homework 8 key1 - (Q=ID MELCROAVJILOOIN When using ANOVA to...

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Unformatted text preview: (Q=ID: MELCROAVJILOOIN) When using ANOVA to compare the means of groups the total variation is divided into two gulps. Which of the folowing statements regarding this division is accurate: the treatment eFfect refers to within-group variation I - the total variation is the product of among-group and within-group variation random error plays no part In the total variation the variation that occurs within-group ls considered to be random error (Q=ID: MST. CP.OAV.fll.m20f) Let MSG denote the mean square among, MSE the mean square within and MST the mean square total. The one-way ANOVA test statistic is calculated using the equation: F = MSG MST MST MSE MSE MST MSG MSE [Q=lD: MST.CP.OAU.|11.003N] Seled: the statement that is ll“ an assumption Of the One-WHY ANOVA F test: random sampling independent sampling J - heterogeneous variances both populations are normally distributed [Q-ID: MST.CP.OAU.01.0MMJ Ifan ANON'A test s to be performed. a quick test to see If the pwlng ofvarlanoes is appropriate is to check that: the largest sample standard deviation is at least twice as large as the smallest sample standard deviation J - the largest sample standard deviation Is no more than twloe as large as the smallest sample standard deviation all of the sample standard devlalions are the same (to the nearest unit of measurement) none of the above - the population standard deviations must be looked at to see If pooling variances is appropriate (Q=ID: MST.CP.OAV.|‘|1.005CI') In any OHE‘WGY NOVA test; there is always MCI-SEW one: :2 factor I - population mean level (Q-ID: NET.CP.OAV.UI.006MJ A fll'le‘wa'f ANO‘U'A test is paforrned to investigate the dlfierence between the average time taker by four dlfl‘e-ent machlne opeators to perform a task. In this test, the aternative hypothes‘e would assert that: J . not all of the four average times for the four dlfferent operators will be the same all of the four average times for the four different operaoors will be different all of the four average times for the four different operators will be the same not all of the four average times for the four different operators will be dlfferert [El-ID: M5T.CP.0AV.01-W7Iif} In a one-wayr MOW. be: oompering three population means. the null hypothesis ls rejected. Given thls, the Bonferroni method: =- should he used, because the Bonferroni method is appropriate when the null hypothesis is rejected in a one-way ANOVA test should be used, because the Bonferroni method ls always used In one-war;r ANOVA I 0 should not be used. because the Bonferronl method ls only appropriate when the null hypothesis Is not rejected In a one-way ANOVA test should not be used. because the Bonferroni method is never used in one-way ANOVA (Cl-lb: PET.CF.TWIIOV.DL001M] 1111: following statements reaps to advantage oi two-way ANGVA over one-way movie. 1111: statement that is film is: TWO-way ANCN'A experlmenis can be more efficient than one-way ANO‘UA experiments. TWO-way ANOVA can investigate the interaction between two factors. I - In two-way ANOVA tests the pooled var-lance tends to become smaller and the test becomes more powerful. =‘ The Bonferonnl method can onlyr be used in two-Way ANOVA. (Q-ID: MSTLPJWAOKDLWZDH In a two-way ANOVA test: 4 - all of the populations are assumed to he Independent the null hypotheses state that both factors are significant both factors must have ime same number of levels at least one factor will always be proven to be significant (cl-ID: HSTJZPJWAOUJLDOBDIJ A two-wearr ANOVA test Is Investigating the effect of gende- [rnale or female) and employment status (unemployed, employed, retired) on level ofdepressiun. In thls test. the factors are: gender, employment status and level of depression «I . gender and employment status male. female. unemployed. employed and retired maleI female, unemployed. employed, retired and level of depression [q-ID: relencpxrwaovmmmon The com statement In relation I30 Interaction In W MINA is: If two factors exhibit Interaction with one another, neither factor can be significant on ll: own. I - All factors exhibit interaction with one anomer. If two factors do not exhibit Interadlon with one another, neither factor can be significant. - None of the above statements are correct. [q-ID: Hencpxrwaovmmson A means alot In a W MOVE vest oontalns thm DafaIBI and I’IOI'lZOI’ItaI lines. This Infomatlon alone SUQQHS: I - neither factor is significant there is Interaction between the two factors being tested both factors are significant - there Is no Interaction between the two factors being taped [q-Io: astonmaomomoson The feature of a two-way ANOVA means plot that suggests that there is hteracflon between the two factors 3: all lines in the means plot are parallel J . not all lines In the means plot are parallel all lines in the means plot are horizontal not all lines in the means plot are horizontal {Q=ID: MST.CP.TWAOV.01.0070f) The following means plot is a result of a two-way ANOVA test that has been performed: Flam! —o—Levell - LMIZ Lml Lodz Lilli) Fawn Select the correct statement from the following: Factor 1 appears to be significant. Factor 2 appears to be significant. J 9 There appears to be interaction between the two factors. All of the above statements are correct. (Q-ID: MST.CP.TWADV.02.DDlDf} An employment agency has three offioere, and any new clients get assigned one of these officers. The agency also has an orientation smerne that Le in a trial phase. The agency would like to know lfthe officer assigned to a dlent, and whethe' or not the client takes part In the orientation scheme. are factors In how long It takes the clle'lt to find employment. A two-way MOW. test Is conducted with a level of significance of 0.05. The null hypothesis for the simlficance of the officer Is rejected. and the null hypothais for the signifioanoe of the orientation scheme is retained. It oan he oonduded that: the officer assigned Is not a significant factor in lime taken to find employment the orientation scheme is not a significant factor in time taken to find employment the officer assigned Is a significant factor In time laken to find employment I - the orientation scheme Is a significant factor In time taken to find employment (Q-ID: MST.CP.TWAOV.DR.|III!EI‘) A halo-way ANWA test is conducted to Investigate the significance of employment status and marital status as famrs in willingnefi to donate to marines. The interaction of these two factors Is also tested. A level of significance ofOIIS Is used. The followhg output table is prepared: Source Sum nl squares DF Mean squares P—value m—n—-— ——n—oa— INTERACI'ION 3. 5519 3. 5519 Error 4.714.134 146 32.29 m—-—-- The null hypothesis that is rejected Is: marital status is not a significant factor in willingness to donate to charities employment sutus is not a significant factor In willingness to donate to charlfles there is no significant interaction between employment m5 and marital status as factors I - none of the null hypotheses are rejected [Q-ID: MST.CP.TWIOV.DR. Doll!” A two-way move test is oonduoted to investigate the significance of gender and body weight as factors in risk of diabetes (three different levels of body weight are defined). A level of significance ofoJJs is used. The followhg output table Is prepared: Source Sum ofsquares DF Mean squares P- value ——-—-- flM—.m E_—-—-- Select the correct statement from the followhg: I - It cannot be concluded that either gender or body weight are significant factors In risk of diabetes. It can be concluded that body weight Is a significant factor in risk of diabetes. but It cannot be concluded that gender is a significant factor In risk of diabetes. - It can be concluded that both gender and body weight are significant factors in risk of diabetes. It can be concluded that gender Is a slgr‘ificant factor in risk of diabetes, but It cannot be concluded that body weight is a significant factor In risk of diabetes. lQ=lD: M5T.CP.TWAD\I'.02.DD40H A pharmacist suwacta that a new drug. Drug it, wil Increase the amount oi vitamin c absorption in the human body. A Modway ANOVA test is conducted to test Drug X and the slgnlficance of gender as a factor. A level of slgnlfloanoe of 0.05 is used. The talentan output table Is prepared: Source Sum ofsquarcs DF Hean squares P- value m—-—-- m—I—m- _——EEI- WWI—_- —m-—-- By only considering the figures In this output l:aI:IIer It can be concluded that: gender is not a significant factor In Vitamin C absorption In Ute body Drug X WIII increase the Vitamin C absorption in the body = Drug X is a significant factor in Vitamin C absorption in the body I - all of the above statements can be concluded from the output table (o=to: "Steinem-roam A company Is trying to detennine whether the department that an employee works In and their position in the department affect the number of hours the employee spends at work each wean The folowing means plot was oonstmtted in a two-way ANOVA bat for the significance of these mo factors. Man This means plot suggms: the effect that employee posltlon wlthin a department has on time spent at work is dependent upon the department that the employee is in employees In Human Resources spend more time than employees In financial Services at work each work the position held by an employee does not affect the amount of time spent at work each week if - the department mat an employee works In does not affect the amount of tlme spent at work each week (Q=ID: MST.CP.TWAOV.02.DII600 A two-way ANOVA test is conducted to Investigate the significance of regular exercise and dietary Intake of salt as factors in heart rate. The following means plot was constructed from samples drawn: Slit ill-III The means plot anaemia: salt intake affects heart rate, but regular exercise does not regular exercise affects heart rate, but salt intake does not «I - both regular exercise and salt Intake significantly affect heart rate, but these factors do not Interact both regular exercise and salt Intake sIgnificantlv affect heart rate, and the effects of these factors do interact T 1 pf 3 m: MST.CP.AOV.DS.BUBU A statistic student is given an assignment question about one-way ANOVA, in which there is a test for the hypothesis that three population means (pl, 11;, H3) are all equal. The null hypothesis, that the means are all equal, Is rejected. The Bonfen'onl method Is used to determine which pairs of means differ. It is determined, with a level of slgnlhcanoe offl.05, that l-‘J. and p3 dilfer. It Is not determined that PI. and p2 differ, or that p; and p3 dilfer. The student looks at this, and has the following reaction: '1 don't understand. It seems as though the tat says that p; and #2 are the same, and that 112 andp; are the same. Then surely It follows that #1 and 113 are the same! How an the test conclude that they are ambient? In the field below, explain to the student why the tst may have the oohdusion that it does. OI'IN Feedback [1 out of 1] vou have answerEd this essay. MSG review the model Sfllfltiflfl MIDI”. Model-aletlon, -| Feedback [1 outol‘ 1] You have anlwered till: way. Please revlewthelnodel anhdan below. Mlullmnn When uellg the Banlerronl methad, the dl'fererloe bemeen pair: at populadan means I: Immigeted. Ether threat a hrpotheele teat er a onnllllenee lntervfl enllnatlan, It mayI or may not be sanctified that there Isa meme m “IE m means In BIB question. men: are we population means mum mt unto new. Speculum. using me Mao-II menu. In a compared in u; u: u compared to ”a, and u 2 Is compared to us It a ”thread that u, and u: filer. but I: Is not determined that the other palm of means Mr. This Is not nontradm, and EMS may well he a valid confluent alta- um the Beam-uni method. Thls Is manual! beam. Jug Illa In other human wand «motions. filing to new: the hymen: that twu means are equal duel not mean that that hrpathule In true. So. forenrnple. mneiclerthe hymen: that pl Ia equal to ”2. TM: hypothesis Is not rejected. Thle dds not mean that the tea means '2 In fact and, or that we can say that than ls allude-It euldenae ta aanthlde that tile means are equal. They may In fact nat be equal. 1t metehr has not been m that theyI are equal. In the language ufslatlstlal nferenoe: there was Inwfl'ldmt cumulus in the samples collected to conclude that the two Emulation means we not allual. and so this candusion about the uowlauun was not made.Thenmkmehrmempubendpgmuymeinpertantnetlm Ismattheaeommesdanatmnuadittdul‘att dlattherelneenougl Mummflllh Mthalevel afelonllaneeef 0.051 em .1. and u, an am. so. II galeral. UIEWIH which thbeurtofthhg can happen lethal: fllerelsslmw not enuufll aathiaalevldulcebueetafllilthat |I1dhiafrum Hz. arthat LI: dlflkehunuy 11m doe-m mull-edit: mammtnn laencelgh monument-a: that U1 dflEreli'nm lla» Tommiammufluly. mmmmmam bathe mammal could wmmm that the mantisaunrlng. It appears. from the males draw. that the first population has slnhtnr higher values than the second population. and thatthe second population Helm hlgher duel than the third population. In thle ease, the eamplee dawn may not plwlde all'l'lclult evlaenoe that the mean ofthe llm population ll In lhtt hlgher than the mean of the second non-lawn. And the same mar be true In moamn the second nowlatlan to the third. "was the samples drawn mailr amide all'l'ltlam m that the mam efthe first population Is Mlle-em: tothe m of the dilld pawlltbn. We!” W2 W3 - 20f? ID:MS. .ROV. .0010 n mnuenienee store manager. Thrifty ‘I1na. needs to dehennlne the quantitis Mm drinks that she should eraerfortne next month. She is hlmm Ill (lemming whether she would slmplv order equal amounts of Fantra. Saline. Cloak and Pempfl, orwtlether she should cider mule alone particular brand. Tina believes that the more grumble a salt d-I'nk company Is. the more popular that company's d-I'nk must be. Theretare, she wants to test whether the mean prallte ol' Fantl'a {up}, Spite {pg}. Cloak (It) and Pempel (up) are equal. ‘I1na has defined the null humans for a one-way analysis «variance (WWI) to be Hg: ”F n- u; - u: - up. 536d: the «“9“!th hymlhesis that Is most amp-late ”11183 W MW ANUUA: - Ha: not all U18 population means are equal Ha: none of the population means are eeual Ha: Dr ‘ vs ‘ Dc e Lu Ha: PF ’ lls 3‘ Pl: ’ HP [1 outof 1] Maren-rad. A reasonable albemeuve hypothesis must be the opposite offline nul. That Is. the null and the ate-name hwothaes nwst he mutually exam and mum. tithe null hypothesis Isoeflned as "0: p, - p5 - llc - upthen the aka-name hyped!“ mun he die appeeh «ma. whlth Is flnptr dlat not all Illa pagulaflan means In “III. — I 3 0‘ 3 ll]: MSTJIPADVDZJIIEIID a monitor-skis trainer has developed lbur tiflerent types of semlnar that he am provlde lor teaming genene oompuoer sklls. He would line to know III the type of seminar dellvered Is a Porter in how well his students learn the materlal. In arfier to find thls out. the tralneroeddes to (£an an MID“! best. He chooses four random samples ofoeapre. To eadl sample, he dulvers one ofthe Murdlllerent eemhar types. and he then adminlmers a test to determlne how well the students Ieamed the material. In the fir: sample. mm 49 “dents, the mean soore was 31 and the fiandard dewallon was 2.111 the second sample. with 4? students. the mean soore was 42 and the fiandard deviatlon was 5. 1n the thlrd sample, with 52 students, the mean some was 93 and the standard devlatlon was 3. In the fourth sample, with 51 students, the mean some was 72 and the standard deviation was 8. a) The trailer can not trust that the MU“ 0' my MWA testlnfi that he WOUEB are sow-ate mouse: - the allies ol' the Ibur samples drawn are not equal the means 0" the four mt oowlatlans 00 I1“ appear to be NUS the standard deviations of the Fair olllerent populations do not appear to be equal rs) Explaln why thIs Issue Is a prouem in ANDVA teotlng. oh no... —| Feedback [1. out of 2] e] Thrslsnot oer-rem The trelner unnnt m that the results r'lfarl'llI move “8an that he mndum are aomrate amuse Illa fl-Irlerrl Haul-alone alllle [our tiller-rt mum-do not lppur to be equal. Ir) You have answered tllls may. Maesemnawthemodelsohrtlonbelow. DIacIrsslon h) Explaln why thls Issue Is a prehlem In ARENA timing. lnh no... - Feedback [1 out of Z] I) Thlslsnot correct. The trailer annot mm that the results ofanly move oedng that he confirm are morale Intense lire standard dart-Hone shire [our Client population. do not app-rte be equal. Is) we have errand-ed thrs essay. Pleaserevlewthemooelaofutlonaelow. Hint”: . I} Because the largest sample slam deviatloo m is morethan twloe as large as the smallest sample standard deflation. there Is erloenoetmt the Downhill mu deviatlons are not equal I‘l thls shratlon. In ordaroo m tllat the results ornriwn are morale, there Is an assumption that the peprrlation sandard devlaflons ofthe dflerenl populatlona are equal. ll} Model eoluuon When moualng an move test. It 5mm lrthe semues from the afferent populations are not efeoual she. It Is also more fume oowlstlon meansou not appear to be equal. In fact. whether ornot the populadon mare equfl iswhat lsbealgmedln MWthlsndI amen-Nehru population Random dMatlonsdorrot appearllobeeqrrflJhlsls Mailman condemn; WA breathe. I: Is assumed that the merit mmlms lbrthe mmtm al have the some steward m. Ifflfls assumption Sm there Is newer-anbeethat the results “MD“! are aomraee. The reason Ibrthls assumption meson the teat sure“: usn‘l In move. Pan: otmrsm mtlme Is a numenul quantity that Is an estlnatle brunet Bel-amen to be a constant venande am all of the woulatlans hel'lg lulled. Thls quantity Is the pooled artlmabe afthe pourlatlon varllllcle. and II onlyr an aomrabe narrator fnr thls parameter If lt ran he assumed that al ofthe populatlun vaflanm (and hence all afthe population standard donations) are equal woos all mutilation: who tested. In Dram. tenet requlreo thatm aohrallr muamllnmsmstln adorwaeerflle woulaten va-Ianoesareal dual. The reason forthrsrsthat MAM“ can handle anal violauonsof theesaurnptlon ofeqralmdrddevladonsnmmmon m formemnrpdon lsshnphrodohedrdlatthelarw nmphnmm reoarded elitism twloeas lwuthemalaatmple standard oeulauon weed. In the fiuflfiofl presented it this Guam. the largest mm ml! Is In the fourth sample. and Is 8. The 5mm standard meow Is In the first smile. am! Is 2. Therdbrethe mample sandarddeidetlon lsrnole thmtrrrloethe smiles sample summon. The-efore thekunnot hegraranteed thatd'reresuhsofANOVAe-eetnrrahe. : 1 ol 3 ID: MST.CP.ROV.01.0010 The database engineers at Gurgle are interested in testing whether the average search times taken using the Gurgle search engine are dependent {In the type 0“ inter-net bI'OWSEf that is Used. 30 serum tlfl’les WEN med for seard'les carried wt on ead'r 0f 3 dmnt I — DNWSEI'S. The data have been summarized in the one-way nuova summary table shown here. LL 1" 8 8 Calculate the test statistic corresponding to the one-millr analysis of variance (ANOUA) summary table. Give your answer to 2 decimal places. Source Degrees at freedom Sum of squares Mean squares Treatment Random error [0 out of 1] Thls Is not correct. F = 1.50 alwbflw ._ 3 Feedback [0 out of 1] This IS not correct. F = 1.50 Calculation The value of F can be calculated using the following formula: hide variables MSG = mean square error among the groups = 0.0009 MSE = mean square error within the groups = 0.0006 F = test statistic corresponding to the one-way ANOVA = unknown _ MSG ‘ MSE = 0.0009 0.0005 1.50 j 2 of3 ID: MST.CP.AOV.04.0010 Select all the asmmptbns required to carry out a one-way analysis of variance (mom): the populations are identically distributed J the populations are Independent the populations are normally distributed J the data are randomlyr selected J the population variances are all equal the population means are all equal -' Feedback [1 out of 2] You are partly correct. flio data are randomly abound: you are correct. lilo populations are Independent: you are correct. the populations are normally Ill-hinted: this optlon should have been selected. the p...
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