Question

An article in the IEEE Transactions on Instrumentation and Measurement ("Direct, Fast, and Accurate Measurement of VT and K of MOS Transistor Using VT-Sift Circuit," 1991, Vol. 40, pp. 951- 955) described the use of a simple linear regression model to express drain current y (in milliamperes) as a function of ground-to-source voltage x (in volts). The data are as follows: y 0.734 0.886 1.04 1.19 1.351.50 1.66 1.811.972.12 Round your answers to 3 decimal places (a) Fit a simple linear regression model to these data ß。 ь, α = 0.01 (b) Test for significance of regression using (c) Find a 99.00% confidence interval estimate on the slope. (d) Test the hypothesis Ho: β0-0 versus HI: β0z0 using α 0.01

Answer 1

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AS FOR GIVEN DATA....

0.734 0.886 1.04 1.19 1.35 1.5 1.66 1.81 1.97 2.12 0.8074 1.0632 1.352 1.666 2.025 2.4 2.822 3.258 3.743 4.24 1.21 0.538756 1.44 0.784996 1.69 1.96 2.25 2.56 2.89 3.24 3.61 1.2 1.3 1.0816 1.4161 1.8225 2.25 2.7556 3.2761 3.8809 4.4944 1.5 1.6 1.7 1.8 1.9 15.514.26 23.3766 24.85 22.300952 Least Squares Regression Equation : yo Bia 10 (23.3766)- (15.5) (14.26) 10(24.85) - (15.5)2 233.766- 221.03 12.736 ะ 1.543758 ะ 1.5438 248.5 -240.25 8.25 Slope term, 1.5438 TL TL 14.26 15.5 10 (1.543758) 0.966824 0.9668 10 Intercept term, Bo-0.9668 Least Squares Regression Eq":0.9668(1.5438)c (15.5)2 10 SSE,-Σ2.2 24.85 = 0.825 TL (Σ y)2 = 22.300952- (14.2- 10 = 1.966192 (15.5)(14.26) = 1.2736 Tl 23.3766 10 7n SST = SS,/,-1.966192

1.2736 0.825-1.96613 SSE = SST-SSR-1.966192-1.96613-0.00006235 We test the hypothesis for significance of A as follows: Significance level, α = 0.1 The sample slope1.543758 test statistic -tobserved - 1.543758 - 0 -502.258811 0.00006235 80.825 Degrees of freedom, df-n-2-8 Using excel we find p-value = P(t < 1502.258811) = 2*t.dist(502.258811, 8,1 0.0000000000 Critical value: tcr Rejection region: If to > te we reject Hio to/2d.f.=n-2-1.860 (from t table) itical Decision: Since |to| > te, we reject Ho There is enough evidence to conclude linear relation between r and y 90% CI for Slope 90% Confidence interval for A is given by: SSE1 1.543758 CI: 1.543758 1.860(0.002792)/ 士 0.825 = 1.543758 ± 1.860 x 0.003074 90% CI for slope : (1.5380 < A < 1.5495) ese Thank you

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