interval :
Lower limit = ŷ - tc*se*√(1 + (1/n) + ((x-x̅)²/(SSxx)))
= 18.6725 - 1.8595*4.0003*√(1 + (1/10) + ((12 - 14.5)²/(682.5))) = 10.8383
Upper limit = ŷ + tc*se*√(1 + (1/n) + ((x-x̅)²/(SSxx)))
= 18.6725 + 1.8595*4.0003*√(1 + (1/10) + ((12 - 14.5)²/(682.5))) = 26.5067
j)
Null and alternative hypothesis:
Ho: β₁ = 0
Ha: β₁ > 0
Test statistic:
t = b/(se/√SSxx) = 6.6025
df = n-2 = 8
p-value = T.DIST.2T(ABS(6.6025), 8) = 0.0001
Conclusion:
p-value < α, Reject the null hypothesis.
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