ENGINEERING MATHEMATICS UNIT-5 : Statistical Techniques = III (Testing a Hypothesis and Statistical Quality Control) Lec-01 Today's Target > Basic terms of Hypothesis testing @ Population (W Sample (iii) Parameter and Statisties (iv) Hypothesis: Null Hypothesis and Alternative Hypothesis (v) Test of significance (vi) Level of significancei@, tn ENGINEERING MATHEMATICS UNIT-5 : Statistical Techniques — Ill i@, Aktu Syllabus Testing a Hypothesis > Introduction of sampling theory, Hypothesis, Null hypothesis, Alternative Hypothesis, Testing a Hypothesis, Level of Significance, Confidence limit, > Test of significance of difference of means :\(i) t-test (ii) Z— test (iil) Chi-square test| |\\ av Statistical Quality Control) XS > Control charts » Control charts for variables (X and R charts) > Control charts for variables (p, np and C charts)Basic terms of Hypothesis testing a =e (i) Population or Universe: —— Any collection of individual | members |objects | set of all experimental data Population Finite (countable) Infinite (Uncountable) Os (ii) SAMPLE : A Finite subset of population > To study population, we select arandom sample in such a way that each member of population has equal chance to include in sample > Sample Size : The number of member | objects \individual in a sample > Sampling : Process of selecting a sample(iii)Parameter and Statistics: Symbols for Population and Samples Parameter Statistic S.NO. Measures used for Population Measures used for Sample 1 Population Size (V) Sample Size (n) 2 Population Mean (i) Sample Mean (Z) Population S. D (a) Sample S.D (s) Population Variance (a*) Sample Variance (s*) PopulationProportion (P) Sample Proportion (p)a (iv) Hypothesis / Statistical Hypotheses a Hypotheses | > Assumptions required to make decisions about population on the basis of sample inf ormatiton + Deciding a hypotheses is to be accepted or rejected Null Hypothesis (H,) Alternative Hypotheses(H;) > Itisadefinite statement It is complementary to the null hypothesis about population parameter > Tested under the assumption that it is truea Note: (1) (Ho) : Accept then (H;) : Reject (2) (Ho) + Reject then (H,): Accept Test of significance (or Null Hypothesis testing) The tests which enable us, to decide whether the Null Hypothesis is eccepted or rejected on the basis of sample information (t-test ~~ (iz test (iii) F -test (iv) chi, square testa Level of significance («) a “> The value of probability above which we do not reject the Null Hypothesis. ™ > Usually, level of significance in testing of hypothesis are 5% ,1% S By Defaule Iu “ n or ae” Re pon 4 VN - b St Ci, is not Kucped)UNIT-5 : Statistical Techniques — III (Testing a Hypothesis| Today’s Target >| t — test for one sample > PYQ > DPP jand Statistical Quality Control) Lec-02Parameter and Statistics: Symbols for Population and Samples Parameter Statistic S.NO. Measures used for Population known as | Measures used for Sample known as Parameter statistic 1 Population Size (N) Sample Size (n) 2 Population Mean (j1) Sample Mean (2) 3 Population S. D (o) Sample S.D (s) 4 Population Variance (6) Sample Variance (s*) 5 PopulationProportion (P) Sample Proportion (p)[Hypothesis Testing Test of significance (t-test 7) (i) z -test | (iii) F —test 4 (iv) chi —squaretest “ —} M FM t= test OR Students t — distribution > Itis a parametric test of hypothesis testing based on student's T distribution. > To test the significance of the mean of a sample. > t-test is applicable when (® sample size is small (n < 30) (i) Population standard deviation (a)is not knowna ie a For one sample For two sample 4 Next Lecture — test for one sample AA > To test the significant difference between the sample mean and population mean. Steps to solve Questions STEP ~ 1: Set Null Hypothesis (Ho) Ho : There is no significant dif ference between sample mean and population mean Ho : = Ho (given) STEP ~2:: Alternative Hypothesis (H;) Ay: nemo (T WAPA\led Fest)a Step -3: Calculate |t| (©) When S.D is not given t= is = Sample mean = Population mean n= Sample size S =Sample $.D (i) WhenS.Dis given xa tot veta Step —4 : Write degree of freedom (n-1) and level of significance « (« = 5% OR x= 1%) Ss — 5: From the t — test table By aefaule Find the value of t,. at the level of significance « for the degree of freedom (n- 1) Step — 6: Conclusion @ ire) (42) then Hg is accepted \® There is no significant dif ference between sample mean and population mean _/% Sample have been taken from correct population. (i) If \t|> t. then Hp is rejected > There is significant dif ference between sample mean and population mean > Sample could not have come from this population.Fiducial limits or confidence limits Ky< (© 95% confidence limits (level of significance is 5%) are | ¥ (© 99% confidence limits (level of significance is 1%) areQ.1 Asample of 18 items has a mean 24 units and standard deviation 3 units.Test the “= hypothesis that it is a random sample from a normal population with mean 27 units. oven Null Hypothesis (H,) FX 24-27 yw =8y Hy: U=2F 5s =8 A\yernanve Hy pothesis (Fh) t= aun = kt yt WERE (1p roarted test ee = = HES, Calla 4 tees H Vigpothests is veyede* a Q.2 The 9 items of a sample have the following values: 45,47,50,52, 48,47,49, 53,51. a “wvw" Does the mean of these values dif fer significantly from the assumed mean 47.5? — hiven i @-*) [Oca AKTU - 2021 45 [=4el 16-8] a n=" 4? =8. | 4.4] n So Jo.4 ondt we Ue M4 52 a4 buy 37 . Wy) V2 K = 44. z ~ Au I uu . és ao) 0.00) iney = 8 53 BS. 4 Ql TR) = S404 34 4 3.61 mae Zaay= [54.34a 5 = [ea Alkeenahve Hy bothests(H,) a on n=1 ~~ Hi M#YF 5 (Two taded test) 5 = | 54-54 ae caliutahon 4 | t| 5 = 9.614 b= Wow Ss Null Hypothesis ( Hy & aii -5#5 4: —— ati Hi Meads “ b=-1.8327 \e\ =|. 8324B degree 4 freedom A=(o Fd =o Level 4 sagns fiean A= Sle = Fag a= 0-05 Fyom t-test table boos, = ay conclusion a [t] = 1.9324 andy o5 = 2,31) “TEL < to, of H, Hy petals acepcedthe following data: a Q.3 The lifetime of electric bulbs for arandom sample of 10 froma large consignment gave a Can we accept the hypothesis that the average lifetime of bulb is 4000 hrs?a laven Se Ce (ce = om = 44 A Ug -0-2 0. 04 n lo n=lo ub od 0-04 a 34 | oe ows} CSU 4 A =4000 ' 2e.% 0-04 7 a od 064 és Twa) = —b: § DG n- M=4 -0-5 025 \ oe 0.01 Peay 5.6 jay 4 m= KY [way B12a $= 0-584 Null Hy po the sis (H.) 4 Ho =4000 Alevnahve we OS Hy1 b+ 4oool (Twotaled ise & Ca\cwlatton ale | e Ate 2.123 Hlth = alas Rema nny skp HWa } 9/44 random sample of size 16 has 53 as mean. The sum of squares of the deviation from a “<= mean is 135.Can this sample be regarded as taken from the population having —— 56 as mean? Obtai 95%and 99% confidence limits of the mean of the population. uven ga [ze mw) | Null Hu pbthesis(IH, ) n-1 nat H, =~ Rs ‘ > nos S=\ie=7 | os u=S6 2 Tlw-w) =BS fr Alternahwe taypathesis (7) sij4 oe : M= 56 Hy. #5 6 (Tod tailed test)caluslahon lel degree 4 freedom -[ = te-| 215 Level 4 sun ie an g= 57. ts A=005 From. tetest tabi tio5 = % conclusion a [E] sou and bos = Rall Sit] > bo.05 “Hy Hypothesis is seyededa When Level { Stgnificanut Ts'b confidenu Lammik 4 mean af 45 Jo a od, Wo tyes =534 VWZKS ween Oe = /°) =L =0.0) 106 a=0.0) Y le = 5402S , 54.5745 tZ = 2.95 aX conf Menu Limit muan of 44 "ly SFT boo S HSER ASN = SO.FETS 55.285a Engineering Mathematics by Gulshan Sir a UNIT-5 : DPP- 02 Topic: t — test for one sample @ 1A sample of 20 items has mean 42 units and $.D.5 units. Test the hypothesis that it is arandom sample from anormal population with mean 45 units. 0.2 Ten individuals are chosen at random from a normal population of students and their marks are found to be 63, 63, 66,67, 68, 69,70, 70,71, 71. In the light of these data, discuss the that mean mark of the population of students is 66.UNIT-5 : Statistical Techniques — III ‘Testing a Hypothesis and Statistical Quality Control) Today’s Target >| t — test for two sample > PYQ > DPP Lec-03t-test for two sample > Totest the significant dif ference between mean of two independent samples. > Let two independent samples are sample — 1 and sample — 2 Sample -1 Sample =2 ) X11 Xz 1X3 me Xm ) Yu Y20Va mm Yn i) Mean |2=% (i) Mean . | = (ii) 5, = PEE Wi) S liv) Size m $30 liv) Size n, $30 (v) Population mean py (v) Population mean p,a Step to sovle Questions cate tt 1: Set Null Hypothesis (Ho) Ho: The means of the two Population are same Ho? My = Ha TPR — 2: Alternative Hypothesis( H;) Ay My # Me are - : Calculate | t | x where xy ¥ = Mean of Sample-1 Y = Mean of Sample-2 ny = Size of Sample-1 ny = Size of sample-2 5S, =5.D of sample ~ 1 S, =5S.D of sample ~2A If S.D S; and Sz are given InS3+m25y ny+ng—2 AG If $.D S$, and S$; are not given + Se STEP — 4: Write degree of freedom (n, + nz — 2) and level of significance « (« = 5% OR «= 1%) By defawlt Find the value of tz at level of significance\for degree of freedom (ns +2 ~ 2) STEP —5 : From thet ~ test tableSTEP ~ 6: Conclusion (© If |t| < te then Ho is accepted There is no significant dif ference between between their means (W If |t|> te then Hg is rejected There is w» significant dif ference between between their meansa Q.1 Samples of sizes 10 and 14 were taken from two normal populations with S.D.3.5 and 5.2 a The sample means were found to be 20.3 and 18.6.Test whether the means of the two populations are the same at 5% level. Null 4 the H, diven We Know that | HaPe thesis (Hy) n,=10 oe nse an, H, S2-—__ ny = a, +0, 2 Bee u = = . - s=35 |e VOR SY+ MNS] pricy ahve Hy potaesis (14) =5, \orlu-2 ae Pye MFA = 103 5 2184 5 = UGa Calulahon 4 itl Degree 4 freedom = nytm—2 cS) b= 3a = Log -2 S ran en Level 4 St sniper) =0.05 t=9-9404B conctuswn a |b] =0-8604 and toc = 2.07 iP Pee tog H Hy pothests Ts accepteda Q.2 The height of 6 randomly chosen sailors in inches are 63, 65,68, 69,71 and 72. Those of 9 a "=" randomly chosen soldiers are 61,62, 65,66,69,70,71,72 and 73. Test whether the sailors are onthe average taller thansoldiers| { y [xx Jo] 9 [9-9 [y3) 63 FS 25] 6) [6-67 [44.4984 baven 65 | 73 % 62 |-5. 67 [32.1484 Sample—! Height + Sailoxs ea * o | 65 F267 fez 89 Sample-2 — Height a Soldiev] | gq 66 [HL 67. [2.7889 nab a1 3 g JAE [1 a firess plu iy | FO [2.33 founss n,=4 TW) 43.33 0984 Te [us 19.7484 TB] 5-33 [23-4094 Ex= eo] =y= oa 4o® = 6d [604 N51.444= 408 6 w= 68 y= he Ny y= 64 Beo.G Ale Know Mat ga [ Ser ew ar Caton lahon lel aytn-2 t= Hoy sz[Soristagge” | sfied 6 + Gz 9, ne s= 4.038} t= GR-6E ST LeL Null Hypothesis Ue) 4.038 cet Hed t= 0.1505 Bs, ttl ga Degree 4 {reedom=n,tn,-2 conclusion a =6t4-2 -® Level 4 significana x x= Se = 0.0 wl job 4 =0-05 up ey Evo the t-kest able’ @ bo, om = Xo | [t] = 0. isa @end tigge= tHb aia) | Z—-test > PYQ > DPPHypothesis Testing Sim Test of significance AD t-test Mi 2 ~ test (ii) F-test __fiv) chi - square test Z-test is applicable when n>30 Population S.D (a) is Known [eaeese | yar es Population S.D (a) is Unknown 30 est est] JAC z-testFor one sample For two'sample Z-test for one sample To test the significant dif ference between the sample mean and population mean. Steps to solve Questions STEP ~ 1: Set Null Hypothesis (Ho) Ho : There is no significant dif ference between sample mean and population mean Ho : = Ho (given) STEP — 2: Alternative Hypothesis (Hy) mgs wene( T uioitailed tert)a Step -3: Calculate Z Zea = 2G on Zens = 234 | z a ¥ = Sample mean 4. = Population mean n= Sample size @ = Population S.D 5 =Samples.D By Befwull Step — 4: Write level of significance « ( = ke OR «= 1%) x= Sle loo-S*) = 4b. 54 = OUFS Zita = 4bCy Step —5:From the Z test table a Find the tabulated value Z;ay at the level of significance « Step —6:Conclusion CO IfZcat < Zrapthen Ho ts accepted > There is no significant difference between sample mean and population mean > Sample have been taken from correct population. (W) 1f2Z cat > Zea then Ho is rejected > There is significant dif ference between sample mean and population mean > Sample could not have come from this population.2-test for two sample ~ Zeqp then Ho ts rejected There is no significant dif ference between between their meansQ.1Arandom sample of 50 items gives the mean 8 and variabce 15 can it be regarded a “=e as drawn from a normal population with } as mean at 5% of level of significance? — boven Null. Hy pethests Urs) Le Fo Hi There Ts no Sranificante Aiffeunr be bueen xy = 8 Somple mean andk-Populahon mean tf = Hi w= t v ais Abeynahve befethiests (4) u=t Hs MAF (Tw failed tert) s=S1Caliwahon dz Ziggy = | OBE Conclusion) cS) Z gs ). 826 aad Level 4 seqmificane unt S Zeq < Tha mon Uypothesis fs accepa Engineering Mathematics _by Gulshan Sir a UNIT-5 : DPP- 04 Topic: 2-test 0.1 Arandom sample of 60 items gives the mean 10 and variabce 20 can it be regarded as drawn from anormal population with 8 as mean at 5% of level of significance?UNIT-5 : Statistical Techniques — III (Testing a Hypothesisland Statistical Quality Control) Lec-05 Today’s Target >| x? — test of goodness of FIT (Part — 1) > PYQ > DPPHypothesis Testing Test of significance AD t-test _ AM) 2 - test x (lid) F-test ? _ iv) chi — square test (x2 - test) ) Hpothesis Testing P tric test ‘arametric tes! Non — Parametric test (Sample drawn from normal population) (No information about population) t—test Ot Dye wot z—test an os (iii) F - testa Chi — Square test (x? - test) x ~ test x2 — test of goodness of FIT les Lg 2 — test of goodness of Fit: Given Data can be fitted in (@ Poisson Distribution (@ Binomial Distribution (iit) Normal Distribution twin Fitting is good or not is test by x? - test. In x? — test|o;, 02 ....On is a set of observed frequencies|and set of expected frequencies such that ya 3 E 2 — testas atest of independence L=# im Question ae (Ey, Ez, Ey ~.. E, are the corresponding|a (Conditions for applying x* — test "(0 Sum of frequencies must be greater than equal to 50. i) The expected frequency E; of any item must be greater than 5, Note + If any item has frequency less than 5 then it should be combined with the next or preceding item until the total frequency exceed 5. Steps to solve questions Step — 1: Null Hypothesis (H) (H,): There is no significant dif ference between observed and expected frequency. Step — 2 : Calculate x? under H, x(t Where 0, = observed frequency E, = Expected frequencya Step —3: Degree of freedom = Number of variables — number of constraints a @=n-k=pn-! Level of significance x= 5% = 0.05 Step —4: From x? - table Calculate x74, at the given level of significance for the given degree of freedom Step-S: CONCWSION (O If 7 < x7 495 then H,accepted > There is no significants difference between observed frequency and expected frequency. > Fit of the data is considered to be good. (i) 7 > o95, then Hy is rejected. > There is significants difference between observed frequency and expected frequency. > Fit of the data is not good.a Q.1A die is thrown 276 times and the results of these throws are given below: a l No. appeared on the die 1j[z;3/4|[s5]e6 Frequency 40 | 32 | 29 | 59 [57 | 59 Test whether the die is biased or not. (A.K.T.U.2019) Nul| Fiypothest s (H, } B O-Ey n= H,: The die is an unbiased die] | Ae ut —6 36 = 4 calculahon 4s uniky Ve 82] MG “ m ; Ad | ue =43 234 .= Dot at & = Toe = a 94 46 ie \64 57} ub M1 wa Sa} 4d RB 14 E. =44 So = Ee = EWR = 2H rb 980a = & fom | Degree 4 Freedom= n-I] conclusion wi x = = (oie) = 6-2 z Mos 230 Ka loFO =o Level 4 suymipican’ sf “ Xr 5 X= 51 erees | H, is wejeutal From yertest table Henw, 2 ae og = [oto | haven die tsa braseh diea Q.2The following table gives the number of accidents that took place in an industry during a various days of the week. Test if accidents are uniformly distributed over the week. Day Mon | Tue | Wed | Thu | Fri | Sat No. of accidents 18) | @ 1d) Null Hypothesis (He) Or by -E; [@.- a) H,: Accidents ave unrtorm| 14 14 oO distribukd over he wee 12 hl 4 \6 a Ta. 2 y caludahon 4 YC under Hy “4 : \\ \y ai 4 Ep seta BY. yy "Sy 1 \ ° 6 4 | ° ® Eo, = Torey = 84 RoCS oe bah | Degree 4 frecdow=n-t | concusioy) cS) BA x fered j - 2 a a Er =e! ,‘ = I.1U28 , Ng y= M070 = &(j-&)> "66 Sion “i ae senfegs®) » 4, is accepred Ae x= 5" Ine 0.05 - 4 Hou, From ‘test FECA dents ave uniform , = \\.o+0 Vy distyibowke ovey tha oo 05 weeka Q.3 Inexperiments on pea breeding, the following frequencies of seeds were obtained: Round and yellow | Wrinkled and yellow] Round and green | Wrinkled and green | Total 315 101 108 32 556 Theory predicts that the frequencies should be in proportions@: Examine the correspondence between theory and experiment. = Null Hypothesis (rts) or | &¢ or-Er [@i-F&) Ho Expenmental Results| | 3'5 32-75 [2-25 [5.0625 [0.016187 support Yncowy, WY ]1o¥.g5 [3.25 [10-5625 ]O-101319 2 ualahony Ae under Hy 108 J\o¥. as 3.45 Jim. 0625 Jo-4gqr. E = Ta 556 = 3g FS 32 | M75 [-2FS [F-56025 [o-917626 y 8 Toe pee ssS peste? 556 &a Ey = ER TDS = 14.25 Degree freedom =n-!l a =4-1 Ey = Bx556= 104.25 & & Level S Efe S35 = 3 5 $ fren) a= ST, = 0105 + oa} From ¥-test table Ni] Wo = 0-4710024 Co.og = 1185ey = Conclusion o XK, = OFM 70024, RK = 7.185Nv cal tab & “Kar § Ke we ~ H Is pccebred & 2 Henu , or 7 Experimental ReGilis, Support theovitiwl vets osEngineering Mathematics _by Gulshan Sir UNIT-S : DPP- 05 ‘Topic: x? — test of goodness of FIT (Part — 1) Q.1 What is x? — test? A die is thrown 90 times with the following results: [akTu 2010} Face : 1 Zz 3 4 5 6 Total Frequency 3 10 12 16 14 18 20 90 Use x? — test to test whether these data are consistent with the hypothesis that die is unbiased. @.2 The demand for a particular spare part in a factory was found to vary from day — to — day. Ina sample study, the following information was obtained: Days: Mon \Tue Wed Thurs Fri Sat No.of parts demanded: 11241125 1110 1120 1126 1115 Test the hypothesis that the number of parts demanded does not depend on the day of the week. 0.3 The theory predicts the proportion of beans in the four groups, G,,G,,G;,G, should be in the ratio 9: 1. Inan experiment with 1600 beans the numbers in the four groups were 882,313,287 (and 118, Does the experimental restié suppert the theory’a @ ENGINEERING MATHEMATICS go UNIT-5 : Statistical Techniques — III (esting a Hypothesis|and Statistical Quality Control) Lec-06 Today’s Target >| 22 — test of goodness of FIT (Part — 2) > > PYQ DPPHpothesis Testing Parametric test Non ~ Parametric test (Sample drawn from normal population) (No information about population) i) t-test @ x? — test AM 2 test (iii) F — test if = test 2X? ~ test of goodness of FIT x? ~ test as atest of independence 1-8 lg 6 Leha x? — test of goodness of Fit: a > Given Data can be fitted in @ Poisson Distribution Ali) Binomial Distribution (#8 Normat Distribution > Fitting is good or not is test by x" — test. > Inz2 - test 04,02....09 is a set of observed frequencies and E,E2, Ey .. E, are the corresponding set of expected frequencies such that > oo x cA Conditions for applying x? test (© Sum of frequencies must be greater than equal to 50. The expected frequency E, of any item must be greater than 5. Note + If any item has frequency less than 5 then it should be combined with the next or preceding item until the total frequency exceed 5.a Steps to solve questions a “~“" Step — 1: Null Hypothesis (H.) (H,): There is no significant dif ference between observed and expected frequency. Step —2: Calculate x? under H, Where 0, = observed frequency E, = Expected frequency Step —3: Degree of freedom = Number of variables — number of constraints d=n-k Level of significancea Step -4: From x? - table a Calculate z*,o, at the given level of significance for the given degree of freedom Step-5: @ If x? < 2% o95 then H,accepted > There is no significants difference between observed frequency and expected frequency. > Fit of the data is considered to be good. (Dx? > o95, — thenH, isrejected. > There is significants dif ference between observed frequency and expected frequency. > Fit of the data is not good.Q.1 A survey of 320 families with 5 children shows the following distribution: No of boys Sboys 4boys 3boys 2boys _ boys Oboys Total & girls: &Ogirl &igirl &2girl &3girl ~&4girl &Sgirl No.of families: 18 56 110 88 40 8 320 Test the hypothesis that data are binomially distributed and male and female births are equal probability. b= q= 1 (G.B.T.U.2010) Nul| Flypothesis (H, ) Hy. Data. ave browne distribu | N= 320 and prokabiity 4 Male and Hinle | Ves bavth ave eqpak ¥=0,1,2 3,45 Let b= Prob. 4 luv] (ges) Bino mak Bist yvoarey) q ~ Prob of fob Faitwee) P(xEX)= an c ¥- \o B= 320%, af) yf) =seons x = 50 SB epoca F reqmencivs Fey EpaN% pry Ey = 50 - 5 iS . > ~aRoy Ser oo easton ae gl Exe 0" Sealed Ea go E, = 3208 5 x8) ALY E> ~ 2 = BOK IHL Eye 320 nSc nay na) = ORE he Ey = 100 E, =10] KL B= 50 \ o Es = a0 Sun he) = 3KOXS Me@ [E-= Caluulation 4 under Hy loo my 100 MAY |BG = 46 ta ||.oFO De ey vee freedom an-k conclusion, 6-1 5 2 Level Asn van (x) | our 7 Bs , k= 5], 20:08], Hd Neyer Keay = Neds wou and Ky, = ll-oF0 2 Here From X= table Data axe not Binomod hy | Dickrbuied ~ = Proboalaite laavth 4 bo, and (yx) one 9 tl. capes > PEta Q.1 Fit a Poisson distribution to the following data and test the goodness of fit: x o 1 2 f 105 65 23° Null Hy 0 Hess (Hy ) R, Fitting data IN poisson Distvibution fs yoo! Mw 4a ° 104 oO \ gS 6S 2 dL 4 3 3 4 M | 4 E}= 200 [Eq 3 Bu 3 si 4 =e gs 1 Meani= = 4% b 22=0 ZO = 61 lo By Poisson dist v'xgnon Pe) = eA 3% wn s0-bl NS Pin) =e” xb-bi) wlcy Expected Frequencies Ee N68 a1)" é ni E; = 200 ne KI) | —! =a od £, = 20086"! v4 ol ~O0KE Ex 200.8 1 E = 108.64 = 104 a i Ey = 66.24 REE ~~ E, = 20.22 SoG 4 yee 4 Eg = gis SS! Coctyedahio 49E vider HyI® & ]o-a le-at Degree { freedom = n—K a 104 o o 4.} 398 -1-I-! a 66 | -1 ) Ve ab= 2 go} 2a] y oe 4 Lear 4 sign’ frcane ) \5 a | og |. bs | |, d= S),=S =0.05 ois lw oe From Yetalole 2 ese 5.94]wi contusion ey WE, = OMISIS and = 5-441 oe ° ve & Te < enn oo & H, is aceyitd A“ Henu Fiking 4 duke Mn n gar sstriien ba se cPQ.2 When the first proof of 392 pages of a book of 1200 pages were read, the distribution Km of printing mistakes were found to be as follows: @ No.of mistakesinapage(x) : 0 1 2 3 45°5 6 No.of pages (f) : 275 72 30 7 8 2 1 Fit a poisson distribution to the above data and test the goodness of fit. Null Hypothesis (Ha) K 4 4% H,". Poisson Distvibwhon 14 a oe 2h, S od fir te tre daba | +e Fe aa a 30 60 - ~ =) Mean = = = 3) 4 Py \ 4 3 20 5 ey 1d A= 0-4 G2) i \ $ Tha Mr [T {n= 184a By Poisson Disty1 f button ~ nik f E,= ax CO abun) fr) = a . Enpeckd Evequnauy on aut E, = Nxe* as e, = feed = NET - Qe 28d co 26 awa eh Bye 52 SWS nV j B= oo = - 6 B= 00h OT = p-o04r 0 tyIS Ep | or-& |Or-e, eee Ga [eee] 30 ]2B.1 1.4 13.61 Tyas SER =.)) 1:9 | 48-0) 245|24a.1 [32-4 flosz. 4] 4.47! 42] 16-F 4. [1948.04] 1B. 121 0-128 14. 2] zs La = 4o.43F Degree 4 Frcedom= ne a4 =4-H- at= Q yo Level 4 Signy can) H=Sly =0-05 sioe = + @ Fvom Xe tante |’ Xe > Xie cy Ge ett] v tb H, Ps rejeded oo concusion Hen yo” v = Yo. ast Poisson ts triewhon fs nob a geod fal z May = 2441a Engineering Mathematics _ by Gulshan Sir a UNIT-5 : DPP- 06 Topic: x? — test of goodness of FIT (Part — 2) @ 1 Test for goodness of fit of a poisson distribution at 5% level of significance to the following frequency distribution: xo o 1 2 3 4 5 “6 7 8 foe 52151 130 102 45 I2~ 5 1 2 @.2 Test for goodness of fit of a poisson distribution at 5% level of significance to the following frequency distribution: x : o 1 2 386%4 5 f : 275 138 75 7 4 0.3 Records taken of the number of male and female births in 800 families having four children are as follows: ‘No.of male births o 1 2 3 4 ‘No.of female births 4 3 2 1, 0 ‘No.of Families 32 178 290 236 64 Test whether the data are consistent with the hypothesis that the Binomial law holds and the chance of male birth is equal to that of female birth, namely p = q = 1/2. [U.P.7.U.2009)UNIT-5 : Statistical Techniques — III (esting a Hypothesis and Statistical Quality Control) Lec-07 Today’s Target >| > > 2 — test of as atest of independence PYQ DPP2? ~ test of goodness of FIT — test as atest of independence L-s L-¢ l=F 2? ~ test as atest of independence Two attributes A and B are given with size r and $ respectively. Steps to solve questions Step —1: Null Hypothesis (Ho) (H,) : Two attributes are independentle data is given is tabular form, called contingency table a The expected frequency for any cell ina contingency table can be calculated by using formula Expected frequency = ee Observed frequency Expected frequency Oy 02 a+b a b 2,2 GtMera e @+ho+ Gtk, 03 % c+d c d ate btd Ne 2 Ct Oa+o 5 a Et Ob+a | ( +d j ate bra 7 Oieand total at bt 27995 -then H, is rejected. Two attributes A and B are independent notQ.1 From the following table regarding the colour of eyes of father and son, test if the colour of son's eye is associated with that of the father. Eye colour of father Null Hybothesi s(H,) (A.K.T.U.2021) Eye colour of son Light ‘Not light Light an 51 Not light 148 230 Con Fing ong table H+ The colous 4 son'seye i nok associated wi to Hak 4 fate Observed Freaueny Oy x “FI Sl 522 oy 4 Mb | 230 | 318 614] 238l | Foo rT band +o wiExpected Fyre =522K61 22.428) 4oo Foo | 522 = 354,02 =. 48 E, = ST 8G, = SEB 3 900 400 B48 = 254,98] =118.02 614 281 400 ur C= TIES f= erst, 8), 0-8), C4 ESA F = 1 & z “4 = Sati 35463)”, 51-162:48 , '48-25148) B5HG2 162-48 254.938 4 R30—118.02) \1B.02 = BY, 42404 F6-ABAOTUB ARG $106. MIF cm Kar 266. UTa Degree 4 freedom a b = (W-I)(s-1) =@-1)(Q-1) =I Level 4 Segnifieand Conclusion X= TS], = O08 a Fyom Y- table zu Tay Baby Pp ha, = 266, 3494F fim =384] 4 eos > bE e “ys veyed lea he “plouy A Sons eye Wwascounsea} wiv Howl} tat’ vee ga Q.2To test the effectiveness of inoculation against cholera, the following table was obtained ‘Attacked Not attacked Total Inoculated 30 7 160 190) Not inoculated 40S 460 | 600 Total 170 620 790, (The figures represent the number of persons.) Use x? — test to defend or refute the statement that the inoculation prevents attack from cholera. Combtingenvy table Oloseyved Freqmeny Null Hg pothesis (rl) Hy: The Woculation dow No’ prevent from cholera attavk (A.K.T.U.2009, 2018) Oo 6. sof tho P40 o, Wo} 4c) 600 Fo 620 T4D® = Wo xito 140 K620 Ee ft eto H40 = 40.84 | =144.00 =, = 6004620 Eee Fe, Se 600 =l2g. | =4%0-59 Fo | 620 | FA a /- Bo-4o 8), 1 ee —|44. (coma) 4 wob4 MA] i < uy (o=P4.n) 4 bo —4t0.84) Pay pa f= 40 4 0. F4540- AT 0-252a X= 4.865 From X- table An, fs tejeued a — ‘tal 2 = Ss Be — Degree 4 freedom Mat 1) ett df = w-)CS1) LON lusion, The inovutahon prevent = @-1)(241) Keay = We 842 from cholera Attack =| » _ en Arar | Le Sqm {rank sat oe, 8251, = 0051 O17 Leowae 3 Inan experiment on the immunisation of cattle from a disease, the following results a are obtained. Derive your inferences on the efficiency of the vaccine. _ (A. K.T.U.2021) Affected Inaf fected Inoculated with vaccine Not incoulated (asEngineering Mathematics Topic UNIT-5 : DPP- 07 by Gulshan Sir 2 _ test of asa test of independence @ 1 Inan experiment on the immunisation of goats from anthrox, the following results were obtained. Derive your inferences on the ef ficiency of the vaccine. Q.2 Byusin, Died anti Inoculated with vaccine 2 Survived Not incoulated Ve oe) \g x — test, find out whether there is any association between income level and ty] Income Public: Govt. school Low 200 201 High 1000 400) @.3 The following data is collected on two characters: Triokers Raeaae Tierate 3 7 iterate oe) Based on this information can you say that there ype of schooling: "tween habit of smoking and literacy.AKTU Full Courses (Paid) =“ Download Gateway Classes Application From Google Play store All Subjects Link in. Description 2c2