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Question 1
Correct
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A 26-year-old doctor has recently been diagnosed with lung cancer. He would like to find out his survival time for the condition.
Which statistical method is used to predict survival rate?Your Answer: Kaplan-Meier estimator
Explanation:The Weibull distribution are used to describe various types of observed failures of the components. it is used in reliability and survival analysis.
Regression Analysis is used to measure the relationship between among two or more variable. It determines the effect of independent variables on the dependent variables.
Student t-test is one of the most commonly used method to test the hypothesis. It determines the significant difference between the means of two different groups.
A time series is a collection of observations of well-defined data obtained at regular interval of time.
Kaplan-Meier estimator is used to estimate the survival function from lifetime data. It can be derived from maximum likelihood estimation of hazard function. It is most likely used to measure the fraction of patient’s life for a certain amount of time after treatment.
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This question is part of the following fields:
- Statistical Methods
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Question 2
Correct
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During the analysis phase, which of the provided options serves to control confounding factors?
Your Answer: Stratification
Explanation:During analytical stage a technique called stratification is used for controlling confounding variables. This technique involves sorting out the data into discernible groups.
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This question is part of the following fields:
- Statistical Methods
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Question 3
Correct
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Provided below is an abstract of a study conducted recently.
A consensus was developed among international experts. A total of 27 experts were invited. 91% of them decided to show up. A systematic review was performed. This comprised of open ended questions and the participants were encouraged to provide suggestions by e-mail. In the second phase google forms were used. Participants were asked to rate survey items on a scale of 5 points. Items that were rated critical by no less than 80% of the experts were included. Items that were rendered important by 65-79% of experts were inducted in the next survey for re rating. Items that were rated below 65% were rejected.
Which of the following methods was used in the study from which the abstract has been taken?Your Answer: The Delphi method
Explanation:The process used in the study is Delphi method. This method kicks off with an open ended questionnaire and uses its responses as a survey instrument for the next round in which each of the participants is asked to rate the items that the investigators have summarized on the basis of the data collected in the first round.
Any disagreement is further discussed in phases to come on the basis of information obtained from previous phases.
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This question is part of the following fields:
- Statistical Methods
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Question 4
Correct
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Among the following which one is not a criterion for the assessment of causality?
Your Answer: Sensitivity
Explanation:For establishing a cause effect relationship, following criteria must be met:
1. Coherence & Consistency
2. Temporal Precedence
3. Specificity
As can be seen, sensitivity (The probability of a positive test) is not among these deciding factors..
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This question is part of the following fields:
- Statistical Methods
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Question 5
Correct
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The incidence and prevalence of a diabetes in a locality are being measured by a health care professional. This is what he found:
Year People with Diabetes New Cases
2017 150,000 2000
2018 150,000 4000
What can be stated regarding the incidence and prevalence of the disease in that area?Your Answer: Incidence increasing, prevalence equal
Explanation:Incidence tells us about the number of new cases that have been reported while prevalence gives us the idea of existing cases.
In this particular instance, the parameter of the study i.e. the total number of cases has not changed thus the prevalence of the disease remains same. Although, more cases have been reported in the second instance as a result of which incidence has increased.
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This question is part of the following fields:
- Statistical Methods
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Question 6
Correct
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Which of the following is true regarding correlation coefficient?
Your Answer: It can assume any value between -1 and 1
Explanation:The degree of correlation is summarised by the correlation coefficient (r). This indicates how closely the points lie to a line drawn through the plotted data. In parametric data this is called Pearson’s correlation coefficient and can take any value between -1 to +1. A correlation of -1.0 indicates a perfect negative correlation, and a correlation of 1.0 indicates a perfect positive correlation.
For example
r = 1 – strong positive correlation (e.g. systolic blood pressure always increases with age)
r = 0 – no correlation (e.g. there is no correlation between systolic blood pressure and age)
r = – 1 – strong negative correlation (e.g. systolic blood pressure always decreases with age)
Whilst correlation coefficients give information about how one variable may increase or decrease as another variable increases they do not give information about how much the variable will change. They also do not provide information on cause and effect.
In contrast to the correlation coefficient, linear regression may be used to predict how much one variable changes when a second variable is changed.
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This question is part of the following fields:
- Statistical Methods
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Question 7
Correct
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The average diastolic blood pressure of a control group was found out to be 80 with a standard deviation of 5 in a study aimed at exploring the efficiency of a novel anti-hypertensive drug. The trial was randomised.
Making an assumption that the data is normally distributed, find out the number of patients that had diastolic blood pressure over 90.Your Answer: 3%
Explanation:Since the data is normally distributed, 95% of the values lie with in the interval 70 to 90. This can be calculated as follows:
Interval= Mean ± ( 2 times standard deviation)
= 80 ± 2(5)
= 80 ± 10
= 70 & 90The rest of the 5% are distributed symmetrically beyond 90 and below 70 which means 2.5% of the values lie above 90.
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This question is part of the following fields:
- Statistical Methods
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Question 8
Incorrect
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A graph was plotted after administration of fentanyl infusion to a patient. The following are the x- and y-axis of the graph:
X-axis: Dose of fentanyl
Y-axis: Mu receptor occupancy, measured using positron emission tomography
Given the data above, what would be the best representation of the graph if the data on the x-axis are converted to logarithms?Your Answer: Wash-in exponential curve to a straight line
Correct Answer: Rectangular hyperbola to sigmoid curve
Explanation:The dose-response curve plots the graph of the dose (drug concentration) versus the response. As doses increase, the response increment diminishes; finally, doses may be reached at which no further increase in response can be achieved. This relation between drug concentration and effect is traditionally described by a hyperbolic curve. When the x-axis is plotted in log scale, the graph yields a sigmoid curve.
Efficacy (Emax) and potency (EC50) can be derived from this curve. Emax is the maximal effect achievable, with increasing concentration of a drug. EC50 is the concentration of the drug, wherein half of the maximal effect is achieved.
When the graph is plotted using a log [response/1-response] against log dose, the sigmoid curve becomes a straight line (Hill plot). A graph that transforms from a straight line to exponential curve is mathematically incorrect. A graph that transforms from either a wash-in or wash-out exponential curve to a straight line comes from an initial set of data plotted against time, to a logarithmic transformation of the initial data set against time.
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This question is part of the following fields:
- Statistical Methods
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Question 9
Incorrect
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A pharmaceutical company has developed a new drug considered a breakthrough in treating ovarian cancer.
The efficacy of this drug can be assessed by which phase of a clinical trial?Your Answer: Phase III
Correct Answer: Phase IIa
Explanation:Phase IIa studies are usually pilot studies designed to demonstrate clinical efficacy or biological activity (‘proof of concept’ studies) whereas phase IIb studies determine the optimal dose at which the drug shows biological activity with minimal side-effects (definite dose-finding studies).
Phase III and Phase IV studies are performed on larger set of participants (usually hundreds to thousands) when safety and efficacy have been established.
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This question is part of the following fields:
- Statistical Methods
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Question 10
Correct
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In medical testing, there are true negative, true positive, false positive and false negative results for some test.
How are the sensitivity of these predictive tests calculated?Your Answer: True positives / (true positives + false negatives)
Explanation:The following terms are used in medical testing:
True negative – The test is negative and the patient does not have the disease.
True positive – The test is positive and the patient has the disease.
False positive – The test is positive but the patient does not have the disease.
False negative – The test is negative but the patient has the disease.The sensitivity of a predictive test = true positives / (true positives + false negatives).
The specificity of a test = true negatives / (false positives + true negatives).
The negative predictive value of a test = true negatives / (false negatives + true negatives).
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This question is part of the following fields:
- Statistical Methods
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Question 11
Correct
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You are asked about the relative risk associated with a drug. You happened to have read a report in which the same drug was compared with a placebo and your initial thoughts after reading the study were that the drug will have significant side effects.
What do you expect the value of relative risk to be?Your Answer: Greater than 1
Explanation:Risk ratio (relative risk) compares the probability of an event in an exposed (experimental) group to that of an event in the unexposed (control) group.
A relative risk of 1 suggests that there is no discernible difference in the outcome whether or not it has been exposed.
A relative risk of less than 1 indicates that probability of occurrence of an event is less if there is exposure.
A relative risk of greater than 1 highlights that an event is most likely to occur if it was provided exposure. Since you believe that exposure (the new drug) would have side effects, the value should be greater than 1.
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This question is part of the following fields:
- Statistical Methods
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Question 12
Correct
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A 68-year-old man presents worried about his risk of motor neurone disease. No symptoms have developed, but his father suffered from motor neurone disease. Recently, his cousin has also been diagnosed with amyotrophic lateral sclerosis. He searched the internet for screening tests for motor neurone disease and found a blood test called ‘neuron’, and requests to have it done. You search this blood test and find a prospective study going on evaluating the potential benefits of this blood test. On average, this test diagnosed patients with the disease 8 months earlier than the patients who are diagnosed on the basis of their clinical symptoms. The patients diagnosed using this neuron test also survived, on average, 48 months from the diagnosis, whereas the patients diagnosed clinically survived an average of 39 months from the diagnosis. Considering the clear benefits, you decide to have it done on the patient.
Which of the following options best relate to the above scenario?Your Answer: Lead-time bias
Explanation:Hypochondriasis is an illness anxiety disorder, and describes excessively worriedness about the presence of a disease. While the woman is concerned about her possibility of developing motor neurone disease, she understands that no symptoms have yet appeared. Hypochondriasis involves patients who refuse to accept that they don’t have the disease, even if the results come back negative.
Late Look Bias occurs when the data is gathered or analysed at an inappropriate time e.g. when many of the subjects suffering from a fatal disease have died. This type of biasness might occur in some retrospective studies of motor neurone disease, but is not applicable to this prospective study.
In procedure bias, the researcher decides assignment of a treatment versus control and assigns particular patients to one group or the other non-randomly. This is unlikely to have occurred in this case, although it is not mentioned specifically. Of all the options, lead time-bias is a better answer.
The Hawthorne Effect refers to groups modifying their behaviour simply because they are aware of being observed. Any differences in the behaviour have not been mentioned in the question, and it is highly unlikely that a change in patient’s behaviour would have affected their length of survival in this case.
The correct option is lead-time bias. Even if the new blood test diagnoses the disease earlier, it doesn’t affect the outcome, as the survival time was still on average 43 months from the onset of symptoms in both groups. With the help of blood test, the disease was only detected 8 months earlier.
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This question is part of the following fields:
- Statistical Methods
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Question 13
Incorrect
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Levels of serum potassium in around 1000 patients that were on ACE inhibitor were measured. The mean value was calculated to be 4.6mmol/L and a standard deviation of 0.3mmol/L was recorded.
Which among the given options is correct?Your Answer: 95.4% of values lie between 4.3 and 4.9 mmol/l
Correct Answer: 68.3% of values lie between 4.3 and 4.9 mmol/l
Explanation:Its known that 68.3% of the total values of a normally distributed variable are found within a range of 1 standard deviation from the mean which makes the range to be 4.3 to 4.9 mmol/L.
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This question is part of the following fields:
- Statistical Methods
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Question 14
Incorrect
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When there is a relation between two variables due to the existence of a confounding factor, this association is termed as:
Your Answer: Direct
Correct Answer: Indirect
Explanation:When the association among any two classes of object is defined by the presence of a third entity it is termed as indirect association. For an instance, the age of the employee may affect the rate of pay, which would have implications on job satisfaction. So, in this example, an indirect relationship between age and job satisfaction exists due to a third party i.e. rate of pay.
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This question is part of the following fields:
- Statistical Methods
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Question 15
Incorrect
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An study on post-operative nausea and vomiting (PONV) among paediatric patients who underwent tonsillectomy showed a decrease in incidence from 10% to 5% following a new management protocol.
Which of the following best estimates the numbers needed to treat (NNT) for one additional patient to benefit from the new management of PONV?Your Answer: 100
Correct Answer: 20
Explanation:The Number Needed to Treat (NNT) is the number of patients you need to treat to prevent one additional bad outcome. For example, if a drug has an NNT of 5, it means you have to treat 5 people with the drug to prevent one additional bad outcome.
To calculate the NNT, you need to know the Absolute Risk Reduction (ARR); the NNT is the inverse of the ARR:
NNT = 1/ARR
Where ARR = CER (Control Event Rate) – EER (Experimental Event Rate).
NNTs are always rounded up to the nearest whole number.
In this case, the NNT can be computed as follows:
ARR = 10% – 5% = 0.05
NNT = 1/0.05 = 20
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This question is part of the following fields:
- Statistical Methods
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Question 16
Incorrect
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A randomized controlled trail has been conducted to compare two drugs used for the early management of acute severe asthma in the emergency department. After being allocated to the randomized groups, many patients have been excluded due to deleterious effect to the drugs.
How the data would be analysed?Your Answer: For each patient who drops out, remove a patient from the other randomised group
Correct Answer: Include the patients who drop out in the final data set
Explanation:Randomized controlled trails will be analysed by the intention-to-treat (ITT) approach. It provides unbiased comparisons among the treatment groups. ITT analyses are done to avoid the effects of dropout, which may break the random assignment to the treatment groups in a study.
ITT analysis is a comparison of the treatment groups that includes all patients as originally allocated after randomization.
In order to include such participants in an analysis, outcome data could be imputed which involves making assumptions about the outcomes in the lost participants.
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This question is part of the following fields:
- Statistical Methods
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Question 17
Incorrect
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In the United Kingdom, a new breast cancer screening test is being conducted compared to the conventional use of mammography. This test predicts that if the breast cancer is diagnosed at an earlier stage, it could improve the survival rate but the overall results remains constant.
This is an example of what kind of bias?Your Answer: Procedure Bias
Correct Answer: Lead time bias
Explanation:Recall bias introduced when participants in a study are systematically more or less likely to recall and relate information on exposure depending on their outcome status.
In procedure bias, the researcher decides assignment of a treatment versus control and assigns particular patients to one group or the other non-randomly. This is unlikely to have occurred in this case, although it is not mentioned specifically.
Self Selection or volunteer bias occur when those subjects are selected to participate in the study who are not the representative of the entire target population. those subjects may be from high socio-economic status and practice those activities or lifestyle that improves their health.
Lead-time bias occurs when a disease is detected by a screening test at an earlier time point rather than it would have been diagnosed by its clinical appearance. In this bias, earlier detection improves the survival time in the intervention group.
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This question is part of the following fields:
- Statistical Methods
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Question 18
Correct
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During the design phase of a study, which among the given is aimed at addressing confounding factors?
Your Answer: Randomisation
Explanation:Randomisation allows for performance of experimental trials in a random order. Using this method gives us control over the confounding variables that are not supposed to be held constant.
For an instance, by employing randomisation we get to control biological differences among individual human beings during experimental trials.
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This question is part of the following fields:
- Statistical Methods
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Question 19
Incorrect
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In a study lasting over a period of two years, in which the mean age of 800 patients was 82 years, the efficacy of hip protectors in reducing femoral neck fractures was discussed.
Both experimental and control group had 400 members. Instances of fractures reported over the two year time duration were 10 for the control group (that were prescribed hip protector) and 20 for the control group.
What is the value of Absolute Risk Reduction?Your Answer: 10
Correct Answer: 0.025
Explanation:ARR= (Risk factor associated with the new drug group) — (Risk factor associated with the currently available drug)
So,
ARR= (10/400)-(20/400)
ARR= 0.025-0.05
ARR= 0.025 (Numerical Value)
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This question is part of the following fields:
- Statistical Methods
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Question 20
Correct
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Patient’s having disease (Test Positive: 60, Test Negative:40)
Patient’s not having the disease (Test Positive:20, Test Negative: 80)
This is a result of a new tumour marker blood test, that was performed on 200 women for breast cancer screening. The director of the screening programme ask you to evaluate the observations and inform them the specificity of this new test.
Which one of the following figure you will relay to the programme director?Your Answer: 80%
Explanation:The positive predictive value is the ratio of patients truly diagnosed as positive to all those who had positive test results. In this case, this is 60/(60+20)=75%.
The negative predictive value is the ratio of patients truly diagnosed as negative to all those who had negative test results. In this case, this is 80/(80+40)=67%.
The sensitivity is the ratio of patients with the disease who test positive i.e. true positive patients to the total number of people with the disease. In this case, this is 60/(60+40)=60%.
The specificity is the ratio of people who don’t have the disease who test negative i.e. true negatives to the total number of people without the disease. In this case, this is 80/(20+80)=80%.
70% is not the result of any screening measurements
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This question is part of the following fields:
- Statistical Methods
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Question 21
Correct
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A study of blood pressure measurements is being performed in patients with chronic kidney disease.
Considering that the results are normally distributed, what percentage of values lie within two standard deviations of the mean blood pressure reading?Your Answer: 95.40%
Explanation:Normal distribution, also called Gaussian distribution, the most common distribution function for independent, randomly generated variables, and describes the spread for many biological and clinical measurements.
Properties of the Normal distribution
symmetrical i.e. Mean = mode = median
68.3% of values lie within 1 SD of the mean
95.4% of values lie within 2 SD of the mean
99.7% of values lie within 3 SD of the mean
The empirical rule, or the 68-95-99.7 rule, tells you where most of the values lie in a normal distribution: Around 68% of values are within 1 standard deviation of the mean.
Around 95% of values are within 2 standard deviations of the mean. Around 99.7% of values are within 3 standard deviations of the mean.
the standard deviation (SD) is a measure of how much dispersion exists from the mean.SD = square root (variance)
The empirical rule, or the 68-95-99.7 rule states where most of the values lie in a normal distribution. Around 68% of values fall within 1 S.D of the mean, about 95% within 2 S.D of the mean, and about 99.7% of values within 3 S.D of the mean. Therefore, 95.4% is the most reasonable answer if results are normally distributed.
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This question is part of the following fields:
- Statistical Methods
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Question 22
Incorrect
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A new study is being carried out on the measurement of a new cardiovascular disease biomarker, and its applications in preoperative screening. The data for this study is expected to be normally distributed.
Which of the following statements is true about normal distributions?Your Answer: The 95% confidence interval tells us how confident we are in the test
Correct Answer: The mean, median and mode are the same value
Explanation:The correct answer is the mean, median and mode of normally distributed data are the same value. This is as a result of the bell shaped curve which is equal on both sides.
The bell-shape indicates that values around the mean are more frequent in occurrence than the values farther away.
In a normal distribution:
1) +/- one standard deviation of the mean accounts for 68% of the data.
2) +/- two standard deviations of the mean accounts for 95% of the data.
3) +/- three standard deviations of the mean accounts for 99.7% of the data. -
This question is part of the following fields:
- Statistical Methods
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Question 23
Correct
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Regarding the treatment of bladder cancer, a study concerned with the usage of a combined or monotherapy was conducted. A forest plot was used for the visual representation of the data.
Which of the following is true regarding forest plots?Your Answer: Forest plots can present data from multiple studies
Explanation:Being the part of a meta analysis, forest plots are more valued as evidence then randomised control trials.
The notion that forest plots can only be used if the results are substantial is not true. They are good indicators of the significance of the data. If the diamond intersects the central line, the data is rendered significant. It also aggregates means and confidence intervals from studies conducted in the past which makes the study much more reliable as errors associated with individual studies tend to have less of an impact in this way.
The suggestion that forest plots are primarily used for qualitative data is factually incorrect. Forest plots require numerical values to function.
All in all, forest plots help us in determining whether or not there is a significant trend in that particular field of study.
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This question is part of the following fields:
- Statistical Methods
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Question 24
Incorrect
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A study was concerned with finding out the normal reference range of IgE levels in adults was conducted. Presuming that the curve follows a normal distribution, what is the percentage of individuals having IgE levels greater than 2 standard deviations from mean?
Your Answer: 0.50%
Correct Answer: 2.30%
Explanation:Since the data is normally distributed, 95.4% of the values lie with in 2 standard deviations from mean. The rest of the 4.6% are distributed symmetrically outside of that range which means 2.3% of the values lie above 2 standard deviations of the mean.
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This question is part of the following fields:
- Statistical Methods
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Question 25
Correct
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A randomized study aimed at finding out the efficacy of a novel anticoagulant, in preventing stroke in patients suffering from atrial fibrillation, relative to those already available in the market was performed. A 59 year old woman volunteered for it and was randomised to the treatment arm. A year later, following findings were reported:
165 out of 1050 patients who were prescribed the already prevalent medicine had a stroke while the number of patients who had a single stroke after using the new drug was 132 out of 1044.
In order to avoid one stroke case, what is the number of patients that need to be treated?Your Answer: 32
Explanation:Number needed to treat can be defined as the number of patients who need to be treated to prevent one additional bad outcome.
It can be found as:
NNT=1/Absolute Risk Reduction (rounded to the next integer since number of patients can be integer only).
where ARR= (Risk factor associated with the new drug group) — (Risk factor associated with the currently available drug)
So,
ARR= (165/1050)-(132/1044)
ARR= (0.157-0.126)
ARR= 0.031
NNT= 1/0.031
NNT=32.3
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This question is part of the following fields:
- Statistical Methods
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Question 26
Incorrect
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Which of the following is true when testing for statistical significance?
Your Answer: A type II error is to accept the alternative hypothesis when it should be rejected
Correct Answer: The probability associated with a type I error is the significance level
Explanation:The null hypothesis states that there is no significant difference between two groups.
The alternative hypothesis states that there is a significant difference between two groups.
A type I error (false positive) occurs when a null hypothesis is rejected when it should be accepted.
A type II error (false negative) occurs when the alternative hypothesis is rejected when it should be accepted.
The probability determines the rejection of a null hypothesis.
The level of significance is set at p <0.05.
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This question is part of the following fields:
- Statistical Methods
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Question 27
Correct
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A breakthrough lipid-lowering therapy for stroke had a number needed to treat (NNT) of 20 for the prevention of the primary end-point. These results can be best described as:
Your Answer: For 1000 patients treated with active therapy, there would be 50 fewer strokes
Explanation:Number needed to treat (NNT) is a time specific epidemiological measure that indicates how many patients would be require for an intervention to prevent one additional bad outcome. A perfect NNT would be 1, where everyone improves with treatment, thus the higher the NNT, the less effective the treatment.
Thus if you treat 1000 patients then you will expect to have 50 fewer strokes.
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This question is part of the following fields:
- Statistical Methods
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Question 28
Incorrect
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What is factually correct regarding correlation and regression?
Your Answer: Correlation is concerned with demonstrating difference between variables
Correct Answer: Regression allows one variable to be predicted from another variable
Explanation:Linear regression, using a technique called curve fitting, allows us to make predictions regarding a certain variable.
Correlation coefficient gives us an idea whether or not the two parameters provide have any relation of some sort or not i.e. does change in one prompt any change in other?
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This question is part of the following fields:
- Statistical Methods
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Question 29
Incorrect
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Drug X, a new intravenous induction drug, is being administered as a bolus at regular time intervals, and the following data were observed:
Time following injection (hours) vs Plasma concentration of drug X (mcg/mL)
2 / 400
6 / 100
10 / 25
14 / 6.25
Which of the following values estimate the plasma half-life (T½) of drug X?
Your Answer: 4 hours
Correct Answer: 2 hours
Explanation:Half life (T½) is the time required to change the amount of drug in the body by one-half (or 50%) during elimination. The time course of a drug in the body will depend on both the volume of distribution and the clearance.
Extrapolating the values from the plasma concentration vs time:
Plasma concentration at 0 hours = 800 mcg/mL
Plasma concentration at 2 hours = 400 mcg/mL
Plasma concentration at 4 hours = 200 mcg/mL
Plasma concentration at 6 hours = 100 mcg/mL
Plasma concentration at 8 hours = 50 mcg/mL
Plasma concentration at 10 hours = 25 mcg/mL
Plasma concentration at 12 hours = 12.5 mcg/mL
Plasma concentration at 14 hours = 6.25 mcg/mL -
This question is part of the following fields:
- Statistical Methods
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Question 30
Incorrect
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In reference to confounding variables, which among the given is not true?
Your Answer: Stratification is a technique used to control for confounding
Correct Answer: In the analytic stage of a study confounding can be controlled for by randomisation
Explanation:Randomisation can be used to provide control over the confounding variables during the design stage of a study however during analytical stage a technique called stratification is used for controlling confounding variables. Since the question asks for the information that is factually incorrect.
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This question is part of the following fields:
- Statistical Methods
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