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Step 1Step 2(a)calculation procedure for regressionsum of (x) = ∑x = 84sum of (y) = ∑y = 3610sum of (x^2)= ∑x^2 = 968sum of (y^2)= ∑y^2 = 1808396sum of (x*y)= ∑x*y = 41238mean(x)= ∑x/n = 10.5mean(y)= ∑y/n = 451.25intercept (bo) = (∑y)(∑x²) - (∑x)(∑xy) / n(∑x²) - (∑x)²intercept (bo) = (3610)(968) - (84)(41238) / 8(968)-(84)^2intercept (bo) = (3494480-3463992) / (7744-7056)intercept (bo) = 44.314slope (b1) = n(∑xy) - (∑x)(∑y) /n(∑x²) - (∑x)²slope (b1) = (8(41238) -(84)(3610))/(8(968) - (84)^2 slope (b1) = (329904-303240)/(7744-7056) slope (b1) = 38.7558regression equation is, y = bo + b1 Xy =44.314+38.7558X    ----------------------------------------------------------------------------------------Step 3(b) the point estimate of the predicted value for x = 10 using the regression equation is,y* = 44.314+38.7558 * 10 = 431.8721    ----------------------------------------------------------------------------------------Step 4(c)calculation procedure for hypothesisstandard error of β1 = sqrt( ( ∑ Y -Yi )^2/ n-2  * (Xi - mean)^2 ))∑ (Y -Yi )^2 = 50210.371 ; n-2 = 8 -  2 = 6 ; ∑ (Xi - mean)^2 =86standard error of β1= sqrt( 50210.371/ 6 * 86)standard error of β1= sqrt(97.3069)standard error of β1 = 9.8644calculated/given areβo= 44.314β1= 38.7558standard error of β1= 9.8644number (n)=8null, Ho: β1 =0 alternate, H1: β1 ≠0level of significance, alpha = 0.05from standard normal table, two tailed t alpha/2 =2.4469since our test is two-tailedreject Ho, if to  < -2.4469 OR if to > 2.4469we use test statistic (t) = β1/standard error of (β1) to =38.7558/ 9.8644to =3.9288| to | =3.9288critical valuethe value of |t alpha| with n-2 = 6 d.f is 2.4469we got |to| =3.9288  & | t alpha | =2.4469make decisionhence value of  | to | > | t alpha| and  here we reject Hop-value :two tailed  ( double the one tail ) - Ha : ( p ≠ 3.9288 ) = 0.0077hence value of p0.05 > 0.0077,here we reject Ho    we conclude that slope is significant  ...