Wetents 44402 63.8 (J Construct a histogram for the water quality data in ‘xereise 6.2.10. Comment on the shape of the histogram, Does, it convey the same information asthe stem-and-lea display” 424 65.7 208 S87 521 S58 S70 68.7 673 673 543 S40 731 813 $99 569 62.2 69.9 669 590 $63 433 S74 45,3 80.1 49.7 42.8 424 $9.6 658 614 640 642 72.6 725 461 S31 S61 672 70.7 426 T1d $47 STA 773 393 764 593 SLI 738 614 73.1 773 485 898 $07 52.0 596 66.1 316 67.2 QI Construct anormal probability plo ofthe octane ata in Exercise 6.2.4. Does it seem reasonable to assume that octane rating is normally distributed? 885 988 8.6 922 927 HEH 875 909 947 $8.3 904 S34 $79 926 B78 899 843 904 91.6 910 930 937 883 918 90.1 912 907 $82 914 965 $92 99.27 $90 6 886 885 94 B43 923 922 898 922 883 933 912 92 BRO 916 $17 942 S74 867 886 898 903 911 853 911 942 88:7 927 900 $6.7 90.1 95 908 927 933 915° 934 893 1003 90.1 993 867 $99 96.1 911 $76 918 910 910 7.2.8 Scientists at the Hopkins Memorial Forest in western Mas- sachusetts have been collecting meteorological and environmen tal data in the forest data for more than 100 years. In the pa Few years, sulfate content in water samples from Birch Brook has averaged 7.48 mg/L with a standard deviation of 1.60 mg/L. ‘a, What is the standard error ofthe sulfate ina collection of 10 water samples? IF 10 students measure the sulfate in their samples, what is the probability th 6.49 and 8.47 mg/L? © What do you need to assume for the probability caloulated (b)to be accurate? their average sulfate will be between Histogram The histognom is raghly Symebic, buawse the highest boas ore soughly in the midlle of the histogaam The stom-and-feaF dingrum conteys the some inbthation buat the shape of dstihction, Sample Qvontiles Standard Normal Quantilos Number of Data Points n= 82 y= 2.80612 + 90.5956 9846 Conelton : Yes, itis reasonable to ossuie that ths 1s narmally Aistribated Let tard wart X reptesent the sulfate content in the water Somples , Ab= 948 ng/L oF bom dL ® aes w) Plo ateKedan= P (Meigs SA = betty & 1a Ee 46) =PZE1%)-P(ZE~1.%) = 0.950 o We assume that sangle prints are ranlam cod the sapling distribution will be Ti pte get the owiete pobublity of prtFeeeneeeseonneaepre rs mber of observed “succes sample of 1 observations where pis the probability of ceach observation. © Dssume the event E; indicate i event as either sucess (21) or or falue (=0) Mean of 0 random E; Beh pro Uep)sp Browse each Ei i far 0 suusss and 0 for a fullae, the total uber of sucess X is the sum of ll events of Es ED) = ft pnd oo 4 An np How, rh E(p)= HED = PEP ‘Therdire,, P=% is an whiaseh atimater ofp 7.44 Consider the probability density function fy +0x), LSet |. Find the maximum likelihood estimator for 6. Ww (b) Let ws find the value of the constart ¢ A= | atid A valid piobdhity. dont Furtion sabifien O) Variance. of each event, Ej Vie = E(E)~CeceaT” Consider Ee = 1" 9 + 0% U-p =p VUE) = p- p= P U-P> The variance of X is VOX) VUE) Ft V LER) =pCI-p)4 v4 ple PD =npl-p) Here , eh vip=v%) = Lv = MUD. pup) nn n Thar fore, the star coer of p is Wip- [ee ad LO): ft F%,6) fW20, Wx J foodeet Thachre Si feodxet J, Urande-y eff 'eer0] xx) = efx 10-21, ]s1 efu-enro(Y 2h] <1 cero) <1 ot ff xdurpoax =H acvol ede] CRTC = abrog]= $ 6 = 3p) 63x o EGR)- 361) 23 pl Sina. sample mean % 15 On unbioael cstinaboy for pupelotion em ph 73-2 50 2 trio 0%) eit n)-Laray ~ Luvs) =()"t =) a tex) Take lop Log L (Xi,0) =-mbay 2. + Zargesioxy (xia) =® So here, cbtain oocinur ikeihod stinatar by chmsing @ 50. thot Las given shore tins its marine Sine the maxinun value of © Garsistent. with the some. is Xn te aged ste chombon gy,fo Intaitively , & will abwoys be smaller than & (urifuly distibutcc on [4,4] thats why. it shoud mot be unbiased eatmater 4 We hae shat EO) = 3h thal. by the defination of ins, 2 na, atm) Reena 2-0 me HA)- nas net Ar “ne A hich means that it approche to zure «© Expectation of wnifumly distibuted roud visible on interval Ce, 0] is A, there, on unbiased estimate wil be 2X becnse EUR) -26(L En) AZHK= Bnd 2m w Fram Y= mox(Xi), we need te fink cusmbative ditnbution function of Let 6 [s,0], then PCY 9) =POwe KDE y) > POLS, Ya Fy heey) PO Ey) PUGLEY) Pep) = (Pou ey)” = (2 (Posy) ={2 Fxcodx: by Te is cosy te Find the derivative of the function Ca)” to get the pnbobility, density function ay Oe yeb,o) 0, otherwise Laking of the bias EM 03 we calculated covliey fin net“ Whee the expeatation of Y is et) fy eee A= an 5 the part, where we wed the pradnbrliig density function te ‘is better estimate than eslimaley Qi in the Sense thet its Voriance is Smalicy . It fillng Tom Vee f> ais * =VR) beause, m(nt2) seanznian o3n fr oy N2 |‘8.1.5 Suppose a freshwater (milligrams per liter) ‘concentration is 0.49 < a, Would 299% Cle longer or shorter? ‘Consider the following statement: There is a 95% chance that pis betwoen 0.49 and 0.82. Is this statement correct? Explain your answer, 100 random samples of water from taken and the caleium ion -asured. A.95% Cl on the mean calcium <0 ed from the same sample data be Consider the following stat In = 100 random samples of water from the ake were taken and the 95% Clon te computed, and this process were repeated 1000 times, 950 of the Cls would cootain the true value ofp. Is this statement correct? Explain your answer, © This statement is coect .By vebability , we trean thot if we repeated this experiment over ood over again, 45% of al) Serpls would prodace, o confide interval (0.49, 0.9) that cantsins the due mean calcinm concentrotion , and ony £% wold the intewal be in envy 8.2.9 The compressive strength of conerete is being tested by a civil engineer who tests 12 specimens and obtains the following data 216 2037 2204 225 2301 263 218 ass 2295 18. Check the assumption that compressive strength is nor ‘mally distributed, Include a graphical display in your answer bb. Construct a 95% two-sided confidence interval on the mean strength fe|Construct 2 95% lower confidence bound on the my sirength, Compare this bound with the lower bound of two-sided confidence interval and discuss why they are diferent ". W% = 0.05 ¥v Toble, 2 ty= 16 (dors n-[-y) The Sawer bound 1s then: Bro ty D phe 2241-4954 (O) The Length of the Cs increases with increase uf confidence Vere) For fixed. simple size nov fixed deviation o, then a 19% C1 coltulated from the Sane sample data, be Luger ) There is & 95% chane thot pr is between 0.4] ond ota This stetement & dot correc We only cbiain ene room somple oud alate om cmfidene interval Because this indovo) either wil or will net contain the tive valnc of Ar, it is mot Yeasenable, te ottack a prabability Tere te this specific. event. The oppo priote, statement is that the observed interval (0.49 0.83) brackets the true value of A with confidence 95%, Or it can be interpreted Hat, we da net uo whether the statonent is tone fir this specif, sample Dut the method used to cbtoin the intenal (0.44, 0.89) yields corect statements 15% of the time, ( Normal Probability Plot lot patter is ringhly finear. then Mn fe values wd ype The oma prabbi we Can vance 2 rally dsb w) eb TE oat. gut, ner, (22lb~ KY°9 wg (2245-K)* se ae 235.543 8 20 05 > ty,= 2.20) (00s 12-121) The canfidene inter 65 Table Kt Bw Tt & 2237. Hef aag>. S165