Question

26.6 273 21.2 23,6 In a certain type of metal test specimen, the normal stress on a specimen is known to be functionally related to the shear resistance. The following is a set of coded experimental data on the two variables Normal Stress, Shear Resistance, u 26.8 25.1 28.9 23.6 27.1 27.7 23.9 25.0 21.7 26.3 28.1 22.5 26.9 21.7 27.4 21.1 22.6 25,8 24.9 (a) Estimate the regression line Myjx = Bo + Bx. (b) Estimate the shear resistance for a normal stress of 24.5. (c) evaluate s2. (d) construct a 99% confidence interval for Bo. (e) construct a 99% confidence interval for Bi. (f) a 95% confidence interval for the mean shear resistance when x = 24.5. (e) a 95% prediction interval for a single predicted value of the shear resistance when x = 24.5.

Answer 1

X	Y	XY	X²	Y²
26.8	26.5	710.2	718.24	702.25
25.4	27.3	693.42	645.16	745.29
28.9	24.2	699.38	835.21	585.64
23.6	27.1	639.56	556.96	734.41
27.7	23.6	653.72	767.29	556.96
23.9	25.9	619.01	571.21	670.81
24.7	26.3	649.61	610.09	691.69
28.1	22.5	632.25	789.61	506.25
26.9	21.7	583.73	723.61	470.89
27.4	21.4	586.36	750.76	457.96
22.6	25.8	583.08	510.76	665.64
25.6	24.9	637.44	655.36	620.01

Ʃx =	311.6
Ʃy =	297.2
Ʃxy =	7687.76
Ʃx² =	8134.26
Ʃy² =	7407.8
Sample size, n =	12
x̅ = Ʃx/n = 311.6/12 =	25.966667
y̅ = Ʃy/n = 297.2/12 =	24.766667
SSxx = Ʃx² - (Ʃx)²/n = 8134.26 - (311.6)²/12 =	43.046667
SSyy = Ʃy² - (Ʃy)²/n = 7407.8 - (297.2)²/12 =	47.146667
SSxy = Ʃxy - (Ʃx)(Ʃy)/n = 7687.76 - (311.6)(297.2)/12 =	-29.533333

a)

Slope, b = SSxy/SSxx = -29.53333/43.04667 = -0.6860771

y-intercept, a = y̅ -b* x̅ = 24.76667 - (-0.68608)*25.96667 = 42.581803

Regression equation :

ŷ = 42.5818 + (-0.6861) x

b)

Predicted value of y at x = 24.5

ŷ = 42.5818 + (-0.6861) * 24.5 = 25.7729

c)

Sum of Square error, SSE = SSyy -SSxy²/SSxx = 47.14667 - (-29.53333)²/43.04667 = 26.884522

Standard error, se = √(SSE/(n-2)) = √(26.88452/(12-2)) = 1.63965

Estimate of variance, s² = SSE/(n-2) = 26.88452/(12-2) = 2.6885

d)

Critical value, t_c = T.INV.2T(0.01, 10) = 3.1693

99% Confidence interval for Intercept:

Lower limit = βₒ - tc*se*√((1/n) + (x̅²/SSxx))

= 42.5818 - 3.1693*1.6397*√((1/12) + (25.9667²/43.0467)) = 21.961

Upper limit = βₒ + tc*se*√((1/n) + (x̅²/SSxx))

= 42.5818 + 3.1693*1.6397*√((1/12) + (25.9667²/43.0467)) = 63.203

e)

99% Confidence interval for slope:

Lower limit = β₁ - tc*se/√SSxx = -0.6861 - 3.1693*1.6397/√43.0467 = -1.478

Upper limit = β₁ + tc*se/√SSxx = -0.6861 + 3.1693*1.6397/√43.0467 = 0.106

f)

Critical value, t_c = T.INV.2T(0.05, 10) = 2.2281

95% Confidence interval :

Lower limit = ŷ - tc*se*√((1/n) + ((x-x̅)²/(SSxx)))

= 25.7729 - 2.2281*1.6397*√((1/12) + ((24.5 - 25.9667)²/(43.0467))) = 24.439

Upper limit = ŷ + tc*se*√((1/n) + ((x-x̅)²/(SSxx)))

= 25.7729 + 2.2281*1.6397*√((1/12) + ((24.5 - 25.9667)²/(43.0467))) = 27.107

g)

95% Prediction interval :

Lower limit = ŷ - tc*se*√(1 + (1/n) + ((x-x̅)²/(SSxx)))

= 25.7729 - 2.2281*1.6397*√(1 + (1/12) + ((24.5 - 25.9667)²/(43.0467))) = 21.884

Upper limit = ŷ + tc*se*√(1 + (1/n) + ((x-x̅)²/(SSxx)))

= 25.7729 + 2.2281*1.6397*√(1 + (1/12) + ((24.5 - 25.9667)²/(43.0467))) = 29.662