Complete your name and other information - both below and on the computer form.

Transcription

1 DPRTMNT OF STTISTIS / DPRTMNT STTISTIK SMSTR TST / SMSTRTOTS BM 10 OTOBR / OKTOBR MINUTS / MINUT: 10 MIN TOTL / TOTL: 51 XMINRS / KSMINTOR: Dr. Legesse Kassa Debusho & Mrs. Janet Van Niekerk XTRNL XMINR / KSTRN KSMINTOR: Ms. Judy oetsee INSTRUTIONS / INSTRUKSIS: i. nswer all questions. Indicate only one alternative. ii. iii. iv. Indicate your answers on side 1 of the computer form. Hand in the question paper and the computer form. omplete your name and other information - both below and on the computer form. i. Beantwoord alle vrae. Kies slegs een alternatief. ii. iii. iv. Gebruik kant 1 van die merkleesvorm om antwoorde aan te dui. Handig die vraestel sowel as die merkleesvorm in. Vul jou naam en ander inligting in - op die vraestel sowel as die merkleesvorm. Surname and Initials / Van en Voorletters Student number / Studentenommer Signature / Handtekening 1

2 Use the following information to answer Questions 1 to Gebruik die volgende inligting om Vrae 1 tot te beantwoord The following data represent the heights (cm) of 10 plants taken at random from a plot. Die volgende data verteenwoordig die hoogte (cm) van plante wat ewekansig gekies is uit n lot. 7.3, 78.9, 8.6, 71.8, 86.1, 85, 7.0, 91.8, 77.3, 88. ssume that the height is normally distributed. anvaar dat die hoogtes normal verdeel is. Question 1 / Vraag 1 [] The standard error of the sample mean is: Die standaardfout van die steekproefgemiddelde is: *Note: Round your final answer to 3 decimal places/rond jou finale antwoord af tot 3 desimale plekke. Question / Vraag [] 95% confidence interval for the mean height of all the plants in the plot is: n 95% vertrouensinterval vir die gemiddelde hoogte van al die plante in die lot. (76.053, 84.47) D (75.171, 85.19) B (75.769, ) None of the above (75.094, 85.06) *Note: Round your final answer to 3 decimal places/rond jou finale antwoord af tot 3 desimale plekke.

3 Use the following information to answer Questions 3 to 5 Gebruik die volgende inligting om Vrae 3 tot 5 te beantwoord cigarette manufacturer wishes to test the claim that the variance of the nicotine content of its cigarettes is 644. Nicotine content is measured in milligrams. ssume that nicotine content is normally distributed. sample of 0 cigarettes has a standard deviation of 1.00 milligram. n Sigaretvervaardiger wil die bewering toets dat die variansie van die nikotieninhoud van sy sigarette 644 is. Nikotien word in milligram gemeet. anvaar dat die nikotieninhoud normaal verdeel is. n Steekproef van 0 sigarette het n standaardafwyking van 1.00 milligram. Question 3 / Vraag 3 [] What will be the null and vs. alternative hypotheses? Wat sal die nul en alternatiewe hipoteses wees? vs. H 0 : σ > 644 H1 : σ = 644 B H0 : σ = 644 H1 : σ 644 H 0 : σ < 644 H1 : σ > 644 D H 0 : σ = 644 H1 : σ > 644 H 0 : σ = 644 H1 : σ < 644 Question 4 / Vraag 4 [] 0 The calculated test statistic, χ is, Die berekende toetsstatistiek, χ0 is, None of the above *Note: Round your final answer to 3 decimal places/ Rond jou finale antwoord af tot 3 desimale plekke. Question 5 / Vraag 5 [1] If it is claimed that the variance has decreased then the p-value of the test is: Indien dit beweer word dat die variansie verminder het, dan is die p-waarde van die toets: B 001 < p-value < < p-waarde < 005 p-value < 01 p-waarde < < p-value < < p-waarde < 05 D 05 < p-value < < p-waarde < 10 None of the above 3

4 Question 6 / Vraag 6 [] Two groups of students are given a problem solving test. The results from independent samples are presented with the respective population variances in the following table: Twee groepe studente doen n probleemoplossingstoets. Die resultate van die onafhanklike steekproewe en die onderskeidelike populasievariansies word in die volgende tabel gegee: Mathematics majors (Group 1) omputer science majors (Group ) n 1 = 36 x = n = 36 x = 79. σ = 4 3 σ = % confidence interval for the true difference in means, i.e. µ 1 µ, is: n 90% vertrouensinterval vir die ware verskil in gemiddelde, i.e. µ 1 µ, is: B ( 4. 3) ( 3. 8) 4. 4 ± D ( 4. 3) ( 3. 8) 4. 4 ± ± ( 4. 3) ( 3. 8) 4. 4 ± ( 4. 3) ( 3. 8) 4. 4 ± Use the following information to answer Questions 7 to 9 Gebruik die volgende inligting om Vrae 7 tot 9 te beantwoord Suppose a clinical trial is conducted to test the efficacy of a new drug for treating gonorrhoea in females. Forty-six patients are given a 4-g daily dose of the drug and are seen 1 week later, at which 6 of the patients still have gonorrhoea. Veronderstel n kliniese eksperiment word uitgevoer om die doeltreffendheid van n nuwe medikasie wat gonorroea in vrouens behandel, te toets. Ses-en-veertig pasiente word daagliks n 4-g dosis van die medikasie gegee en is weer getoets 1 week later, waar 6 pasiente nogsteeds gonorrhoea het. Question 7 / Vraag 7 [1] The standard error of the point estimate for p (the probability of a failure with the drug) is: Die standaardfout van die puntberaming vir p (die waarskynlikheid van n mislukking van die medikasie) is: ( 0. 13)( D ( 0. 13)( 1 B ( 13)( ( 0. 13)( None of the above 4

5 Question 8 / Vraag 8 [] 99% confidence interval for p, the population proportion of patients with gonorrhea after 1 week on the treatment is, n 99% vetrouensinterval vir p, die populasie proporsie van pasiënte wat gonorrea na 1 week van behandeling het is, ( 13)( 13 ± D 1 B ± ( 13)( ( 13)( 13 ± ( 13)( 13 ±. 575 ( 13)( 13 ±. 575 Question 9 / Vraag 9 [] onsider the following situation. H : p = 0 is being test against H : p > 0. The test statistic value, i.e. z 0, was found to be The p-value of the test is, Veronderstel die volgende situasie. H : p = 0 word getoets teen H : p > 0. Die waarde van die 0. toetsstatistiek, i.e. z 0, is Die p-waarde van die toets is, 1. p-value < < p-value < p-waarde < < p-waarde < 15 Question 10 / Vraag 10 [] study is being conducted to determine whether the age of the customer is related to the type of movie he or she rents. sample of renters gives the observed frequencies shown in a table below: n Studie word uirgevoer om te bepaal of die ouderdom van die kliënt verwant is aan die tipe fliek wat hy of sy uitneem. n Steekproef van kliënte gee die waargenome frekwensies soos in die tabel hieronder: Type of movie ge Documentary omedy Mystery Ouderdom Dokumentêr Komedie Misterie and over

6 The expected frequency ( 4 ) of the customers in age group and who rent comedies under the null hypothesis that the type of movie a customer rents independent of his or her age is, Die verwagte frekwensie ( 4 ) van die kliënt in ouerdomsgroep wat n komedie uitneem onder die nulhipotese dat die tipe fliek wat die kliënt uitneem onafhanklik is van sy of haar ouderdom is, Question 11 / Vraag 11 [1] Suppose we wish to test the hypothesis H 0 : µ = vs. H 1 : µ. We find a two sided p-value of 03 and a 95% confidence interval for the population mean µ of (1.5, 4.0). These two results are: Veronderstel ons wil die hipotese H 0 : µ = vs. H 1 : µ toets. Ons bereken n tweekantige p- waarde van 03 en n 95% vertrouensinterval vir die populasiegemiddelde µ : (1.5, 4.0). Hierdie twee resultate is: not compatible because the 95% confidence interval for µ excludes zero. nie ooreenstemmend nie want die 95% vertrouensinterval vir µ sluit 0 uit. B compatible because the test used a 5% level of significance and the confidence interval is based on 95% confidence. ooreenstemmend want die toets gebruik n 5% betekenispeil en die vetrouensinterval is gebaseer op 95% vetroue. not compatible because a two-sided p-value is less than 05 and the 95% confidence interval for µ does not include nie ooreenstemmend nie want die tweekantige p-waarde is minder as 05 en die 95% vertrouensinterval vir µ sluit 0 uit. D not compatible because a two-sided p-value is less than 05, so we reject H 0 however the 95% confidence interval for µ does include. nie ooreenstemmend nie want die tweekantige p-waarde is minder as 05 so ons verwerp H0 alhoewel die 95% vertrouensinterval vir µ insluit. compatible because a two-sided p-value is less than 05, so we reject H 0 and the 95% confidence interval for µ does include. ooreenstemmend want die tweekantige p-waarde is minder as 05, so ons verwerp H0 en die 95% vertrouensinterval vir µ sluit in. 6

7 Use the following information to answer Questions 1 to 13 Gebruik die volgende inligting om Vrae 1 tot 13 te beantwoord n experiment was designed to compare two varieties of spring barley. Twenty plots were used, ten being randomly allocated to variety and ten to variety B. Unfortunately one plot of variety B was destroyed. The accompanying table summarizes the yields (t/ha) from the remaining plots. n ksperiment was ontwerp om twee varïeteite van spring barley te vergelyk. Twintig lotte is gebruik, tien was ewekansig aan varïeteit toegeken en tien aan varïeteit B. Ongelukkig is een lot van varïeteit B vernietig. Die bygaande tabel is n opsomming van die opbrengste (t/ha) van die oorblywende lotte. Variety (Group 1) Variety B (Group ) x s n 10 9 ssume that the plots are independent random samples and the yields from both varieties are normally distributed. anvaar dat die lotte onafhanklike ewekansige steekproewe is en die opbrengste van albei varïeteite is normaal verdeel. Question 1 / Vraag 1 [] The value of the calculated test statistic to test the hypothesis (where σ 1 and 0 : σ1 = σ H versus H σ are the population variances of varieties and B, respectively) is: Die waarde van die berekende toetsstatistiek om die hipotese (waar σ 1 en 0 : σ1 = σ H teen H 1 : σ1 = σ 1 : σ1 = σ σ die populasievariansies van varïeteit en B onderskeidelik is) te toets, is: None of the above Question 13 / Vraag 13 [] The critical or rejection region at a 5% level of significance is, Die kritieke- of verwerpingsgebied teen n 5% betekenispeil is, >3.3 or/of < 31 D >3.3 or/of < 4 B >4.10 or/of < 4 >4.10 or/of < 31 >1.96 or/of <

8 Question 14 / Vraag 14 [] Suppose that X ~ N( µ, σ ) where both µ and σ are unknown. sample of size n = 35 was taken and it was found that x = 6. 1 and s = % confidence interval for µ (the population mean) is given by: Veronderstel dat X ~ N( µ, σ ) en dat beide µ en σ onbekend is. ŉ Steekproef van grootte n = 35 is geneem en daar is gevind dat x = 6. 1 en s = ŉ 99% vertrouensinterval vir µ (die populasiegemiddelde) is: B ± D 6.1± ± ± ± Question 15 / Vraag 15 [1] In Question 14, when a 95% confidence interval is constructed instead of a 99% confidence interval (with all other quantities being the same) the maximum error of the estimate i.e., the width of the confidence interval, will... In Vraag 14, indien ŉ 95% vertrouensinterval in plaas van ŉ 99% vertrouensinterval bereken word (terwyl alle ander waardes onveranderd bly) sal die maksimum fout van die beraming, oftewel die wydte van die vertrouensinterval,... Get smaller / Get larger / Stay the same / Kleiner word Groter word Dieselfde bly Get smaller or get larger / Groter of kleiner word Be undefined / Nie gedefiniëer wees nie Use the following information to answer Questions 16 to 19 Gebruik die volgende inligting om Vrae 16 tot 19 te beantwoord The degree of clinical agreement among physicians on the presence or absence of generalized lymphadenopathy was assessed in 3 randomly selected participants from a prospective study of men with cquired Immunodeficiency Syndrome (IDS) or an IDS-related condition (R). The total number of prominent lymph nodes was assessed by each of three physicians. The summary statistics from two of the three physicians are presented in the following table: Die vlak van kliniese ooreenstemming tussen dokters oor die aanwesigheid of afwesigheid van algemene limf-adenopatie was geassesseer in 3 ewekansig gekose deelnemers van n beplande studie van mans met Verworwe Immuniteitsgebroke Sindroom (VIGS) of n VIGS-verwante toestand (VVT). Die totale hoeveelheid prominente limfkliere was deur elk van drie dokters ondersoek. Die opsommende statistieke van twee van die drie dokters word in die volgende tabel gegee: 8

9 Number of prominent lymph nodes /Hoeveelheid prominente limfkliere Doctor Doctor B Difference Mean SD n The aim of the study was to determine whether there is a systematic difference between the assessments of Doctor versus Doctor B. Die doel van die studie was om vas te stel of daar n sistematiese verskil tussen die ondersoeke van Dokter en Dokter B is. Question 16 / Vraag 16 [1] The null and alternative hypotheses are, Die nul en alternatiewe hipoteses is, vs. H 0 : µ d > 0 H1 : µ d = 0 B H 0 : µ d = 0 H1 : µ d > 0 H 0 : µ d < 0 H1 : µ d = 0 D H 0 : µ d = 0 H1 : µ d 0 H 0 : µ d 0 H1 : µ d > 0 Question 17 / Vraag 17 [] The value of the calculated test statistic is: Die waarde van die berekende toetsstatistiek is: Note: Round your final answer to 3 decimal places / Nota: Rond jou finale antwoord af tot 3 desimale plekke Question 18 / Vraag 18 [] n appropriate conclusion to the above test, given that the p-value < 001, and assuming a nondirectional instead of a directional alternative, is: n Gepaste gevolgtrekking tot die toets, gegee dat die p-waarde < 001 is en n nie rigting gewende alternatief in plaas van n rigting gewende alternatief word aanvaar, is: B There is sufficient evidence (1% level of significance) to indicate that there is no systematic difference between the assessments of Doctor versus Doctor B. Daar is bevredigende bewyse (1% betekenispeil) om aan te dui dat daar geen sistematiese verskil tussen die ondersoeke van Dokter teenoor Dokter B is nie. There is insufficient evidence which indicates that there is systematic difference between the assessments of Doctor versus Doctor B. Daar is nie bevredigende bewyse wat daarop dui dat daar n verskil tussen die ondersoeke van Dokter teenoor Dokter B is nie. The evidence from the samples indicates that the systematic difference between the assessments of Doctor versus Doctor B is highly significant. Die bewyse van die steekproewe dui aan dat die sistematiese verskil tussen die ondersoeke van Dokter teenoor Dokter B hoogs betekenisvol is. 9

10 D There is no sufficient evidence to make comparison between the assessments of Doctor and Doctor B. Daar is geen bevredigende bewyse om die ondesoeke van Dokter teenoor Dokter B te vergelyk nie. None of the above. Geenvan die. Question 19 / Vraag 19 [1] Given a p - value < 001, this result indicates that... Gegee n p-waarde < 001, dit kan gebeur dat ons n maak. B D type I error can be made with probability not exceeding 001 / n Tipe I fout kan gemaak word met n waarskynlikheid van nie meer as 001. type II error can be made with probability not exceeding 001 / n Tipe II fout kan gemaak word met n waarskynlikheid van nie meer as 001. type II error can be made with unknown probability/ n Tipe II fout kan gemaak word met n onbekende waarskynlikheid. type I error can be made with unknown probability/ n Tipe I fout kan gemaak word met n onbekende waarskynlikheid. Use the following information to answer Questions 0 and 1 Gebruik die volgende inligting om Vrae 0 en 1 te beantwoord Plasma-glucose levels are used to determine the presence of diabetes. Suppose the mean plasmaglucose concentration in 35 to 44 year olds is 4.86 (in µ g / ml unit) with standard deviation of 54 µ g / ml. study of 100 sedentary people in this age group is planned to test whether they have a higher or lower level of plasma-glucose than the general population. The expected difference in the mean plasma-glucose concentration is 0 µ g / ml units. Plasma-glukosevlakke word gebruik om die aanwesighed van diabetes vas te stel. Veronderstel die gemiddelde plasma-glukose konsentrasie in 35 tot 44 jariges is 4.86 (in µ g / ml eenhede) met n standaardafwyking van 54 µ g / ml. n Studie van 100 passiewe mense in hierdie ouderdomsgroep word beplan om te toets of hulle n hoër of laer plasma-glukosevlak het as die algemene populasie. Die verwagte verskil in die gemiddelde plasma-glukose konsentrasie is 0 µ g / ml eenhede. Question 0 / Vraag 0 [] The power of such a study if a two sided test is to be used with 5% significance level is: Die onderskeidingsvermoë van die toets as dit n tweekantige toets is met 5% betekenispeil, is: None of the above Note: Round your final answer to 4 decimal places / Nota: Rond jou finale antwoord af tot 4 desimale plekke 10

11 Question 1 / Vraag 1 [] How many people would need to be studied to achieve 80% power? Hoeveel mense word in die studie benodig om n onderskeidingsvermoë van 80% te bereik? None of the above Question / Vraag [] Suppose there are two samples of size 1 and 15, with a rank sum of 0 in the sample of size 1. ssume there are no ties. The appropriate test statistic for the Wilcoxon rank sum test is, Veronderstel daar is twee steekproewe van groottes 1 en 15, met n rang som van 0 in die steekproef van 1. anvaar daar is geen van dieselfde waarnemings nie. Die gepaste toetsstatistiek vir die Wilcoxon rang som toets is, None of the above Question 3 / Vraag 3 [] In a study, 8 adults with mild periodontal disease are assessed before and after 6 months of implementation of a dental-education program intended to promote better oral hygiene. fter 6 months, periodontal status improved in 15 patients, declined in 8, and remained the same in 5. In n studie is 8 volwassenes met peridontiese siekte geassesseer voor en 6 maande na die implementering van tandheelkunde-opvoedingsprogram met die doel om beter peridontiese mondhigiëne te bevorder. Na 6 maande het 15 pasiënte se peridontiese toestand verbeter, 8 pasiënte se toestand het versleg en 5 pasiënte se toestand het dieselfde gebly. The critical or reject region of the sign test at a 5% level of significance is: Die kritieke of verwerpingsgebied van die teken toets met n 5% betekenispeil is: >18.85 or <9.15 >19.69 or <8.31 >18.35 or <9.65 >19.19 or <8.81 None of the above 11

12 Use the following information to answer Questions 4 and 5 Gebruik die volgende inligting om Vrae 4 en 5 te beantwoord Suppose patients are graded on the degree of change in periodontal status on a 7-point scale, with +3 indicating the greatest improvement, 0 indicating no change, -3 indicating the greatest decline. The data are given in the table below. Veronderstel pasiënte word gegradeer op die verandering in hul peridontiese toestand op n 7-punt skaal, met +3 wat die grootste verbetering aandui, 0 dui geen verandering aan en -3 dui die grootste verswakking aan. Die data word in die volgende tabel gegee. hange score: Number of patients: Question 4 / Vraag 4 [] nonparametric test that can be used to determine whether a significant change in periodontal status has occurred or not over time is: n Nie-parametriese toets wat gebruik kan word om te bepaal of daar n betekenisvolle verskil in peridontiese toestand oor tyd plaasgevind het is: B D Sign test Teken toets Wilcoxon signed rank test Wilcoxon teken rang toets Paired t-test Gepaarde t-toets Wilcoxon rank sum test Wilcoxon rang som toets None of the above. Geenvan die. Question 5 / Vraag 5 [1] The calculated test statistic value for the test in Question 4 is, Die berekende toetsstatistiekwaarde vir die toets in Vraag 4 is,

13 Question 6 / Vraag 6 [] To study the effect of the type of soil on the amount of growth attained by a new hybrid plant, seedlings were planted in three different types of soil and their subsequent amounts of growth classified into three categories. The observed results are reported in Table 1. The expected frequencies under statistical independence between the quality of plant growth and soil types are also given in Table. Om die effek van die tipe grond op die hoeveelheid groei van n nuwe hybrid plant te ondersoek, is saailinge in drie verskillende tipes grond geplant en hul groei is in drie kategorieë geklassifiseer. Die waargenome resultate word in Tabel 1 gegee. Die verwagte frekwensies onder statistiese onafhanklikheid tussen die kwaliteit van die plantgroei en die tipe grond word in Tabel gegee. Soil type Growth lay Sand Loam Poor verage Good Total Table 1/Tabel 1. Soil type Growth lay Sand Loam Poor verage Good Total Table /Tabel. The χ - test statistic for testing statistical independence between the quality of plant growth and soil types is: Die χ - toetsstatistiek om statistiese onafhanklikheid te toets tussen die kwaliteit van plantgroei en die tipe grond is: None of the above 13

14 Question 7 / Vraag 7 [1] If we know that F 8,4,975 = 8.98 and F 4,8,975 = 5.05 then what is the value of F 4,8,05? Veronderstel dat F 8,4,975 = 8.98 en F 4,8,975 = 5.05, wat is die waarde van F 4,8,05? None of the above Question 8 / Vraag 8 [1] Find the upper.5 th percentile of the χ distribution with 10 degrees of freedom. Vind die boonste.5 de persentiel van die χ verdeling met 10 grade van vryheid None of the above Question 9 / Vraag 9 [] What is the exact two-sided 95% confidence interval for the parameter of the binomial distribution if the sample mean is calculated to be 60 from a sample of size 10? Wat is die eksakte tweekantige 95% vertouensinterval vir die parameter van die binomiaalverdeling indien die steekproefgemiddelde bereken is as 60 van n steekproef van grootte 10? (7;89) (11;76) (30;90) (35;85) None of the above Question 30 / Vraag 30 [] What is the exact two-sided 99% confidence interval for the parameter (λ) of the poisson distribution if the amount of events over 100 minutes was observed as 8? Wat is die eksakte tweekantige 99% vertrouensinterval vir die parameter (λ) van die poisson verdeling indien die hoeveelheid gebeurtenisse oor 100 minute waargeneem is as 8? (3.45;15.76) (057;1858) (.57;18.58) (0345;1576) None of the above 14