a) Find the value of the line ar correlation coefficient rb) Find the critical values ofr from table A-6 using x= 0.05c) Determine whether there is sufficient evidence to support a claim of a linear correlationBetween the two variables.d) Find the regression equation,e) Letting the first variable be the predictor (x) variable, find the indicated predicted value. Q2. Blood Pressure MeasurementsRight ArmLeft Arm102101947979175169182146144Find the best predicted systolie blood pressure in the left arm given that the systolie bloodpressure in the right arm is 100 mm Hg.

Hello. Since your question has multiple sub-parts, we will solve first three sub-parts for you. If you want remaining sub-parts to be solved, then please resubmit the whole question and specify those sub-parts you want us to solve. Given : Let, x = Right Arm y = Left Arm sample size, n = 5 X Y Xi-X¯2 Yi-Y¯2 Xi-X¯×Yi-Y¯ 102 175 121 139.24 129.8 101 169 100 33.64 58 94 182 9 353.44 56.4 79 146 144 295.84 206.4 79 144 144 368.64 230.4 ∑Xi = 455 ∑Yi = 816 ∑Xi-X¯2 = 518 ∑Yi-Y¯2 = 1190.8 ∑Xi-X¯×Yi-Y¯ = 681 X¯ = ∑XinX¯ = 4555 X¯ = 91 Y¯ = ∑Yin Y¯ = 8165 Y¯ = 163.2 Regression equation : Y^ = a+ b×X b = ∑Xi-X¯×Yi-Y¯ ∑Xi-X¯2 b = 681518 b = 1.3147 a = Y¯ -b×X¯ a = 163.2 -1.3147×91 a = 43.56Y^ = 43.56 + 1.3147×X For X = 100 Y^ = 43.56 + 1.3147×100 Y^ = 175.03 175.03 is the best predicted systolic blood pressure in the left arm given that the systolic blood pressure in the right arm is 100mm HG. a) linear correlation coefficient, r = ∑Xi-X¯×Yi-Y¯ ∑Xi-X¯2×∑Yi-Y¯2 linear correlation coefficient, r = 681 518×1190.8 linear correlation coefficient, r = 0.8671 b) α = 0.05 DF, v = n - 2 = 5 -2 = 3 Critical value = 0.878c) H0 : ρ = 0 Ha : ρ ≠ 0 Test statistic : t = r×n-21-r2 t = 0.8671×5-21-0.86712 t = 3.015 P-value = 0.05699 here, r = 0.8671 < critical value = 0.878 Hence, r is not significant. There is not sufficient evidence to support a claim of a linear correlation between the two variables.