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|Ʃx =||Ʃy =||Ʃxy =||Ʃx² =||Ʃy² =|
|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|
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
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
Null and alternative hypothesis:
Ho: β₁ = 0
Ha: β₁ > 0
t = b/(se/√SSxx) = 6.6025
df = n-2 = 8
p-value = T.DIST.2T(ABS(6.6025), 8) = 0.0001
p-value < α, Reject the null hypothesis.
I only need help with E, F and J please! Statistics Assignment 15 1. A local...
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