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.
5
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
12
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
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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.