TMA4245 Statistics. Corrected 30 May and 4 June Norwegian University of Science and Technology Department of Mathematical Sciences.

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1 Norwegia Uiversity of Sciece ad Techology Departmet of Mathematical Scieces Corrected 3 May ad 4 Jue Solutios TMA445 Statistics Saturday 6 May 9: 3: Problem Sow desity a The probability is.9.5 6x x dx =.9.5 6x 6x dx = [3x x 3 ].9.5 =.47. b The likelihood fuctio is give by Lβ = ββ + x i x i β = β β + x i x i β, i= i= i= ad the log likelihood which has derivative l Lβ = l β + lβ + + l x i + β l x i, i= i= l L β = β + β + + l x i. l L is decreasig o, ad the sum of two first terms teds to whe β + ad to whe β, so that l L will have a sigle zero the third term is egative for β > ad be positive left of the zero ad egative right of the zero. This meas that L has its maximum at this zero. Solvig for the zero, β i= l x i + + i= i= l x i β + =,

2 TMA445, 6 May Solutios Page of 5 we get β = i= l x i ± 4 + i= l x i i= l x i = i= l x i ± i= + l x i 4. We choose the larger zero sice l L has oly oe zero for positive argumets the other we foud must be egative, ad get the maximum likelihood estimator i= + l i 4 i= l i = + l 4 l. For = ad i= l x i = 4. the estimate is.545. /.4 + /4+/.4 / = The discussio of actual attaimet of maximum at the zero ad of which zero to be chose, is ot required. Problem Temperature i March ad April a We ca get a impressio of ormality of or by meas of a histogram Figur a, ad a more accurate assessmet by a ormal quatile quatile plot, which should show liear relatioships betwee ordered observatios agaist ormal quatiles, or a ormal probability plot Figure b could be used, which should show liear relatioships. To assess whether ad are idepedet, we ca plot the values of agaist Figure c. No special patter should emerge, for example the poits beig close to a o-horizotal lie which idicates correlatio. Also a plot of the values of the i ad the i agaist i i the same graph could reveal depedece Figure d. b µ/s/ has the t-distributio with degrees of freedom, so P t α/ < µ S/ < t α/ = α, with degrees of freedom for t α/. Solvig the double iequality for µ, we get 99% cofidece bouds x ± t α/ s/ for µ. Here, =, x = x i / = 9./ =.758, α =., t.5 = 3.6, s = x i x i = = 6.379, givig bouds.758 ± / =.758 ±.65, ad a cofidece iterval.5, 3..

3 TMA445, 6 May Solutios Page 3 of a Histogram 3 4 Probability Normal Probability Plot.5.5 Data, Probability b Normal probability Normal Probability Plot Data, c y agaist x Mothly averages ear Mothly averages d x i ad y i agaist i ear Figure : a Histograms ad b ormal probability plots. Data come from a ormal distributio at the left but ot at the right. c Plots of y agaist x ad d of the x i ad y i agaist i. ad are idepedet at the left, but ot at the right. c We cosider the ull hypothesis H : µ a µ m = 5 or H : µ a µ m 5 ad the alterative hypothesis H : µ a µ m < 5. The observatios come aturally i pairs, ad there is reaso to believe that the March ad the April temperatures of a year are depedet. Therefore we choose a paired test, which is performed as a sigle sample test usig the differeces d i = y i x i. Uder the ull hypothesis, the test statistic T = D 5/S D / has the t distibutio with degrees of freedom. A small value of T is idicative of H, ad the critical value is t.5 =.796. With our data, d = ȳ x = / = 4.86 ad s D = d i d i = = So t = / 7.39/ =.8, which is ot i the critical regio, ad we do ot reject H. At the.5 sigificace level there is ot evidece to state that µ a µ m < 5.

4 TMA445, 6 May Solutios Page 4 of 5 Problem 3 Chemical factory a The probability that the lamp lights up whe the procedure is performed oce, is P > 5 = P /4 > 5 /4 = P Z >.5 = P Z <.5 =.56, where Z has the stadard ormal distributio. If the procedure is performed three times, let the amouts of by-product be,, 3. The the probability that the lamp lights up at least oce is P > 5 > 5 3 > 5 = P 5, 5, 3 5 = P i 5 3 = P i > 5 3 =.56 3 =.85. Let be the umber of times the lamp lights up whe the procedure is performed times. The has the biomial distributio with parameters = ad p =.56, ad P 5 = P 4.5 = P p p p P Z >.8 = P Z.8 =., usig the ormal approximatio with cotiuity correctio. The exact biomial probability is.4. ou are ot pealized if you have ot applied the cotiuity correctio. If you use 5 istead of 4.5 i the ormal approximatio, you get.74, ad if you use 4 arisig from P 5 = P 4, you get.3 ot very good approximatios. b Let be the umber of times the procedure is performed, ad let,,..., be the amouts of by-product. The total amout of by-product, = i= i, has the ormal distributio with mea ad variace 4. We wat to fid such that. P 5 = P = P Z 5 4, that is, 5 /4 z. =.36, or 5 5 z., ad we have a quadratic iequality 5 + z. 5 i. The left had side is a dowwardpoitig parabola as a fuctio of havig zeros z. ± z / 5, that is, 5.4 ad 4.77, meaig that 4.77 ad.8, that is, sice is a iteger.

5 TMA445, 6 May Solutios Page 5 of 5 Problem 4 Barbecue toight? a A Ve diagram is show to the right. P A B > implies that A B, ad A ad B are ot disjoit. P AP B =.4.4 =.6. = P A B, so A ad B are ot idepedet. Note that C = A B. P C = P A B = P A B = P A + P B P A B = =.4. C A B A C = A A B = A A B A A =, so A ad C are disjoit this is is also obvious from the descriptio of the evets. The coditioal probability of barbeque give o rai is P C A = P C A /P A = P C/ P A =.4/.4 =.67. b ˆα = Ȳ ˆβ x. Sice Ȳ is a liear combiatio of the i, i, ad also ˆβ is a liear combiatio of the i, also ˆα ca be writte as a liear combiatio of the i, which are mutually idepedet ad have ormal distributios, thus ˆα has the ormal distributio. E ˆα = EȲ ˆβ x = Ei β x = α + βxi β x = α + β x β x = α, ad Var ˆα = VarȲ ˆβ x = VarȲ + x Var ˆβ = σ /+ x σ / x i x = / + x / x i x σ which ca be show to be equal to σ x i / x i x. The assumptios are that the i are idepedet variables havig the ormal distributio with mea α + βx i ad variace σ. The Figure idicates that relatioship betwee E ad x might ot be liear, as most poits havig a small or large x i lie above the estimated regressio lie ad most other poits lie below. This could also be due to depedece betwee the i. The assumptio of costat variace seems to be OK. c A estimate of the temperature at : if the temperature at 3: is 5 C, is ˆα+ ˆβ 5 = = 4.9. For makig a 95% predictio iterval for a ew observatio correspodig to x, we cosider the variable Ŷ = ˆα + ˆβx α βx ɛ. Sice ˆα ad ˆβ are liear combiatios of,...,, Ŷ is a liear combiatio of,..., ad ɛ, which are all idepedet, so Ŷ has a ormal distributio. Its expected value is EŶ = Eˆα + ˆβx α βx ɛ = ad its variace VarŶ = VarȲ + ˆβx x ɛ = σ + / + x x / i= x i x ˆβ ad Ȳ are idepedet, leadig to a statistic Ŷ / ˆσ + / + x x / i= x i x, which has the t-distributio with degrees of freedom. So Ŷ P t.5 < ˆσ + / + x x / i= x i x < t.5 =.95, with degrees of freedom for t.5. Solvig the double iequality for, we get 95% predictio bouds ŷ ± t.5ˆσ + / + x x / i= x i x for y. Here, = 8, x = 5.5, ŷ = 4.9, t.5 =. = 8 degrees of freedom, givig bouds 4.9±..4 + / /5.7 = 4.9±4.39, ad a predictio iterval.5, 9.3.