Tutorials for MRCP candidates
Multiple Choice Questions

The suggested answer is A. Increase sample size

Type II error is defined as the probability of falsely accepting the null hypothesis. Conceptually, it means how likely to miss a significant difference when it is really there. Both type I and type II errors depend on the relationship between the effect size and the sample size. For any required effect size, Type II error decreases as sample size increases.

The suggested answer is E

Standard Deviation measures where the observations will lie. Standard Error estimates where the means of repeated sampling using the same sample size on a similar population will lie. 95% Confidence interval of the mean therefore represent where 95% of the mean values will lie in repeated samplings of a similar population using the same sample size.

The suggested answer is D.

The problem is that the Ethics Committee has a role in ensuring good ethics in research, and bad research is unethical research, as it wastes resources and everyone's time, as well as putting research subjects at discomfort and risk for no good reason. Ethics committee therefore do also have a role to ensure feasibility and validity of research. The suggested answer is therefore only correct if the question pertains to the most important role, or principle role, or something like that.

The suggested answer is C

A, D, and E control bias. B ensures the result represents effectiveness rather than efficacy. Type II error depends on the relationship between effect and Standard Error, and Standard Error decreases with increasing sample size. Type II error therefore decreases with increasing sample size.

The suggested answer is E

Obtaining informed consent recognises the participant's right to make an informed decision whether to participate or not. The best description is therefore autonomy, although obtaining informed consent also reduces risks of complaint and litigation.

The suggested answer is C, it is a phase III trial.

Phase I is the initial testing for toxicity and dosages in the human. Phase II is a single group trial to evaluate whether the drug is sufficiently efficacious to be worth a controlled trial. Phase III is the definitive controlled trial against an existing treatment or a placebo. Phase IV is sometime called the post marketing trial, and is mostly for evaluating rare side effects.

The suggested answer is B

Use of "intention to treat" means that all the cases recruited, regardless of whether they actually completed the intervention, will be included in the assigned group for analysis. The main reason is that it evaluates "effectiveness", or how it may work in real life situation, rather than "efficacy", how it works if the treatment is actually received in full. It therefore underestimates efficacy because the treatment group will contain some cases that did not complete the treatment.

The suggested answer is D, 20

The number needed to treat is the inverse of risk difference. The risk of the two groups are 0.1 (10%) and 0.05 (5%). The difference is risk is 0.05, and the NNT = 1/0.05 = 20.

Please note that the 50% reduction is Relative Risk, where RR = R1/R2 = 0.05/0.1 = 0.5 = 50%. Relative Risk is usually used in epidemiological studies to evaluate how one parameter may affect another. NNT are used in controlled trials of two groups, and is derived from the risk difference.

The suggested answer is B

Case controlled studies identify cases with an outcome of interest, then match this case with one or more cases similar in every respect except the causation of interest, then evaluate whether the incidence of the causation is different in those with or without the outcome. This is a strategy to examine uncommon outcomes

We are not sure, but D is the only one that is not incorrect

The results may or may not have demonstrated the usefulness of program A, but it failed to demonstrate that the results are different to that from program B, as this difference may arise by chance in more than 1 in 20 (p>0.05). Student's t test must be used with caution in small sample size, but the sample is never too large for the t test if the underlying distribution is normal, so the distribution of individual values is important.

Confidence Intervals is an alternative presentation of statistical results, and its appropriateness is not dependent on the outcome of a t test. All except D are not correct.

By the way, this is a very badly written question, as what the p value is is undefined (?price, protein concentration?).   The guess is that it refers to the probability of Type I Error, and should be stated as such.   At least it should be stated as the p value in testing the null hypothesis.  somebody ought to complain to the College.   

The suggested answer is E

The maintenance of hope is not by drug trials. Define tumour response rate is usually carried out in phase II studies. Measure survival rate is a particular research model, usually carried out in phase III studies. Establishing the maximum tolerated dose is a function of Phase I study. I am not sure about best schedule of administration, but I think this is in another form of study.

The suggested answer is A

NNT is inverse of risk difference. The lowest NNT is therefore the highest risk difference. Drug A is 10%-5%=5%, drug B is 2%-0.5%= 1.5%, drug C is 20%-16%=4%, drug D=15%-11%=4%, and drug E is 10%-6%= 4%. Drug A therefore has the highest difference in risk, therefore the lowest NNT.

The suggested answer is C

The A line is to the left of B at 50% survival, so it has a shorter median survival time. At 19 months the percentage of survival is better

The suggested answer is A

Sensitivity = true pos / (true pos + false neg) = 9/(9+1) = 1/10 = 0.9 = 90%
Specificity = true neg / (true neg + false Pos) = 91/(91+9) = 91/100 = 0.91 = 91%

The suggested answer is A

Post test probability is a function of pre test probability and Likelihood Ratio, as follows.

Pre test Odds = Pre test probability / (1-Pre test probability)
Post test Odds = Pre test Odds x Likelihood Ratio
Post test probability = Post test Odds / (1 + Post test Odds)

The suggested answer is E

There are two methods to work this out
Method 1 : Reorganisation of the formulae

Prevalence is the proportion of the population that is outcome positive
Sensitivity is the proportion of those outcome positive that are true positives
True positive TP = Sensitivity x Prevalence

Those that are outcome negative are those not outcome positive
Outcome negative = (1-Prevalence)
Specificity is the proportion of those outcome negative that are true negatives
True negative TN = Specificity (1-Prevalence)


False Positive is the proportion of those outcome negative that are not true negatives
False Positive FP = (1-Specificity)(1-Prevalence)


PPV = TP / (TP + FN) = Sen x Prev / (Sen x Prev + (1-Spec)(1-Prev))
= 0.95 x 0.01 / (0.95 x 0.01 + 0.2 x 0.99)
= 0.0095 / (0.0095 + 0.198) = 0.0095 / 0.2075 = 0.046 or 4.6%

Method 2 : uses the known formulae associated with Bayesian probability

Likelihood Ratio for positive test = Sensitivity / (1-Specificity) = 0.95 / 0.2 = 4.74
Pre-test Odd = Pre-Test Probability / (1 - Pre-Test Probability) = 0.01 / 0.99 = 0.0101
Post-test Odd = Pre-test Odds x LR = 0.0101 x 4.74 = 0.048
Post-Test Probability = Post-Test Odds / (1 + Post-test Odds) = 0.048 / 1.048 = 0.046 or 4.6%

The suggested answer is B

The same approach to the last question is used, the following is a truncated calculation.

Sensitivity = 100% = 1
Specificity = 95% = 0.95
Prevalence = 0.1% = 0.001
PPV = Sen x Prev / (Sen x Prev + (1-Spc)(1-Prev))
= 0.001 / (0.001 + 0.05 x 0.999)
= 0.001 / (0.001 + 0.04995) = 0.0196 or 2% the nearest

None of the multiple choice answers is correct, but D is the nearest

This is a question on prediction. At the technical level the givens are as follows

1. prevalence or pre-test probability is 5%, in other words True Positive (TP) + False Negative (FN) = 0.05
2. Test Positive is 10%, in other words TP + false Positive (FP) = 0.1
3. The proportion of true positives in those test positive is 3 times those with false negatives in those test negative, in other words TP / (TP+FP) = 3FN / (FN+TN)

Step 1. From the given, 5 simultaneous equations can be defined

1. TP + FN = 0.05; from the first given equation
2. FP + TN = 0.95; logically from 1
3. TP + FP = 0.1; from the second given equation
4. FN + TN = 0.9; logically from 3
5. TP / 0.1 = 3FN / 0.9; from the third given equation, 3, and 4

Step 2 progressive solving of the simultaneous equation

6. TP = 0.05 - FN; from 1
7. TP = 0.3 FN / 0.9 = FN/3; from 5
8. FN/3 = 0.05 - FN; from 6 and 7
9. FN/3 + FN = 0.05, 4FN/3=0.05, FN = 0.05 x 3 / 4 = 0.0375; from 8, in other words 3.75% of the population is false negative
10. TP = 0.0125; from 6, in other words, 1.25% of the population is true positive
11. FP = 0.0875; from 3 and 10; in other words 8.75% of the population is false positive
12. TN = 0.8625; from 2 and 11, in other words 86.25% of the population is true negative

Step 3. The question is the proportion of those with Alzheimer that are ApoE4 negative

13. Proportion of the population with Alzheimer = 0.05; from the first given equation
14. proportion of population that is test negative and with Alzheimer = 0.0375; from 9
15. Answer = 0.0375 / 0.05 = 0.75, in other words, 75%

The suggested answer is E. However this is correct only if E refers to Pearson's Correlation Coefficient. Although Pearson is most famous for this coefficient, he was the Professor of Statistics at Cambridge and had many things named after him. Somebody should ask the College to be more specific.

Student's paired t test requires paired measurements of the same thing, so cannot be done if drug concentration was not measured before administration. 

Chi Squares test usually concern proportions, so is not appropriate for measurements. 

Unpaired t test is for comparison of two groups which is absent in this model. 

Log regression (I assume this means logistic regression) is a multivariate modelling of a binary outcome which drug concentration is not.

 Pearson's correlation coefficient measures correlation between two measurements that are normally distributed, so would be appropriate in this case if one assumes that body weight and drug concentrations are both normally distributed.

This is not a statistics question

None of the answers are truly correct, but B is the nearest to it.  We suspect that there is an error in transcribing the question

SEM = SD/sqrt(n), so A and C are not correct.  In fact SEM decreases with sample size.   SEM=SD when n=1, and SEM=0 when n is infinity.
SD = sqrt(sum of squares / (n-1)). , therefore SD decreases with n. B is correct up to a point as the sums of squares also increases. After a large number of cases are collected however the SD usually stabalises..
The mean is central tendency and SD is the spread. They are different things, so D is not correct.
Student t assumes an underlying normal distribution, so it is a parametric test.

There is possibly a transcrption error here, as none of the answers seem correct. The combinations presented refer to True and False Positives and Negatives. A is true positive, B is false negative, C and E are true negative, and D false positive.

Specificity is the proportion of negative tests amongst those that are disease negative

There may have been multiple transcription errors, as neither the question nor answers make sense

p value as such remains undefined. By common usage, the p value when used statistically in a test of the null hypothesis is an informal short hand way of saying the probability of Type I error, Fisher's p, or the probability of error in rejecting the null hypothesis.

If I am correct in guessing what the examiner call p value, then the answer is the probability the result being null, the smaller the p the less likely the result to be null, the less likely the difference found arise out of chance, so the more likely a significant difference exists.

Therefore, its not so much whether the result is correct, but whether the null hypothesis is correct.

Whether the result is correct depends on the diligence with which observations are made and recorded, and has nothing to do with statistics.

When MRSC can find the right formula the correct answer will be posted. In the mean time, the following is our best try, but may not be correct

In any year, in a population of 100,000, and assuming that there is no death from any other causes, there are 15 cases that are new from the current year, and at least 7.5 cases from each of the previous 5 years, so the minimum number of cases must be 15 + 5 x 7.5 = 52.5. The prevalence must exceed this value, as the survival rate from all the years other than the earliest of the 5 must be in excess of 7.5. Base on this logic, 75 in E is the only possible answer.

The only worry I have with this answer is that it looks too much like a secondary school arithmetic exercise and not a postgraduate epidemiological problem, so I am not certain whether I am correct. However, I cannot think of any other way of calculating the prevalence, as the shape of the survival curve is not stated, and even assuming the survival time to be normally distributed there is no Standard Deviation available from which to calculate probabilities.

The p=0.13 is not accompanied by explanation, so it is difficult to know exactly what it is referred to. By common usage this refers to Type I error, and means that there is a 13% chance of error in rejecting the null hypothesis. In other words, there is a 13% chance that the difference observed was by chance and not real. In other words, there is no significant difference between the groups, so A and D are therefore incorrect

Cumulative incidence is the proportion of positive cases amongst the population of interest at any particular time. The cumulative incidence was 2.5% in 10 years, so it cannot be 5% per year. E is therefore incorrect

I am uncertain of the term relative risk reduction. At end point, the two incidences are roughly 2.2% and 2.5% The risk reduction is about 0.3%, relative to the higher incidence 0.3/2.5 is roughly 0.12, so although I am not sure, I do not think B is correct

By 10 years, both groups have cumulative incidences of over 2% so less than 98% are symptom free. So C is also not correct.

I do not understand this question at all, unless there are transcription errors.