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

(1)

1

DEPARTMENT OF STATISTICS /DEPARTEMENT STATISTIEK SEMESTER TEST /SEMESTERTOETS 2

BME120

OCTOBER/OKTOBER 112011 MINUTES /MINUTE:120 MIN TOTAL /TOTAAL:51

EXAMINERS /EKSAMINATORE:Dr. Legesse Kassa Debusho & Mrs. Janet Van Niekerk EXTERNAL EXAMINER /EKSTERNE EKSAMINATOR:Ms. Judy Coetsee

INSTRUCTIONS /INSTRUKSIES: i. Answer all questions. Indicate only one alternative.

ii. Indicate your answers on side 1 of the computer form. iii. Hand in the question paper and the computer form.

iv. Complete your name and other information - both below and on the computer form. i. Beantwoord alle vrae. Kies slegs een alternatief.

ii. Gebruik kant 1 van die merkleesvorm om antwoorde aan te dui. iii. Handig die vraestel sowel as die merkleesvorm in.

iv. 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

(2)

2

Use the following information to answer Questions 1 to 2 Gebruik die volgende inligting om Vrae 1 tot 2 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. 72.3, 78.9, 82.6, 71.8, 86.1, 80.5, 72.0, 91.8, 77.3, 88.2

Assume that the height is normally distributed. Aanvaar dat die hoogtes normal verdeel is. Question 1 / Vraag 1 [2]

The standard error of the sample mean is:

Die standaardfout van die steekproefgemiddelde is:

A B C D E

0.499 0.707 2.235 7.067 15.793

*Note: Round your final answer to 3 decimal places/Rond jou finale antwoord af tot 3 desimale plekke.

Question 2 / Vraag 2 [2]

A 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.

A (76.053, 84.247) D (75.171, 85.129) B (75.769, 84.531)

E

None of the above Geen van die bogenoemde C (75.094, 85.206)

*Note: Round your final answer to 3 decimal places/Rond jou finale antwoord af tot 3 desimale plekke.

(3)

3

Use the following information to answer Questions 3 to 5 Gebruik die volgende inligting om Vrae 3 tot 5 te beantwoord

A cigarette manufacturer wishes to test the claim that the variance of the nicotine content of its cigarettes is 0.644. Nicotine content is measured in milligrams. Assume that nicotine content is normally distributed.

A sample of 20 cigarettes has a standard deviation of 1.00 milligram.

‘n Sigaretvervaardiger wil die bewering toets dat die variansie van die nikotieninhoud van sy sigarette 0.644 is. Nikotien word in milligram gemeet. Aanvaar dat die nikotieninhoud normaal verdeel is. ’n Steekproef van 20 sigarette het ‘n standaardafwyking van 1.00 milligram.

What will be the null and alternative hypotheses? Wat sal die nul en alternatiewe hipoteses wees?

A ₀₆₄₄ 2 ₀₆₄₄ 1 2 0 : . H : . H σ > vs. σ = B ₀₆₄₄ 2 ₀₆₄₄ 1 2 0 : . H : . H σ = vs. σ ≠ C ₀₆₄₄ 2 ₀₆₄₄ 1 2 0 : . H : . H σ < vs. σ > D ₀₆₄₄ 2 ₀₆₄₄ 1 2 0 : . H : . H σ = vs. σ > E ₀₆₄₄ 2 ₀₆₄₄ 1 2 0 : . H : . H σ = vs. σ < Question 4 / Vraag 4 [2] The calculated test statistic, χ2₀is, Die berekende toetsstatistiek, χ2₀is,

A B C D E

12.236 12.880 29.503 31.056 None of the above Geen van die bogenoemde

*Note: Round your final answer to 3 decimal places/ Rond jou finale antwoord af tot 3 desimale plekke.

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:

A 0.001 < p-value < 0.005 0.001 < p-waarde < 0.005 D 0.05 < p-value < 0.10 0.05 < p-waarde < 0.10 B p-value < 0.01 p-waarde < 0.01

E None of the above _{Geen van die bogenoemde} C 0.025 < p-value < 0.05

(4)

4 Question 6 / Vraag 6 [2]

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) Computer science majors (Group 2) 36 1 = n 6 83 1 . x = 36 2 = n 2 79 2 . x = 3 4 1= . σ σ₁ =3.8

A 90% confidence interval for the true difference in means, i.e. µ₁−µ₂ , is: ‘n 90% vertrouensinterval vir die ware verskil in gemiddelde, i.e. µ₁−µ₂ , is:

A 36 8 3 36 3 4 645 1 4 4 2 2 ) . ( ) . ( . . ± × + D 36 8 3 36 3 4 671 1 4 4 2 2 ) . ( ) . ( . . ± × + B 36 8 3 36 3 4 96 1 4 4 2 2 ) . ( ) . ( . . ± × + E 36 8 3 36 3 4 000 2 4 4 2 2 ) . ( ) . ( . . ± × + C 36 8 3 36 3 4 96 1 4 4. ± . × . + .

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: A 46 87 0 13 0. )( . ) ( D 46 87 0 13 0. )( . ) ( B (0.13)(0.87) 46 1 ×

E _{None of the above}

Geen van die bogenoemde C 46 87 0 13 0. )( . ) (

(5)

A 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,

A 46 87 0 13 0 645 1 13 0. ± . × ( . )( . ) D 46 87 0 13 0 575 2 13 0. ± . × ( . )( . ) B . . (0.13)(0.87) 46 1 645 1 13 0 ± × × _E 46 87 0 13 0 575 2 13 0. ± . ×( . )( . ) C 46 87 0 13 0 96 1 13 0. ± . × ( . )( . ) Question 9 / Vraag 9 [2]

Consider the following situation. H₀:p=0.2 is being test against H₁:p>0.2. The test statistic value, i.e. z₀, was found to be -1.19. The p-value of the test is,

Veronderstel die volgende situasie. H₀:p=0.2word getoets teenH₁:p>0.2. Die waarde van die toetsstatistiek, i.e. z₀, is -1.19. Die p-waarde van die toets is,

A B C D E p-value < 0.05 p-waarde < 0.05 0.117 0.10 < p-value < 0.15 0.10 < p-waarde < 0.15 0.234 0.883 Question 10 / Vraag 10 [2]

A study is being conducted to determine whether the age of the customer is related to the type of movie he or she rents. A 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: Age Ouderdom Type of movie Documentary Dokumentêr Comedy Komedie Mystery Misterie 12 – 20 21 – 29 30 – 38 39 – 47 48 and over 14 15 9 7 6 9 14 21 22 38 8 9 39 17 12

(6)

6

The expected frequency (E₄₂) of the customers in age group 39 – 47 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 (E₄₂) van die kliënt in ouerdomsgroep 39 – 47 wat ’n komedie uitneem onder die nulhipotese dat die tipe fliek wat die kliënt uitneem onafhanklik is van sy of haar ouderdom is, A B C D E 13.43 16.47 19.93 24.27 29.90 Question 11 / Vraag 11 [1]

Suppose we wish to test the hypothesis H₀:µ=2 vs. H₁:µ≠2. We find a two sided p-value of 0.03 and a 95% confidence interval for the population mean µ of (1.5, 4.0). These two results are: Veronderstel ons wil die hipotese H₀:µ=2 vs. H₁:µ≠2 toets. Ons bereken ‘n tweekantige p-waarde van 0.03 en ‘n 95% vertrouensinterval vir die populasiegemiddelde µ: (1.5, 4.0). Hierdie twee resultate is:

A 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.

C not compatible because a two-sided p-value is less than 0.05 and the 95% confidence interval for µ does not include 0.

nie ooreenstemmend nie want die tweekantige p-waarde is minder as 0.05 en die 95% vertrouensinterval vir µ sluit 0 uit.

D not compatible because a two-sided p-value is less than 0.05, so we reject H₀ however the 95% confidence interval for µ does include 2.

nie ooreenstemmend nie want die tweekantige p-waarde is minder as 0.05 so ons verwerp 0

H alhoewel die 95% vertrouensinterval vir µ 2 insluit.

E compatible because a two-sided p-value is less than 0.05, so we reject H₀ and the 95% confidence interval for µ does include 2.

ooreenstemmend want die tweekantige p-waarde is minder as 0.05, so ons verwerp H₀ en die 95% vertrouensinterval vir µ sluit 2 in.

(7)

7

An experiment was designed to compare two varieties of spring barley. Twenty plots were used, ten being randomly allocated to variety A 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 Eksperiment was ontwerp om twee varïeteite van spring barley te vergelyk. Twintig lotte is gebruik, tien was ewekansig aan varïeteit A 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 A (Group 1) Variety B (Group 2)

x 4.08 4.70

2

s 0.344 0.210

n 10 9

Assume that the plots are independent random samples and the yields from both varieties are normally distributed.

Aanvaar dat die lotte onafhanklike ewekansige steekproewe is en die opbrengste van albei varïeteite is normaal verdeel.

The value of the calculated test statistic to test the hypothesis H₀:σ₁2 =σ2₂ versus H₁:σ₁2 ≠=σ2₂ (where σ₁2 and σ2₂ are the population variances of varieties A and B, respectively) is:

Die waarde van die berekende toetsstatistiek om die hipotese H₀:σ₁2 =σ2₂ teen H₁:σ₁2 ≠=σ2₂ (waar σ₁2 en σ2₂ die populasievariansies van varïeteit A en B onderskeidelik is) te toets, is:

A B C D E

The critical or rejection region at a 5% level of significance is, Die kritieke- of verwerpingsgebied teen ‘n 5% betekenispeil is,

A >3.23 or/of < 0.31 D >3.23 or/of < 0.24 B >4.10 or/of < 0.24

E >4.10 or/of < 0.31 C >1.96 or/of < -1.96

(8)

Suppose that _X ~_N(_µ,_σ2)_{where both}_µ_and_σ2_{are unknown. A sample of size}_n₌₃₅_{was taken} and it was found that x =6.21 and s=1.84. A 99% confidence interval for

µ

(the population mean) is given by:

Veronderstel dat _X ~_N(_µ,_σ2)_{en dat beide}_µ_en_σ2_{onbekend is. ŉ Steekproef van grootte} 35

=

n is geneem en daar is gevind dat x=6.21 en s=1.84. ŉ 99% vertrouensinterval vir

µ

(die populasiegemiddelde) is: A 35 84 . 1 457 . 2 21 . 6 ± × _D 6.21±2.33×1.84 B 6.21±2.457×1.84 E 35 84 . 1 33 . 2 21 . 6 ± × C 35 84 . 1 576 . 2 21 . 6 ± × 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, ...

A B C D E

Get smaller / Kleiner word

Get larger / Groter word

Stay the same / Dieselfde bly

Get smaller or get larger / Groter of kleiner word

Be undefined / Nie gedefiniëer wees nie

The degree of clinical agreement among physicians on the presence or absence of generalized lymphadenopathy was assessed in 32 randomly selected participants from a prospective study of men with Acquired Immunodeficiency Syndrome (AIDS) or an AIDS-related condition (ARC). 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 32 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:

(9)

9

Number of prominent lymph nodes /Hoeveelheid prominente limfkliere

Doctor A Doctor B Difference

Mean 7.91 5.16 2.75

SD 4.35 3.93 2.83

n 32 32 32

The aim of the study was to determine whether there is a systematic difference between the assessments of Doctor A versus Doctor B.

Die doel van die studie was om vas te stel of daar ‘n sistematiese verskil tussen die ondersoeke van Dokter A en Dokter B is.

The null and alternative hypotheses are, Die nul en alternatiewe hipoteses is,

A H₀:µ_d >0 vs. H₁:µ_d =0 B H₀:µ_d =0 vs. H₁:µ_d >0 C H₀:µ_d <0 vs. H₁:µ_d =0 D H₀:µ_d =0 vs. H₁:µ_d ≠0 E H₀:µ_d ≤0 vs. H₁:µ_d >0 Question 17 / Vraag 17 [2]

The value of the calculated test statistic is: Die waarde van die berekende toetsstatistiek is:

A B C D E

1.036 2.654 5.497 10.691 13.377

Note: Round your final answer to 3 decimal places / Nota: Rond jou finale antwoord af tot 3 desimale plekke

An appropriate conclusion to the above test, given that the p-value < 0.001, and assuming a non-directional instead of a non-directional alternative, is:

‘n Gepaste gevolgtrekking tot die bogenoemde toets, gegee dat die p-waarde < 0.001 is en ‘n nie rigting gewende alternatief in plaas van ‘n rigting gewende alternatief word aanvaar, is:

A There is sufficient evidence (1% level of significance) to indicate that there is no systematic difference between the assessments of Doctor A versus Doctor B.

Daar is bevredigende bewyse (1% betekenispeil) om aan te dui dat daar geen sistematiese verskil tussen die ondersoeke van Dokter A teenoor Dokter B is nie.

B There is insufficient evidence which indicates that there is systematic difference between the assessments of Doctor A versus Doctor B.

Daar is nie bevredigende bewyse wat daarop dui dat daar ‘n verskil tussen die ondersoeke van Dokter A teenoor Dokter B is nie.

C The evidence from the samples indicates that the systematic difference between the assessments of Doctor A versus Doctor B is highly significant.

Die bewyse van die steekproewe dui aan dat die sistematiese verskil tussen die ondersoeke van Dokter A teenoor Dokter B hoogs betekenisvol is.

(10)

10

D There is no sufficient evidence to make comparison between the assessments of Doctor A and Doctor B.

Daar is geen bevredigende bewyse om die ondesoeke van Dokter A teenoor Dokter B te vergelyk nie.

E None of the above.

Geenvan die bogenoemde.

Given a p -value < 0.001, this result indicates that...

Gegee ‘n p-waarde < 0.001, dit kan gebeur dat ons ‘n … maak. A A type I error can be made with probability not

exceeding 0.001 / ‘n Tipe I fout kan gemaak word met ‘n waarskynlikheid van nie meer as 0.001.

B A type II error can be made with probability not exceeding 0.001 / ‘n Tipe II fout kan gemaak word met ‘n waarskynlikheid van nie meer as 0.001.

C A type II error can be made with unknown probability/ ‘n Tipe II fout kan gemaak word met ‘n onbekende waarskynlikheid.

D A 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 20 and 21 Gebruik die volgende inligting om Vrae 20 en 21 te beantwoord

Plasma-glucose levels are used to determine the presence of diabetes. Suppose the mean plasma-glucose concentration in 35 to 44 year olds is 4.86 (in µg/mL unit) with standard deviation of 0.54

mL / g

µ . A 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 0.20 µg/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 µg/mLeenhede) met ‘n standaardafwyking van 0.54 µg/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 0.20 µg/mLeenhede. Question 20 / Vraag 20 [2]

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:

A B C D E

Note: Round your final answer to 4 decimal places / Nota: Rond jou finale antwoord af tot 4 desimale plekke

(11)

11 Question 21 / Vraag 21 [2]

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?

A B C D E

14 10 9 5 None of the above

Geen van die bogenoemde

Suppose there are two samples of size 12 and 15, with a rank sum of 220 in the sample of size 12. Assume there are no ties. The appropriate test statistic for the Wilcoxon rank sum test is,

Veronderstel daar is twee steekproewe van groottes 12 en 15, met ‘n rang som van 220 in die steekproef van 12. Aanvaar daar is geen van dieselfde waarnemings nie. Die gepaste toetsstatistiek vir die Wilcoxon rang som toets is,

A B C D E

220 0 0.733 2.513 None of the above Geen van die bogenoemde

In a study, 28 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. After 6 months, periodontal status improved in 15 patients, declined in 8, and remained the same in 5.

In ‘n studie is 28 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:

A B C D E >18.85 or <9.15 >19.69 or <8.31 >18.35 or <9.65 >19.19 or <8.81

None of the above Geen van die bogenoemde

(12)

12

Use the following information to answer Questions 24 and 25 Gebruik die volgende inligting om Vrae 24 en 25 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.

Change score: +3 +2 +1 0 -1 -2 -3 Number of patients: 4 5 6 5 4 2 2

A 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:

A Sign test Teken toets

B Wilcoxon signed rank test Wilcoxon teken rang toets C Paired t-test

Gepaarde t-toets D Wilcoxon rank sum test

Wilcoxon rang som toets E None of the above.

Geenvan die bogenoemde.

The calculated test statistic value for the test in Question 24 is, Die berekende toetsstatistiekwaarde vir die toets in Vraag 24 is,

A B C D E

(13)

13 Question 26 / Vraag 26 [2]

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 2.

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 onafhank-likheid tussen die kwaliteit van die plantgroei en die tipe grond word in Tabel 2 gegee.

Growth

Soil type

Clay Sand Loam

Poor Average Good 16 31 18 8 16 36 14 21 25 Total 65 60 60 Table 1/Tabel 1. Growth Soil type

Clay Sand Loam

Poor Average Good 13.35 23.89 27.76 12.32 22.05 25.62 12.32 22.05 25.62 Total 65 60 60 Table 2/Tabel 2.

The χ2- test statistic for testing statistical independence between the quality of plant growth and soil types is:

Die χ2 - toetsstatistiek om statistiese onafhanklikheid te toets tussen die kwaliteit van plantgroei en die tipe grond is:

A B C D E

(14)

14 Question 27 / Vraag 27 [1]

If we know that F8,4,0.975 = 8.98 and F4,8,0.975 = 5.05 then what is the value of F4,8,0.025? Veronderstel dat F8,4,0.975 = 8.98 en F4,8,0.975 = 5.05, wat is die waarde van F4,8,0.025?

A B C D E

Find the upper 2.5th percentile of the χ2distribution with 10 degrees of freedom. Vind die boonste 2.5de persentiel van die χ2verdeling met 10 grade van vryheid.

A B C D E

What is the exact two-sided 95% confidence interval for the parameter of the binomial distribution if the sample mean is calculated to be 0.60 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 0.60 van ‘n steekproef van grootte 10?

A B C D E

(0.27;0.89) (0.11;0.76) (0.30;0.90) (0.35;0.85) None of the above Geen van die bogenoemde

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?

A B C D E

(3.45;15.76) (0.0257;0.1858) (2.57;18.58) (0.0345;0.1576) None of the above Geen van die bogenoemde