1

EP400: Additional Paper

Examination

Wednesday 22 June 2011: 10.00 am – 1.15 pm

Candidates are advised to spend the first FIFTEEN minutes of this exam reading the

question paper and planning their answers.

Candidates should answer ALL questions.

Use a SEPARATE answer book for each question and put a page number at the bottom of

each page used in the answer book.

A hand held calculator may be used when answering questions on this paper. The calculator

may be pre-programmed before the examination. The make and type of machine must be

stated clearly on the front cover of the answer book.

A formulae sheet and statistical tables are provided for use at the end of the paper.

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Question 1

Group B streptococcus (GBS) infection is a major bacterial cause of death among newborn

babies in Malawi. Most women with GBS do not have any symptoms, but if a pregnant

woman has GBS in her vagina or gastro-intestinal tract, then there is a risk that her baby will

acquire the bacteria and develop infection of the blood (septicaemia) at birth. Maternal GBS

infection has also been associated with premature labour and still birth.

A study was conducted to investigate the association of maternal HIV-infection with the

presence of maternal GBS infection in the labour ward of the Queen Elizabeth Central

Hospital in Blantyre, Malawi. Of 4,037 attendees over a 12 month period who were asked to

participate 1,838 were recruited into the study. The major reasons stated for non-recruitment

were that the woman was too sick or had an obstetric emergency requiring immediate care.

Participants had an HIV test, and presence of GBS was assessed by a novel rapid urinary

antigen test.

A total of 396 participants tested HIV positive and 1442 tested HIV negative. A positive

urine test for GBS was identified in 77 HIV positive women and 313 HIV negative women.

a) Do you think selection bias in this study could be a problem?

Give reasons for your answer. [15%]

b) Considering your answers to question a) provide two alternative approaches to

recruitment for this study and describe their strengths and weaknesses. [10%]

c) i) Calculate the odds ratio for the association between GBS prevalence and HIV

infection, showing your working. [2.5%]

ii) Provide a description of this result. [2.5%]

The urine test for detection of GBS was compared against the gold-standard (culture) in a

randomly selected subgroup of the study population.

Results by HIV status are as follows:

Comparison of rapid urinary test to bacterial culture broken down by HIV status of women.

HIV negative women:

Culture

Urine test Positive Negative

Positive 40 10

Negative 6 135

HIV positive women:

Culture

Urine test Positive Negative

Positive 21 6

Negative 2 70

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d) i) Calculate the sensitivity and specificity of the urine test against culture by HIV

status, showing your working. [8%]

ii) Comment on your findings, in terms of misclassification bias, and the implications

for the urinary test as a screening test for GBS. [5%]

e) i) Calculate the positive predictive value and negative predictive value of the urine test

against culture by HIV status, showing your working. [8%]

ii) Using the information that you have calculated, estimate the total number of GBS

carriers who would be positive for GBS by culture in the whole study population by

HIV status. State your assumption. [20%]

iii) Estimate the odds ratio for the association between GBS test positive on culture

testing and HIV infection. Show your calculation. [5%]

The investigators concluded that they did not find evidence of increased presence of GBS in

the HIV-infected mothers. They next examined the relationship of GBS prevalence with

maternal age. The table below shows the breakdown of GBS prevalence by age group.

GBS n (%)

Present

Maternal age

15-22 132 (19.4%)

23-27 131 (23.5%)

28+ 127 (21.1%)

Total 390 (21.2%)

f) State one reason why the investigators may have chosen these age groupings. [5%]

g) i) Previous studies of GBS prevalence and age using gold standard culture techniques

have shown higher prevalence at young maternal ages. Briefly describe the results

in the table above. [2%]

ii) Describe how misclassification in the urinary test for GBS might explain the

discrepancy between this study and other published work. [8%]

iii) Describe three other possible reasons for this discrepancy. [9%]

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Question 2

Stroke is a major cause of disability worldwide. A stroke occurs when blood flow to a part of

the brain is interrupted, causing brain cells to die. This results in permanent damage that is

often fatal. People who survive a stroke can experience major changes in their lives including

their ability to carry out day to day tasks following the event.

The risk of stroke increases with age. In a typical European high-income country, at ages 65-

84 years, the incidence of stroke (fatal and non-fatal) is around 600 per 100,000 per year, and

the population mortality rate from stroke is around 400 per 100,000 per year.

a) You have been asked to design a study to identify risk factors for stroke in a European

high-income country. Based entirely upon the information contained in the above

paragraph, discuss the main advantages and disadvantages of the following study designs

to do this, suggesting how each design might be optimised to deal with potential

disadvantages:

i) A case-control design in which cases are people aged 65-84 years who have had a

stroke. [20%]

ii) A prospective cohort design in which the outcome is having a fatal or non-fatal

stroke at ages 65-84 years. [15%]

The association of total cholesterol levels in the blood with mortality rate from stroke was

investigated by conducting a combined (pooled) analysis of individual-level data (i.e. data on

exposures and outcomes in individuals) from 61 prospective studies conducted mainly in

Europe and North America. Person years at risk were accumulated for study subjects between

the ages of 40 and 89 years. In total, the analyses were based on 892,000 men and women

among whom there were 11,663 deaths from stroke.

b) Describe the key variables that would be needed by researchers to conduct this analysis of

the association of stroke mortality with total cholesterol levels based on pooling of

individual-level data. [10%]

c) What sort of practical and/or organisational problems would the team have had to

overcome in order to do this successfully? [5%]

The authors used Cox regression to analyse the association between total cholesterol levels

and stroke mortality, adjusted for age, sex, and study and by baseline blood pressure. The

results of one of the analyses conducted are shown in the following Figure.

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Hazard ratios of stroke mortality (showing 95% confidence intervals) by total

cholesterol levels measured at entry to follow-up within categories of systolic blood

pressure (SBP)

Figure Notes : (i) The area of each square in the Figure is inversely proportional to the

variance of the corresponding log hazard ratio. (ii) Final hazard ratios are adjusted for age,

sex and study only.

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d) i) What do you think is the most likely reference category shown in the Figure and

why? [5%]

ii) Describe the results shown in the Figure. [10%]

iii) Briefly outline two additional statistical analyses (and any corresponding statistical

tests) that would help your interpretation. [15%]

e) The hazard ratios shown in the figure were adjusted for age and sex. Why were they

additionally adjusted for study? [10%]

f) You have shown this figure to a colleague who argued that to get the key message across

it would be best to estimate a single set of hazard ratios for the different levels of

cholesterol in a model that adjusted for SBP as well as age, sex and study. Would this be

appropriate? Justify your answer. [10%]

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Question 3

Adults who are overweight or obese (i.e. have body mass index over 25 kg/m2) are at higher

risk than adults with normal weight of developing many diseases. Face-to-face weight loss

interventions are often seen as time consuming and inconvenient. Listening to podcasts

(audio files for MP3 players or computers) could be an effective way to provide weight loss

information. To evaluate the effect of podcasts on weight loss in adults, researchers

conducted a study over a period of 12 weeks.

The figure below shows a participant recruitment and analysis flow diagram for the study.

a) What type of study was this? [3%]

b) Why is it important to report the information on i) exclusions and ii) loss to follow-up?

Give examples in the context of this trial. [14%]

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c) The authors reported that “participants were randomly assigned to receive a weight loss

podcast or to a control podcast about art history.”

(i) Explain what is meant by “randomly assigned”? [5%]

(ii) Why is randomisation important in trials? [5%]

(iii) Discuss the aim of randomization in this trial, and whether it is likely to have

achieved its aims. [5%]

d) How would you prevent the investigators from knowing in advance to which type of

podcast (i.e. intervention or control) each participant will be allocated? Please state why

this is important. [6%]

e) Participants received 2 podcasts per week for 12 weeks. The authors report that their

analysis was done using “intention-to-treat”.

(i) Explain what intention-to-treat means in the context of this study. [5%]

(ii) What are the advantages of intention-to-treat analyses? [5%]

(iii) Do you agree that their analysis was intention-to-treat? Give reasons for your

answer. [10%]

(iv) How would adherence be measured in this trial? [5%]

f) (i) Give three reasons why blinding is used in trials. [3%]

(ii) Discuss whether participants were blinded in this trial. [3%]

(iii) How could bias in outcome assessment (weight and BMI) be minimized in this

trial? [3%]

g) The investigators measured the change in weight and BMI between the start and end of

the study in the two treatment arms. The results are shown in the table overleaf. Describe

and interpret the results. [10%]

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Control group

(n = 37)

(mean±SD)

Intervention

group (n = 41)

(mean±SD)

P-value for difference

between groups

Weight (kg)

Baseline 89.0 ± 13.6 91.9 ± 15.0

12 weeks 88.7 ± 13.9 89.0 ± 13.6

Difference −0.3 ± 2.1 −2.9 ± 3.5 p<0.001

BMI (kg/m2)

Baseline 31.4 ± 4.1 31.8 ± 3.2

12 weeks 31.3 ± 4.3 30.8 ± 3.4

Difference −0.1 ± 0.7 −1.0 ± 1.2 p<0.001

h) The STATA output below shows an analysis of the results comparing mean BMI at 12

weeks follow-up.

(i) Interpret these results. [3%]

(ii) Discuss how these results differ from those shown in the table above which were

published by the authors. [3%]

(iii) Give a possible explanation for the difference. [5%]

i) The authors did not “pre-specify” an analysis plan in their research protocol. Comment on

why it is important to pre-specify an analysis plan using the data shown as an example.

[7%]

END OF QUESTIONS

(Formulae and statistical tables follow)

Two-sample t test with equal variances

------------------------------------------------------------------------------

BMI | Obs Mean Std. Err. Std. Dev. [95% Conf. Interval]

---------+--------------------------------------------------------------------

Control | 37 31.3 .7069156 4.3 29.86631 32.73369

Intervent| 41 30.8 .5309908 3.4 29.72683 31.87317

---------+--------------------------------------------------------------------

combined | 78 31.03718 .4343137 3.835755 30.17235 31.90201

---------+--------------------------------------------------------------------

diff | .5 .8735946 -1.239915 2.239915

------------------------------------------------------------------------------

diff = mean(x) - mean(y) t = 0.5723

Ho: diff = 0 degrees of freedom = 76

Ha: diff < 0 Ha: diff != 0 Ha: diff > 0

Pr(T < t) = 0.7156 Pr(|T| > |t|) = 0.5688 Pr(T > t) = 0.2844

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SUMMARY OF STATISTICAL FORMULAE

MSc/Postgraduate Diploma Epidemiology

June 2011 examinations

This summary sheet includes formulae from all EP modules and so will include formulae which

some students are not familiar with; students are only expected to be able to apply formulae

covered in modules they have studied. Please note however that more basic formulae are not

included here and students are expected to know these.

1) Single Sample:

a) Proportion, p ,

npSE

1, estimated as

n

pppSE

1, for confidence intervals

95% confidence interval for : pSEp 96.1

npSE 00 1 , for significance tests

Test hypothesis 0 : pSE

pz 0

b) Mean, ,x n

xSE

, estimated as n

sxSE

i) Large Sample

95% confidence interval for : xSEx 96.1

Test hypothesis 0 : xSE

xz 0

ii) Small Sample

95% confidence interval for : xSEtx 05.0,

where 1 n and 05.0,t is the 2-tailed 5% point of a t-

distribution with degrees of freedom (df)

Test hypothesis 0 : xSE

xt 0 , df 1 n

11

2) Two Independent Samples:

a) Difference in proportions, 21 pp (where 1

11

n

rp and

2

22

n

rp )

95% confidence interval for 21 : 2121 96.1 ppSEpp

where 21 ppSE estimated as:

2

22

1

11 11

n

pp

n

pp

Test hypothesis 21 : 21

21

ppSE

ppz

pooled

where 21 ppSEpooled estimated as:

21

111

nnpp

and the common proportion, 21

21

nn

rrp

A slightly more conservative test uses a continuity correction, where

21

2121

21 11

ppSE

nnppz

pooled

,

or analyse as a 22 contingency table (see 6 below)

b) Difference in means, 21 xx

i) Large Samples 2

2

2

1

2

121

nnxxSE

, estimated as

2

2

2

1

2

121

n

s

n

sxxSE

95% confidence interval for 21 : 2121 96.1 xxSExx

Test hypothesis 21 : 21

21

xxSE

xxz

12

ii) Small Samples (where 21 )

21 xxSE estimated as 21

11

nns

where

11

11

21

2

22

2

112

nn

snsns

95% confidence interval for 21 : 2105.0,21 xxSEtxx

where 221 nn and 05.0,t is the 2-tailed 5% point of

a t-distribution with degrees of freedom (df)

Test hypothesis 21 :

21

21

11

nns

xxt

, df 221 nn

3) Paired Samples

a. Difference in means, 21 xx

Take differences in paired values; analyse differences using formulae for single sample mean [1(b)].

b. Difference in proportions, 21 pp , N

srppSE

21

95% confidence interval for 21 : 2121 96.1 ppSEpp

Test hypothesis 21 :

sr

srX paired

2

21

, df = 1

where r and s are the number of discordant pairs, and N is the total number of pairs

4) r x c contingency table

Test hypothesis of no association :

E

EOX

2

2 , df = (r-1) ×(c-1)

Where: O = observed number in a cell

E = expected number in a cell under the null hypothesis

r = number of rows, c = number of columns

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5) 2 x c contingency table

Assign score to each column of table

Test hypothesis of no linear trend :

21

2

2

212

11

nns

xxX , df = 1

Where 1x = mean score for subjects in row 1 of table

2x = mean score for subjects in row 2 of table

1n = number of subjects in row 1 of table

2n = number of subjects in row 2 of table

s = standard deviation of scores combining subjects in rows 1 and 2

6) 2 x 2 contingency table

Test hypothesis of no association :

a b a+b dbcadcba

NbcadX

2

2

c d c+d df = 1

a+c b+d N

A slightly more conservative test uses a continuity correction,

dbcadcba

NNbcadX

2

21

2 , df = 1

7) Mantel-Haenszel χ2 test for several 2 x 2 tables :

Test hypothesis of no association :

i

ii

aV

aEaX

2

2 , df = 1

where

i

iiiii

n

cabaaE

and

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ii

iiiiiiiii

nn

dbcadcbaaV

Mantel-Haenszel Odds Ratio =

iii

iii

ncb

nda where the summation is over each of

the strata.

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8) Linear regression : xy

Equation of fitted line y = a + b x

95% confidence interval for : bSEtb 05.0,

where 2 n and 05.0,t is the 2-tailed 5% point of a t-distribution with degrees

of freedom (df)

Test hypothesis of no linear association: bSE

bt , df = 2n

Alternatively, the same test in terms of the correlation coefficient r: 21

2

r

nrt

d.f. = 2n .

9) Likelihood Ratio Test The likelihood ratio statistic (LRS) for testing for an association is calculated as: = ( − ) , where L1 is the log likelihood of the model with the exposure variable, and L0 is the log likelihood of the model without the exposure

variable. The LRS is then referred to the χ2 distribution, with the degrees of freedom equal

to the number of parameters that were excluded from the model.

10) Population attributable risk & population attributable risk fraction

r0 is risk (or rate) in unexposed group, r1 is risk (or rate) in exposed group;

r is risk (or rate) in total study population,

p is proportion of exposed in the population,

p1 is the proportion of exposed among cases

RR is risk ratio (rate ratio, odds ratio)

PAR = r – r0, or PAR = p(r1 – r0)

PAF = PAR/ r

So PAF = (r – r0)/ r or PAF = p(RR–1)/ [p(RR–1) + 1]

Also:

PAF = [p1 (RR – 1)] / RR. For matched case control studies, this formula is used with RR the

matched odds ratio. This formula is also used when adjusting for confounding, with RR the

adjusted rate ratio (or odds ratio for exposure in a case control study) obtained by

stratification or regression methods.

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11) Risk Ratio and Odds Ratio: Error Factor (EF) for use in calculation of 95% confidence

intervals:

Exposure Outcome

Yes No

Yes a b

No c d

95% confidence limits for the risk ratio, RR, in cohort or cross-sectional studies, are given by

(RR/EF) to (RRxEF) where EF is the error factor:

EF = exp (1.96 x dccbaa

1111 )

95% confidence limits for the odds ratio, OR, for cross-sectional or unmatched case control

studies, are given by (OR/EF) to (ORxEF) where EF is the error factor:

EF = exp (1.96 x dcba

1111 )

For 1:1 matched case control studies, the 95% confidence limits for the odds ratio are given

by (OR/EF) to (ORxEF), where OR is the matched odds ratio and

EF = exp (1.96 x sr

11 ), where r and s are the numbers of discordant pairs.

12) Rates and the Rate Ratio:

95% confidence limits for a rate is given by:

(R/EF) to( RxEF) where EF is the error factor:

EF = exp (1.96 x √ (1/e)), where e is the number of events observed.

95% confidence limits for the rate ratio RR are given by (RR/EF) to (RRxEF) where EF is the

error factor:

EF = exp (1.96 x

21

11ee

) where e1 and e2 are the number of events in the

exposed and unexposed groups.

13) Vaccine efficacy: When the two groups being compared are vaccinated and unvaccinated

individuals in a cohort study or randomized trial, vaccine efficacy is defined as:

100 x (1-RR), where RR is the ratio of the incidence rate in the

vaccinated group to the incidence rate in the unvaccinated group.

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Table A1 Areas in tail of the standard normal distribution.

Tabulated area: Proportion of the area of the standard normal distribution that is above z

Second decimal place of z

z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09

0.0 0.5000 0.4960 0.4920 0.4880 0.4840 0.4801 0.4761 0.4721 0.4681 0.4641 0.1 0.4602 0.4562 0.4522 0.4483 0.4443 0.4404 0.4364 0.4325 0.4286 0.4247 0.2 0.4207 0.4168 0.4129 0.4090 0.4052 0.4013 0.3974 0.3936 0.3897 0.3859 0.3 0.3821 0.3783 0.3745 0.3707 0.3669 0.3632 0.3594 0.3557 0.3520 0.3483 0.4 0.3446 0.3409 0.3372 0.3336 0.3300 0.3264 0.3228 0.3192 0.3156 0.3121 0.5 0.3085 0.3050 0.3015 0.2981 0.2946 0.2912 0.2877 0.2843 0.2810 0.2776 0.6 0.2743 0.2709 0.2676 0.2643 0.2611 0.2578 0.2546 0.2514 0.2483 0.2451 0.7 0.2420 0.2389 0.2358 0.2327 0.2296 0.2266 0.2236 0.2206 0.2177 0.2148 0.8 0.2119 0.2090 0.2061 0.2033 0.2005 0.1977 0.1949 0.1922 0.1894 0.1867 0.9 0.1841 0.1814 0.1788 0.1762 0.1736 0.1711 0.1685 0.1660 0.1635 0.1611 1.0 0.1587 0.1562 0.1539 0.1515 0.1492 0.1469 0.1446 0.1423 0.1401 0.1379 1.1 0.1357 0.1335 0.1314 0.1292 0.1271 0.1251 0.1230 0.1210 0.1190 0.1170 1.2 0.1151 0.1131 0.1112 0.1093 0.1075 0.1056 0.1038 0.1020 0.1003 0.0985 1.3 0.0968 0.0951 0.0934 0.0918 0.0901 0.0885 0.0869 0.0853 0.0838 0.0823 1.4 0.0808 0.0793 0.0778 0.0764 0.0749 0.0735 0.0721 0.0708 0.0694 0.0681 1.5 0.0668 0.0655 0.0643 0.0630 0.0618 0.0606 0.0594 0.0582 0.0571 0.0559 1.6 0.0548 0.0537 0.0526 0.0516 0.0505 0.0495 0.0485 0.0475 0.0465 0.0455 1.7 0.0446 0.0436 0.0427 0.0418 0.0409 0.0401 0.0392 0.0384 0.0375 0.0367 1.8 0.0359 0.0351 0.0344 0.0336 0.0329 0.0322 0.0314 0.0307 0.0301 0.0294 1.9 0.0287 0.0281 0.0274 0.0268 0.0262 0.0256 0.0250 0.0244 0.0239 0.0233 2.0 0.02275 0.02222 0.02169 0.02118 0.02068 0.02018 0.01970 0.01923 0.01876 0.01831 2.1 0.01786 0.01743 0.01700 0.01659 0.01618 0.01578 0.01539 0.01500 0.01463 0.01426 2.2 0.01390 0.01355 0.01321 0.01287 0.01255 0.01222 0.01191 0.01160 0.01130 0.01101 2.3 0.01072 0.01044 0.01017 0.00990 0.00964 0.00939 0.00914 0.00889 0.00866 0.00842 2.4 0.00820 0.00798 0.00776 0.00755 0.00734 0.00714 0.00695 0.00676 0.00657 0.00639 2.5 0.00621 0.00604 0.00587 0.00570 0.00554 0.00539 0.00523 0.00508 0.00494 0.00480 2.6 0.00466 0.00453 0.00440 0.00427 0.00415 0.00402 0.00391 0.00379 0.00368 0.00357 2.7 0.00347 0.00336 0.00326 0.00317 0.00307 0.00298 0.00289 0.00280 0.00272 0.00264 2.8 0.00256 0.00248 0.00240 0.00233 0.00226 0.00219 0.00212 0.00205 0.01999 0.00193 2.9 0.00187 0.00181 0.00175 0.00169 0.00164 0.00159 0.00154 0.00149 0.00144 0.00139 3.0 0.00135 0.00131 0.00126 0.00122 0.00118 0.00114 0.00111 0.00107 0.00104 0.00100 3.1 0.00097 0.00094 0.00090 0.00087 0.00084 0.00082 0.00079 0.00076 0.00074 0.00071 3.2 0.00069 0.00066 0.00064 0.00062 0.00060 0.00058 0.00056 0.00054 0.00052 0.00050 3.3 0.00048 0.00047 0.00045 0.00043 0.00042 0.00040 0.00039 0.00038 0.00036 0.00035 3.4 0.00034 0.00032 0.00031 0.00030 0.00029 0.00028 0.00027 0.00026 0.00025 0.00024 3.3 0.00023 0.00022 0.00022 0.00021 0.00020 0.00019 0.00019 0.00018 0.00017 0.00017 3.6 0.00016 0.00015 0.00015 0.00014 0.00014 0.00013 0.00013 0.00012 0.00012 0.00011 3.7 0.00011 0.00010 0.00010 0.00010 0.00009 0.00009 0.00008 0.00008 0.00008 0.00008 3.8 0.00007 0.00007 0.00007 0.00006 0.00006 0.00006 0.00006 0.00005 0.00005 0.00005 3.9 0.00005 0.00005 0.00004 0.00004 0.00004 0.00004 0.00004 0.00004 0.00003 0.00003

17

Table A2 Percentage points of the t distribution.

One-sided P value

0.25 0.1 0.05 0.025 0.01 0.005 0.0025 0.001 0.0005

Two-sided P value

d.f. 0.5 0.2 0.1 0.05 0.02 0.01 0.005 0.002 0.001

1 1.00 3.08 6.31 12.71 31.82 63.66 127.32 318.31 636.62 2 0.82 1.89 2.92 4.30 6.96 9.92 14.09 22.33 31.60 3 0.76 1.64 2.35 3.18 4.54 5.84 7.45 10.21 12.92 4 0.74 1.53 2.13 2.78 3.75 4.60 5.60 7.17 8.61 5 0.73 1.48 2.02 2.57 3.36 4.03 4.77 5.89 6.87 6 0.72 1.44 1.94 2.45 3.14 3.71 4.32 5.21 5.96 7 0.71 1.42 1.90 2.36 3.00 3.50 4,03 4.78 5.41 8 0.71 1.40 1.86 2.31 2.90 3.36 3.83 4.50 5.04 9 0.70 1.38 1.83 2.26 2.82 3.25 3.69 4.30 4.78 10 0.70 1.37 1.81 2.23 2.76 3.17 3.58 4.14 4.59 11 0.70 1.36 1.80 2.20 2.72 3.11 3.50 4.02 4.44 12 0.70 1.36 1.78 2.18 2.68 3.06 3.43 3.93 4.32 13 0.69 1.35 1.77 2.16 2.65 3.01 3.37 3.85 4.22 14 0.69 1.34 1.76 2.14 2.62 2.98 3.33 3.79 4.14 15 0.69 1.34 1.75 2.13 2.60 2.95 3.29 3.73 4.07 16 0.69 1.34 1.75 2.12 2.58 2.92 3.25 3.69 4.02 17 0.69 1.33 1.74 2.11 2.57 2.90 3.22 3.65 3.96 18 0.69 1.33 1.73 2.10 2.55 2.88 3.20 3.61 3.92 19 0.69 1.33 1.73 2.09 2.54 2.86 3.17 3.58 3.88 20 0.69 1.32 1.72 2.09 2.53 2.84 3.15 155 3.85 21 0.69 1.32 1.72 2.08 2.52 2.83 3.14 3.53 3.82 22 0.69 1.32 1.72 2.07 2.51 2.82 3.12 3.50 3.79 23 0.68 1.32 1.71 2.07 2.50 2.81 3.10 3.48 3.77 24 0.68 1.32 1.71 2.06 2.49 2.80 3.09 3.47 3.74 25 0.68 1.32 1.71 2.06 2.48 2.79 3.08 3.45 3.72 26 0.68 1.32 1.71 2.06 2.48 2.78 3.07 3.44 3.71 27 0.68 1.31 1.70 2.05 2.47 2.77 3.06 3.42 3.69 28 0.68 1.31 1.70 2.05 2.47 2.76 3.05 3.41 3.67 29 0.68 1.31 1.70 2.04 2.46 2.76 3.04 3.40 3.66 30 0.68 1.31 1.70 2.04 2.46 2.75 3.03 3.38 3.65 40 0.68 1.30 1.68 2.02 2.42 2.70 2.97 3.31 3.55 60 0.68 1.30 1.67 2.00 2.39 2.66 2.92 3.23 3.46 120 0.68 1.29 1.66 1.98 2.36 2.62 2.86 3.16 3.37

0.67 1.28 1.65 1.96 2.33 2.58 2.81 3.09 3.29

18

Table A3 Percentage points of the 2 distribution.

In the comparison of two proportions (2 × 2 2 or Mantel–Haenszel 2 test) or in the assessment of a trend, the percentage points give a two-sided test. A one-sided test may be obtained by halving the P values. (Concepts of one- and two-sidedness do not apply to larger degrees of freedom, as these relate to tests of multiple comparisons.)

P value

d.f. 0.5 0.25 0.1 0.05 0.025 0.01 0.005 0.001

1 0.45 1.32 2.71 3.84 5.02 6.63 7.88 10.83 2 1.39 2.77 4.61 5.99 7.38 9.21 10.60 13.82 3 2.37 4.11 6.25 7.81 9.35 11.34 12.84 16.27 4 3.36 5.39 7.78 9.49 11.14 13.28 14.86 18.47 5 4.35 6.63 9.24 11.07 12.83 15.09 16.75 20.52 6 5.35 7.84 10.64 12.59 14.45 16.81 18.55 22.46 7 6.35 9.04 12.02 14.07 16.01 18.48 20.28 24.32 8 7.34 10.22 13.36 15.51 17.53 20.09 21.96 26.13 9 8.34 11.39 14.68 16.92 19.02 21.67 23.59 27.88 10 9.34 12.55 15.99 18.31 20.48 23.21 25.19 29.59 11 10.34 13.70 17.28 19.68 21.92 24.73 26.76 31.26 12 11.34 14.85 18.55 21.03 23.34 26.22 28.30 32.91 13 12.34 15.98 19.81 22.36 24.74 27.69 29.82 34.53 14 13.34 17.12 21.06 23.68 26.12 29.14 31.32 36.12 15 14.34 18.25 22.31 25.00 27.49 30.58 32.80 37.70 16 15.34 19.37 23.54 26.30 28.85 32.00 34.27 39.25 17 16.34 20.49 24.77 27.59 30.19 33.41 35.72 40.79 18 17.34 21.60 25.99 28.87 31.53 34.81 37.16 42.31 19 18.34 22.72 27.20 30.14 32.85 36.19 38.58 43.82 20 19.34 23.83 28.41 31.41 34.17 37.57 40.00 45.32 21 20.34 24.93 29.62 32.67 35.48 38.93 41.40 46.80 22 21.34 26.04 30.81 33.92 36.78 40.29 42.80 48.27 23 22.34 27.14 32.01 35.17 38.08 41.64 44.18 49.73 24 23.34 28.24 33.20 36.42 39.36 42.98 45.56 51.18 25 24.34 29.34 34.38 37.65 40.65 44.31 46.93 52.62 26 25.34 30.43 35.56 38.89 41.92 45.64 48.29 54.05 27 26.34 31.53 36.74 40.11 43.19 46.96 49.64 55.48 28 27.34 32.62 37.92 41.34 44.46 48.28 50.99 56.89 29 28.34 33.71 39.09 42.56 45.72 49.59 52.34 58.30 30 29.34 34.80 40.26 43.77 46.98 50.89 53.67 59.70 40 39.34 45.62 51.81 55.76 59.34 63.69 66.77 73.40 50 49.33 56.33 63.17 67.50 71.42 76.15 79.49 86.66 60 59.33 66.98 74.40 79.08 83.30 88.38 91.95 99.61 70 69.33 77.58 85.53 90.53 95.02 100.43 104.22 112.32 80 79.33 88.13 96.58 101.88 106.63 112.33 116.32 124.84 90 89.33 98.65 107.57 113.15 118.14 124.12 128.30 137.21 100 99.33 109.14 118.50 124.34 129.56 135.81 140.17 149.45

19

E400: Additional Paper

Examiner’s Report

Question 1

a) This question was answered well in general though many only got two of the three

areas expected. These were a) the recognition that this was a hospital based study and

therefore poorly representative of all deliveries b) that the participation rate was very

low potentially leading to bias and c) that this low rate was particularly associated

with emergencies and illness in the mother – which might well indicate an association

with HIV or GBS.

b) This question was specifically asking for approaches to recruitment – not alternative

study designs. It was expected that these approaches would be at antenatal clinic or in

the community. Both having the disadvantage of more difficult logistics and follow

up but the advantage of better generalisability. Some candidates suggested

recruitment after delivery but in that situation the GBS status of the mother is likely to

have changed.

c) The odds ratio is calculated in the usual way – those who got this wrong mostly got

the ratio the wrong way round. The description of what it means is straightforward.

d) i) Sensitivity and specificity are calculated in the usual way. Some students got this

wrong by using the formulae for predictive value here. Most got this section correct.

ii) The examiner here was looking for the candidate to comment on the level of

sensitivity and specificity in the two groups and to say whether misclassification

would be differential or non-differential. Many students did not pay regard to the

question about screening and failed to say that the sensitivity was not as high as one

would hope for in a screening test.

e) i) This was well answered by those who knew the formula – some students failed to

do the calculation for negative as well as positive predictive value – hence missing out

on half the marks.

iii) This question was answered very poorly by virtually all students. What would

have helped them all was if they had drawn a 2x2 table for each of the HIV

positives and negatives and filled in or worked out each cell.

20

Culture

Urine test Positive Negative Total

Positive PPV x row total Row total minus

true positives

Total test positives

(given)

Negative Row total minus

true negatives

NPV x row total Column total minus

positive row total

(above)

Total Total for answer Total (given)

Since predictive values for the sample are used the key assumption is that the

prevalence in the sample is the same as in the whole population.

iii) This odds ratio requires the numbers from ii) above and is then straightforward.

Using the 2 x 2 table:

HIV

GBS culture Positive Negative

Positive

Negative

f) Why did the investigators choose these groupings – most likely to get the same

denominator in each group which maximises power. Some candidates suggested

because other studies had used these groupings to allow comparison – this answer

also gained marks.

g) i) This was a simple description of the numbers.

ii) The question here specifically asked how misclassification could lead to this. Many

people gave other reasons why it might happen but gained no marks since it was

misclassification that was asked. The key issue is if the test performs differently at

different ages.

iii) This is where other reasons were asked for – chance is clearly one possibility but

also differences in age distribution and co-morbidity between populations. Also the

possibility of misclassification of age by the women was a possibility but not

suggested by many candidates.

21

Question 2

This question concerned risk factors for stroke. Students were asked to summarise the

advantages and disadvantages of case-control and cohort studies to investigate risk

factors for stroke, show understanding of the requirements to conduct a pooled analysis

of individual data from 61 prospective studies, interpret a figure of the association of

cholesterol stratified by systolic blood pressure, and describe additional analyses and

interpretation

Part a All students were able to identify some of the advantages and disadvantages of case

control and cohort studies in general. The question specifically stated that the answers should

be based on the information provided on fatal and non fatal stroke rates. A full answer would

have explained how these advantages or disadvantages specifically related to a study of

stroke and further the advantages and disadvantages with respect to fatal or non fatal stroke.

For the E400 additional paper students must provide much fuller answers that in an

introductory study module such as EP101.

For example, an answer that an advantage of a case-control study is that it is quicker to carry

out for a rare disease compared to a cohort study would have got higher marks if the answer

had been illustrated with the numbers given in the descriptive information, i.e. annual stroke

incidence (fatal and non fatal) of 0.6% and 0.4% for fatal stroke. Conversely this information

could be used as a disadvantage of cohort studies with respect to the large sample size

required for adequate power and, dependent on the age of cohort participants at baseline, the

need for long follow-up as stroke is very rare at young ages. Information bias (recall or

observer bias) is a potential problem for case-control studies; a full answer requires that

students consider how these biases might arise with respect to (i) fatal strokes and (ii) non

fatal strokes. For fatal strokes there may be information bias from using proxy informants or

medical records while for non fatal strokes recall error is likely to be higher in stroke patients

because of the effect of the stroke itself causing brain cells to die (this information was

provided in the descriptive text and therefore no specialist medical knowledge about a stroke

was needed). An answer just stating that reverse causality can be a problem in case control

studies without any explanation of why that could be a problem for stroke would only get

partial marks. While most students identified that reverse causality was not a problem in a

prospective cohort few students stated that the reason was because the exposure was

measured before the outcome.

A few students incorrectly stated that problems of selection bias could be dealt with by a

matched case control study.

Part b Nearly all students gained some marks by stating the key information to be included

was death from stroke, person years of follow-up, and cholesterol level. Higher marks were

given for answers that gave the specific information required for a person years analysis at

the individual level i.e. individual dates of entry and exit plus date of death and cause of

death i.e. whether stroke or not. A few students confused the analysis with a meta analysis of

22

trials based only on a literature review so received no marks as the question clearly stated that

the analysis was a pooling of individual level data.

Part c Most students correctly answered that the quality of the data was important but did

not describe or explain the important considerations when data are pooled from 61 different

centres e.g. variations in measurement of cholesterol, reliability of ascertainment of deaths,

in particular stroke deaths.

Part d (i). Most students identified the reference group as those with a SBP <145 mmHg but

many did not identify that the reference group was both a SBP <145 mmHg AND a

cholesterol of 4.5 mmol/L. Very few students gave a reason as to why these levels were

considered the reference group.

Part d (ii). This section was answered well. Most students correctly described an increasing

risk of fatal stroke with increasing categories of relationship with blood pressure, and many

also observed that this relationship varied with the level of cholesterol. An answer obtaining

high marks included a full description of the results stratified by cholesterol and a discussion

of the 95% CI and the precision of the results

Part d (iii). Most students mentioned interaction and/or trend but few students described

what the interaction was (blood pressure with cholesterol) or what trend they would be testing

(i.e. trend of mortality rate with cholesterol within categories of SBP (or conversely, trend

with SBP within categories of cholesterol). Only a few students clearly described how the

tests would be carried out.

Part e This question had the poorest answers and many students got no marks. Possibly,

students thought it was more difficult than it was. The answer required was about adjusting

for other study centre characteristics which were not measured but might confound the

association between stroke mortality and cholesterol or blood pressure. Some students

discussed age and sex adjustment but received no marks since the question asked about

centre adjustment.

Part f Many students understood that it would not be appropriate to summarise the data in

the Figure and gave the reason why.

23

Question 3

This question concerned an intervention study in overweight and obese individuals, which

evaluated over a 12 week period the benefit of using a podcast to provide information about

weight loss. To answer the question well required an understanding of the principles of

randomized trials. Some parts of the question asked for an example to be given in the context

of the weight loss trial, some people lost marks because they did not provide an example (this

highlights the importance of carefully reading the question to understand what is being asked

for).

Part a) asked for the type of study design, this was a Randomised Controlled Trial.

Part b) asked why it is important to provide information about exclusions and loss to follow

up in the report of a trial. Inclusion and exclusion criteria define the study population, this

information should be reported because it is important to know to what population the results

of this study can be generalized, and for comparison with results from other studies. Students

who discussed generalisability or external validity got high marks here. Loss-to-follow

impacts on the overall quality of the study. If the loss-to-follow up is different in the two

groups, this can be a source of bias (attrition bias). In fact in this trial the loss to follow up

was quite low (~7%).

Part c(i) asked for a definition of randomization. Most answers explained this well. It was

important to show that you understand what random means for example by explaining that

the allocation cannot be predicted, and that each participant has a known and usually equal

chance to be allocated to either group.

Part c(ii) asked why randomization is important. Randomisation is important because it

ensures there is no systematic bias in allocation of trial participants to treatment groups.

It is not correct to say that randomization ensures the treatment groups will be similar (e.g. in

a particular trial the groups may be unbalanced yet this does not mean that randomisation has

failed). Randomization ensures that any imbalance between the groups is due to chance and

not due to the influence of the investigator or the participant. This applies equally to

characteristics that we know are related to outcome, and to any other factors we do not know

about or have not measured that are also predictive of outcome.

Part c(iii) asked what the aim of randomization was in this trial, and whether this aim was

likely to have been achieved. The aim of randomization was to ensure that allocation of each

participant to the Podcast group or the control group would be unpredictable, and could not

be influenced by the investigator or the participant, so that any imbalance between the groups

would be purely due to chance.

Part d) asked how you would achieve concealment in this trial and why it is important to do

so. Concealment is important because if the investigator knows the next group allocation in

the randomization list, this could influence which person they choose to enrol next, and

thereby create selection bias. Concealment could be achieved in many ways, for example by

telephone randomization, the randomization list is kept secret at a separate location, when the

next person is enrolled a phone call is made to find out the allocation, and then the person is

given access to that podcast. Or, Podcasts could be stored on memory sticks that appear

24

identical, sealed so they could not be read prior to being assigned to next enrolled person, and

bearing an enrolment number. Some answers referred to blinding as a means of concealment,

blinding would be difficult in this type of trial (the group will be obvious to the participant if

they listen to the podcast). Concealment can always be achieved. The question asks for an

example of how to perform allocation concealment, and ideally put it in the context of this

trial which uses .mp3 podcasts.

Part e(i). Most answers gave a textbook description of Intention-To-Treat. However, for full

marks it was necessary to give an answer in the context of this trial, describing how

participants in the weight-loss podcast group and control group would be analysed in the

group they were randomized to, regardless of whether they actually listened to the podcast or

accidentally listened to the wrong podcast, and all subjects would be included regardless of

any protocol violation.

Part e(ii). Students were expected to point out that the main advantage of ITT analysis is that

it assesses effectiveness, as it incorporates the non-compliance and group switching that may

happen in practice. Although it would have been correct to state that a per-protocol analysis

would instead assess efficacy, this information was not expected for this question and did not

carry any marks.

Part e(iii). This part asked if the trial analysis was by intention to treat. This was not a pure

ITT because of the exclusion of seven participants who did not receive the intervention and

nine in the control group. However, some students mistakenly stated that this was a per-

protocol analysis. It is not, because in per-protocol the 4 (intervention group) and 3 (control)

participants who discontinued the intervention would have been excluded. Remember that if

you add extra information to your answers that is not required, you risk losing points if they

are wrong

Part e(iv) asked how could adherence have been measured in this trial. For full marks it was

necessary to explain what would need to be measures in the context of the podcast

intervention (did participants listen to the podcast, did they listen to all of it, how often etc)

and suggest a practical method. It was also important to also acknowledge the difficulty in

confirming full adherence, e.g. even if you were able to record the number of podcast

downloads it may be difficult to determine whether participants actually listened to them.

Part f (i). This question asked for three reasons why blinding is used in trials. Here, it was

important to give three distinct examples of types of bias that could occur in the absence of

adequate blinding. For example, if the participant knows which group they are in this may

influence their behaviour (in this trial they might be more likely to drop out in the

intervention group if they feel there are expectations they cannot meet that they should lose

weight); if the investigator knows, this may influence how participants are cared for

(performance bias) – for example if all participants are encouraged to exercise, this advice

could be given more forcefully to those in the intervention group; if the persons assessing

outcome know the randomised allocation, this may influence their assessment, for example,

taking more care to measure weight accurately if they know the person was in the

intervention group.

Part f(ii). This part of the question asked whether participants were blinded in this trial.To

get full marks, you would need to distinguish between blinding participants to the study

25

hypothesis (possible) as well as to the intervention (not possible - a participant will know

whether a podcast will be about losing weight).”

Part f(iii). The best way to reduce bias would be to blind the outcome assessor to the group

allocation and to the study hypothesis but this would be difficult to ensure in this type of trial.

It would be important to standardise weighting procedures and make sure these procedures

were adhered to.

Part g) asked you to describe and interpret the results provided. When describing the results,

it is useful to also give the key numbers in your answer (e.g. the changes in weight and BMI

of the groups) and then interpret the evidence. This section was answered well in general, but

many answers failed to express the strength of evidence (weak, strong or very strong

evidence against the null hypothesis). It is also important to comment on comparability of the

two groups at baseline. 78 overweight or obese individuals were included in the analysis, 41

in the intervention group, who had used the podcast, and 37 in the control group, mean

weight and mean BMI were similar at baseline. The average decrease in weight after 12

weeks was 2.9kg in the intervention group and 0.3kg in the controls. There was strong

evidence against the null hypothesis that the mean change in weight was the same in both

groups. The average decrease in BMI among participants in the intervention group was

1.0kg/m2 in the intervention group and 0.1kg/m

2 in the control group. The p-value indicated

strong evidence against the null hypothesis that the mean change in BMI was the same in

both groups.

Part h(i) asked for an interpretation of the Stata output. Most answered correctly that in this

analysis there was no evidence of a difference in mean BMI at 12 weeks between the groups

Part h(ii) asked how the results in the Stata output differ from those in the main table.

Different outcome measures were chosen, in the Stata analysis, mean BMI at 12 weeks is

compared, without adjustment for baseline BMI, whereas in the Table, the change in weight

and the change in BMI from baseline is compared.

Part h(iii) asked why the two approaches lead to different answers. There was also slight

imbalance in mean BMI at baseline, adjusting for baseline BMI makes some correction for

this.

Part i. It is important to specify the outcomes and analysis methods in an analysis plan, to

guard against the possibility of choosing, after analysis, the outcome and analysis that gives

the most favourable result, for example, selectively reporting the result for change in BMI

from baseline.