Question

Question 2 (30 points) The following table presents shear strengths (in kN/mm) and weld diameters (in mm) for a random sample of spot welds. Diameter Strength 4.2 50.7 5.0 76.4 6.0 102.5 4.3 58.3 5.8 81.0 6.3 98.3 70.5 4.8 5.2 94.7 6.1 102.6 4.9 82.6 a) (6 points) Construct a scatterplot of strength versus diameter. Verify that a linear model is appropriate. b) (6 points) Compute the least-squares line for predicting strength using diameter. Write the equation of the line. c) (6 points) Predict the strength for a diameter of 5.5 mm d) (6 points) Interpret the least squares slope coefficient, in context. e) (6 points) Interpret the coefficient of determination, in context.

I do not need help with a. Thank you for all of your help!

Answer 1

Answer:- a) To Construct a scattenpool of strength' ls diameter following is indicatos brometer &e strength Diamelon 412 5.0 6.0 403 5.8 6.3 4.8 5.2 6.149 15 strength 50.7 7614 102.5 58.3 81 98.3 70.5 99.] $026826 from these table draw a Seatter plot Strength Cy) 넴영봉영 은 흥 ㅎ 40 4.5 6.0 6.5 5.0 5.5 diameter 'x? There is almost linear relationship between the two variables. su, there is a positive Sincai Kelationship between the test varfables,

у х Pg No il - х” 17.64 2570.49 212.94 25 5836.96 382 36 10506.25 615 18.49 3398.89 250.69 33.64 6561 469.8 39.69 9662.89 619.29 23.04 4970,25 338.4 27.04 8968.09 492.44 3702) 10526.76 625.86 24.01 6822.76 404.74 Ex} 281.76; £4? 69824 Exy= 44112 Now calculate Correlation coefficient ag" is - nExy-Exzy nEx² (Ex)² nEy²-({y} (1044411-2) – (52.6*81706) 170x281.70)-(52.6) (10469824)-(817.6)2 = 0.899 Hypothesis: Ho :p=o Hi P70

Test statistic :- t=an-2 Vi-g2 - 0.899 x 510-1 11-018992 t = 6.158 10-1 Degace of freedom df = n-1 . df =9 P-value df=q t = 6.158 P value = 0.0001 Cfrom P-value caleu lados) Decisions : P-value = 0.0001 < 0.05 P-value <a so we repeet null hypothesis Conclusion:- There is suffecient Evidence. tu say. that there is a lfneal relationship ...between the two variables.

@ D) To Compute the least - Squares lire for pareciding strength using diameter. slope - nExy-Ex Ey nEx2 - (Ex)² = (lo x4411.2) (52.6 x 81706) (10 x281.76)– (5206)2 = 21.759 n intercept = Eye - (slope x Smi) -876 - (21.754x576) = -32.692 The least Square regression line is ŷ = -32.692 +21.759 x c) To poredict the storingth for a diameter - of s.smm. is = -32.692 +(21.759x5.5). ~ 86.98 Km Imm

D to interpsiet the least Squanes slope a Coefficfent, If the diametern increases by Imm, then the shear strength is predicted to be increased by 21.759 kn/mm e) Te interpret the coefficient of determination R²=22=0.8992 - 0.80820) ~ 80.82% 80.82% of the dependent variable can be poredicted by the independent Vorfable