Question

0000000 0000000 13. An engineer is attempting to model the potential energy of a spring based placement from equilibrium of the spring. The experimental data is provided below. on the dis- equilibrium position displacement, x M x (displacement) y (potential energy) 2.23 2 3 4 3.62 6.77 13.47 21.06 (a) Plot the data by hand. Does there appear to be a linear relationship relationship? or a non-linear (b) Calculate the estimated regression equation by hand and plot the least squares regression line on your graph in (a). (c) Calculate SSR, SSE, SST, and R2 by hand. (d) At a 0.05, is there evidence of a significant linear relationship between x and y? Com- plete a six-step hypothesis test.

Answer 1

(a)

Scatterplot of y vs X 20 15 10 0 4 3 2 1 X LC LO un

From this graph we can see there seems to be a linear relationship between y and x.

(b) Letthe regression line of y on x be  

y = abx

x^2 у^2 y ху 4.9729 2.23 1 2.23 1 2 4 3.62 13.1044 7.24 6.77 20.31 9 45.8329 13.47 16 181.4409 53.88 5 21.06 25 443.5236 105.3 Total 15 47.15 55 688.8747 188.96 XI

x x/n 15/5 3

y/n 47.15/5 9.43

55 x2 -x2 n sdx 32 1.414(approx.) 5 X Ln

Ey 2 = sdy 688.8747 6.989(approx.) 9.432 5 n LO

188.96 ΣΧΥ y)- xy = 5 cov(x, y) 9.502 (3 x 9.43) Ε Ε η

Cov(X, y) 4.751 sd2 byx X

a= y-byxx = -4.823

Hence the regression line is  

y 4.823+ 4.751x

Fitted Line Plot y =4.823 4.751 x S 2.48530 R-Sq R-Sq(adj 92.4% 20 89.9% 15 10 5 1 2 3 5 X LC

(c)

Predicted y(y'(y'-y)^2 2.23 y 0.072 5.299204 3.62 4.679 1.121481 6.77 7.0756 9.43 13.47 14.181 0.505521 21.06 18.932 4.528384 18.53019 Total 47.15 47.15

$SST=n.var(y)=n.sd_y^2=244.2502$

$SSE=\sum (y-\hat{y})^2=18.53019$

$\therefore SSR=SST-SSE=225.72001$

Cov(X, y)0.96132(aaprox.) R = sd,sdy

$\therefore R^2=0.92413(approx.)$

(d) Here we have to test   $H_0:b_{yx}=0$    against   $H_1:b_{yx}\neq 0$    at level of significance   $\alpha =0.05$ .

Our test statistic is   $T=\frac{b_{yx}-b}{var(b_{yx})}$

We know   $T \sim t_{3}$

$var(b_{yx})=[var(y)-b_{yx}^2var(x)]/var(x)=1.853019$