Solutions for 02402 exam 15. December 20111

Question XII.1 (28) The percentage explained variation r 2 and the residual standarddeviation s e are? These two numbers can be read off directly from the R-output ,cf. the Rnote Section 10.1, so the correct answer is 1: Percentage explained variation:98.6%, s e = 0.000571Question XII.2 (29) From the theory of the tested materials the slope of the lineis expected to be β = 0.155. Is this in correspondance with the observed slope, ifa significance level of 5% is used? (Both answer and argument must be correct) Theconclusion of the hypothesis test is in the answers given based on the confidence interval,so we must identify the right confidence interval, cf. book page 311, (for β). The degreesof freedom is 6 − 2 = 4, so the t-percentile to use is t 0.025 = 2.776 (Table 4 or in R:qt(0.975,4). So, since S xx = 0.005767, the only correctly specified confidence intervalis given in answer 4, which is the correct answer: (And the answer is ”no” since 0.155is outside of the confidence band that has an upper limit of 0.149)No since a 95% konfidence interval for the slope becomes: 0.1280 ± 2.776 ·√ 0.000571 20.005767Question XII.3 (30) A 95% confidence interval for the thermal conductivity, if thedensity is 0.200 (g/cm 3 ), becomes? We use here the confidence band formula for apoint on the line (box on page 313) with a = 0.025036, b = 0.128031 and x 0 = 0.200.The degrees of freedom is 6 − 2 = 4, so the t-percentile to use is t 0.025 = 2.776 (Table4 or in R: qt(0.975,4). So the only correct use of this formula is given as the correctanswer 1:√1 (0.200 − 0.2288)20.0250356 + 0.128031 · 0.200 ± 2.776 · 0.000571 · +6 0.0057715