Solutions for 02402 exam 15. December 20111
Solutions for 02402 exam 15. December 20111 Solutions for 02402 exam 15. December 20111
so:∂f∂D = π 8 · D3and∂f∂d = −π 8 · d3and hence: (according to error propagation rule)and hence the correct answer is 5.Exercise IIV ar(I p ) ≈( π8 · D3 ) 2· σ2D +( π8 · d3 ) 2· σ2dA strength calculation on an old tube in a construction is to be performed. Because ofcorrosion and age diameters are quite ’indeterminate’. Therefore several measurementsare made of as well outer as inner diameter. The measurements of outer respectivelyinner diameter are independent of each other. The results are listed below: (all dimensionsin mm)Outer diameter, x: 44.9, 44.2 , 44.6, 44.8 , 44.0, 45.1Inner diameter, y: 32.4, 32.5, 31.5, 32.2, 32.6, 31.7From the data we get:(¯x; s x ) = (44.6; 0.424) mm and (ȳ; s y ) = (32.15; 0.451) mmQuestion II.1 (3) The outer diameter of the tube is as new 45 mm. The followingtest is performed:H 0 : µ x = 45H 1 : µ x < 45At a 5% level of significance the result of this study is? (As well conclusion as argumentmust be correct) The t-statistic becomes:t =44.6 − 450.424/6 = −2.31and using a t-distribution with df=5 the P-value becomesP (t ≤ −2.31) = 0.0345found in R by pt(-2.31,5). Using Table 4 of the book it would be found that theP-value is between 0.025 and 0.05. If the data was put into R, the result could also befound as:2
x=c(44.9,44.2,44.6,44.8,44,45.1)>t.test(x,mu=45,alt="less")One Sample t-testdata: xt = -2.3094, df = 5, p-value = 0.03448alternative hypothesis: true mean is less than 4595 percent confidence interval:-Inf 44.94902sample estimates:mean of x44.6Anyway, answer 1 is clearly the correct answer: The outer diameter is significantlysmaller than the original, since the P-value is approximately 0.035Question II.2 (4) The original diameter difference is 13 mm. It is assumed that σ x =σ y . The t-statistic for the test of the null hypothesis H 0 : µ x − µ y = 13 mm becomes?t =44.6 − 32.15 − 13√s2p (1/6 + 1/6)=44.6 − 32.15 − 13√0.1916 · (1/6 + 1/6) = −2.18sinceOr in R:s 2 p = (s 2 D + s 2 d)/2 = 0.1916> x=c(44.9,44.2,44.6,44.8,44,45.1)> y=c(32.4,32.5,31.5,32.2,32.6,31.7)> t.test(x,y,mu=13,var.equal=TRUE)Two Sample t-testdata: x and yt = -2.1769, df = 10, p-value = 0.05453alternative hypothesis: true difference in means is not equal to 1395 percent confidence interval:11.88705 13.01295sample estimates:mean of x mean of y44.60 32.15So correct answer: 2.3
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x=c(44.9,44.2,44.6,44.8,44,45.1)>t.test(x,mu=45,alt="less")One Sample t-testdata: xt = -2.3094, df = 5, p-value = 0.03448alternative hypothesis: true mean is less than 4595 percent confidence interval:-Inf 44.94902sample estimates:mean of x44.6Anyway, answer 1 is clearly the correct answer: The outer diameter is significantlysmaller than the original, since the P-value is approximately 0.035Question II.2 (4) The original diameter difference is 13 mm. It is assumed that σ x =σ y . The t-statistic <strong>for</strong> the test of the null hypothesis H 0 : µ x − µ y = 13 mm becomes?t =44.6 − 32.15 − 13√s2p (1/6 + 1/6)=44.6 − 32.15 − 13√0.1916 · (1/6 + 1/6) = −2.18sinceOr in R:s 2 p = (s 2 D + s 2 d)/2 = 0.1916> x=c(44.9,44.2,44.6,44.8,44,45.1)> y=c(32.4,32.5,31.5,32.2,32.6,31.7)> t.test(x,y,mu=13,var.equal=TRUE)Two Sample t-testdata: x and yt = -2.1769, df = 10, p-value = 0.05453alternative hypothesis: true difference in means is not equal to 1395 percent confidence interval:11.88705 13.01295sample estimates:mean of x mean of y44.60 32.15So correct answer: 2.3
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