Solutions for 02402 exam 15. December 20111

lm(formula = y ~ x)Residuals:1 2 3 4 52.2358 0.3042 -2.3774 -1.8090 1.6462Coefficients:Estimate Std. Error t value Pr(>|t|)(Intercept) 10.038 3.448 2.911 0.06193 .x 34.544 3.419 10.104 0.00206 **---Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1Residual standard error: 2.361 on 3 degrees of freedomMultiple R-squared: 0.9715, Adjusted R-squared: 0.9619F-statistic: 102.1 on 1 and 3 DF, p-value: 0.002065Question X.1 (23) A 95% confidence interval for the slope in the model behind theanalysis is? We use that we can read off the standard error of the slope in the R-output:(cf. page 311 in the book and the R-note)SE b = 3.419 = s e ·1√SXXSo to apply the confidence limit formula from page 311, we just need to finde the propert-percentile, t 0.025 with 5 − 2 = 3 degrees of freedom:t 0.025 (3) = 3.182In R: qt(0.975,3). So the correct answer is 3: 34.544 ± 3.182 · 3.419Question X.2 (24) For a random tablet of some other brand, where the surface area/volume= 1.25 mm 2 /mm 3 it is found that % dissovled= 47. Is this in correspondance with thegiven model? Since we ”predict” the value for a single new observation we should usethe prediction interval, page 314, (and NOT the confidence interval) with x 0 = 1.25.And only in answer 3 the correct prediction interval is given, So the correct answer is3: Yes, since the prediction interval 53.218±3.182·2.361·√1 + 1 + (1.25−0.96)2 contains5 0.477the valueExercise XIA company works with a 3-shift operation, that is, it is running 24 hours with 3 differentshifts of people working 8 hours. It is expected that the ’down time’ on the machinesis higher on evening and night shifts than it is on day shifts. In order to verify this the’downtime’ is registered on four machines of the same type in a period of time. Themeasured data is shown in the table below. The observations are listed as ’down time’in hours in the given period of time:12