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34 Cards in this Set
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41) Comparing New instrument of BP measurement with gold std - choice of test
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Bland-Altman plot
- From wikipedia 'Bland and Altman make the point that any two methods that are designed to measure the same parameter (or property) should have good correlation when a set of samples are chosen such that the property to be determined varies considerably. A high correlation for any two methods designed to measure the same property is thus in itself just a sign that one has chosen a wide spread sample. A high correlation does not automatically imply that there is good agreement between the two methods. One primary application of the Bland Altman plot is to compare two clinical measurements that each provide some errors in their measure. it can also be used to compare a new measurement technique or method with a gold standard even so the interest of the Bland-Altman plot is contested in this particular case because the error pertains to the sole new measure.' ----------- |
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[May09][Oct09][Mar10]
Non-normally distributed pain scores, what is the best way to describe the spread of data? A. Interquartile range B. Standard deviation C. confidence interval D. standard error E. variance coefficient |
A. Interquartile range
- (CEACCP 7(4): 127-130, answer on p129). ----------- |
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ST03
The standard error of the mean is A. dependent for its validity on a normal distribution in the population B. an indication of the likelihood of making a type II error C. about 2, if the standard deviation is 15 and the sample size 50 D. NOT necessary for calculating the confidence interval for the mean E. the variance of the population of sample means |
C. about 2, if the standard deviation is 15 and the sample size 50
SEM = SD/ root n ie. 15/7 = 2 - A - Incorrect. Because of central limit theorem: the shape of a a sampling distribution will approximate a normal distribution when the number of samples is high, even when the underlying population is not a normal distribution. P14 Myles and Gin. B - false C - true see above D - false: 95% CI = mean +/- 1.96 x SEM E - This is incorrect. SD is a measure of variability of results from the mean, whilst SEM is a measure of precision and relates the sample mean to the population mean. It is the standard deviation of the means of multiple samples, but not the variance, since this is the square of SD. (Myles and Gin). ----------- |
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ST04a ANZCA version [2001-Apr] Q38, [2002-Aug] Q26, [2003-Apr] Q38, [Mar06] Q60, [Jul06]
The power of a statistical test can be expected to decrease, if there is an increase in A. the sample size B. the size of the treatment effect C. the chance of making a Type I error D. the variability of the population E. none of the above |
D. the variability of the population
- Calculated using: n (sample size) > 2 ((Z1-α/2 + Z1-β)sq) (σsq) / (Δsq) Where: - α = Type I error (rejecting null hypothesis incorrectly) set at 0.05. - β = Type II error (accepting null hypothesis incorrectly) set at 0.2. - σ = Variance which can be estimated from pilot studies and minimised by ↑ measurement precision. - Δ = Effect size which is the clinically signficant difference one wishes to detect. - from Myles and Gin: When working out sample size to maximise power of a study variance is the only variable the investigator cannot choose. Estimates can be obtained from pilot studies or othe data. "...if the variance is underestimated, on completion of the study at the given sample size, the power will be diminished and a statistically significant difference may not be found." This suggests that a larger variability reduces power. The more variable a population the harder it is going to be to find a difference between two groups when you make an intervention in one of them. Eg if everyone who had a GA got sick it would be easier to see if an antiemitc worked. But if only some people get sick it's hard to know if the drug works or not so you need a more powerful test to avoid saying "oh well, some people don't get sick anyway" ----------- |
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ST07 [Jul07][Apr08]
In a clinical trial, 3 out of 10 patients develop a complication in the control group, and 1 of 10 patients develops the complication in the treated group. To assess whether this is a statistically significant difference the most appropriate statistical test to use would be the A. Chi-square Test B. Chi-square Test with Yates correction C. Student's t-test D. Fisher's Exact Test E. Mann-Whitney Test |
D. Fisher's Exact Test
Used in categorical data where sample sizes are small, where the Chi-Squared test is not valid. Chi squared is used with larger samples - it is invalid where _expected value_ less than 10. The Yates correction can be used for Chi-Square in larger samples but where one cell has an expected value that is <10. It is not very useful. ~~ ----------- |
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ST10b ANZCA version [2002-Mar] Q13
When a new diagnostic test is evaluated in a group of subjects in whom the diagnosis is known, the following results are obtained Disease known to be present Disease known to be absent New test result positive 50 20 New test result negative x 80 If a subject from this population tests positive, the probability of having the disease is approximately A. 0.8 B. 0.7 C. 0.6 D. 0.5 E. cannot be calculated because 'x' is unknown |
A. 0.7
- PPV = TP/(FP+TP) = 50/70 = 0.71 ----------- |
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ST10a [Apr08]
When a new diagnostic test is evaluated in a group of subjects in whom the diagnosis is known, the following results are obtained Disease known to be present Disease known to be absent New test result positive 2 4 New test result negative 6 8 The specificity of the new test is closest to A. 25% B. 33% C. 57% D. 67% E. 75% |
D. 67%
Specificity is the rate of not having the disease if you test negative TN/TN+FP ----------- |
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ST16
A placebo should be used in a clinical trial when A. Observer bias is possible B. A type one error is possible C. An acceptable standard treatment is not known to exist D. A placebo effect is anticipated E. Human patients are used a s subjects |
C. An acceptable standard treatment is not known to exist
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ST18
Which of the following alternatives is correct regarding the range of values that odds ratios (OR) and relative risks (RR) can take? A. OR (0 to positive infinity); RR (0 to positive infinity) B. OR (negative infinity to positive infinity); RR (negative infinity to positive infinity) C. OR (0 to 1); RR (0 to 1) D. OR (0 to positive infinity); RR (negative infinity to positive infinity) E. OR (0 to positive infinity); RR (negative 1 to positive 1) |
A. OR (0 to positive infinity); RR (0 to positive infinity)
both OR and RR are used to measure effect size RR = no. exposed / no. control (used in RCTS) OR = P1 (1-P2) / P2 (1-P1) (used in case control and retrospective trials OR is inferior to RR (used when the denominator not known) ie. the experimentor has looked for two groups to compare rather than a cohort of a population ----------- |
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ST19 [Apr08][Oct08][Sep11][Mar12][Aug12]
A diagnostic test has a sensitivity of 90% and a specificity of 99% in detecting a certain disease. From this we can conclude that A. the false positive rate of this test is 1% B. the false negative rate of this test is 1% C. the positive predictive value of this test is 90% D. the negative predictive value of this test is 90% E. this test would be a useful screening test for this disease |
A. the false positive rate of this test is 1%
FPR = FP/TN+FP or 1 - Specificity FNR = 1 - Sensitivity ----------- |
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ST20 [Apr07][Jul07][Oct09][Mar10]
In a trial, 75 patients with an uncommon, newly described complication and 50 matched patients without this complication are selected for comparison of their exposure to a new drug. The results show Complication present Complication absent Exposed to new drug 50 25 NOT exposed 25 25 From this data A. the relative risk of this complication with drug exposure CANNOT be determined B. the odds ratio of this complication with drug exposure CANNOT be determined C. the relative risk of this complication with drug exposure is 2 D. the odds ratio of this complication with drug exposure is 1.33 (recurring) E. none of the above |
A. the relative risk of this complication with drug exposure CANNOT be determined
~~ This is a retrospective case-control study. Because it is case-control (the key word is "matched" ), then chances of developing the complication when given the drug cannot be determined. IF the trial was 100 patients who were then randomly assigned to the drug or non-drug, and then the complication rates examined, THEN you COULD estimate risk ratios an excellent article: http://www.childrens-mercy.org/stats/journal/oddsratio.asp "Both the odds ratio and the relative risk compare the relative likelihood of an event occurring between two distinct groups. The relative risk is easier to interpret and consistent with the general intuition. Some designs, however, prevent the calculation of the relative risk." ~~ Wiki: Patients with complication were identified and matched with controls (without the complication) and THEN their exposures were looked at. When you define your sample size by identifying patients with the disease, you can't define an incidence rate and therefore can no longer calculate a risk ratio (ie ratio of disease incidence in those exposed to those not - or a/(a+b)/c/(c+d). So you calculate an odds ratio (sort of a way of estimating risk ratio). In this case: OR=ad/bc = (50x25)/(25x25)=2 This looks at how much greater the odds of having the disease are if you're exposed compared to if you're not. ie say the odds of having the outcome if you're exposed are 2 to 1, vs 1 to 1 if you weren't, then odds ratio is 2/1 or 2. ----------- Patients with complication were identified and matched with controls (without the complication) and THEN their exposures were looked at. When you define your sample size by identifying patients with the disease, you can't define an incidence rate and therefore can no longer calculate a risk ratio (ie ratio of disease incidence in those exposed to those not - or a/(a+b)/c/(c+d). So you calculate an odds ratio (sort of a way of estimating risk ratio). In this case: OR=ad/bc = (50x25)/(25x25)=2 This looks at how much greater the odds of having the disease are if you're exposed compared to if you're not. ie say the odds of having the outcome if you're exposed are 2 to 1, vs 1 to 1 if you weren't, then odds ratio is 2/1 or 2. If this is all too complicated just remember: if you define the incidence rate of the disease at the start (by selecting a group WITH the disease) you CANT CALCULATE A RISK RATIO! See pg 38 and 74 in Myles and Gin |
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ST21
Forty patients are randomly divided into two groups - one to receive induction agent A and another to receive induction agent B. The next day they are asked to rate their anaesthetic experience on a scale of 1 (very bad) to 5 (very good). The most appropriate test to compare the anaesthetic experience of the two groups is the A. unpaired t-test B. Mann-Whitney test C. Chi-square test D. Kruskal-Wallis test E. paired t-test |
B. Mann-Whitney test
- TRUE - ordinal non parametric data (relative ranking of categories), not normally distributed It is non-parametric as the numbers are qualitative (A pain score is non-parametric. 2 is not twice as much pain as 1, and scores of 2.81 don't exist) A. Unpaired student t-test - Normal distribution C. Chi-squared - Categorical Data, whereas this is ORDINAL data (i.e. a range of values are given) D. Kruskal-Wallis test - The Kruskall Wallis test is a non-parametric test for three or more samples E. Paired student t-test - Normal distribution ----------- |
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ST22 [Sep11][Aug12]
Recognised weaknesses of systematic reviews include all of the following EXCEPT A. publication bias B. duplicate publication C. study heterogeneity D. inclusion of outdated studies E. systematic review author bias |
E. systematic review author bias
Meta-analysis has been criticized, and some of its potential weaknesses identified. These include publication bias (negative studies are less likely to be submitted, or accepted, for publication), duplicate publication (and therefore double-counting in the meta-analysis), heterogeneity (different interventions, different clinical circumstances) and inclusion of historical (outdated) studies. Myles & Gin p.114-115 ----------- |
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ST24
In a group of subjects, the proportion vomiting is 80%. With treatment, this can be reduced to 60%. The number needed to treat (NNT) with this treatment is A. 3 B. 4 C. 5 D. 6 E. 7 |
C. 5
NNT = 1 / ARR ARR = the absolute difference in outcome rates between the control and treatment groups ie. 0.8 - 0.6 = 0.2 therefore, NNT = 1 / 0.2 = 5 ----------- |
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ST26 [Apr07][Apr08][Oct08]
Correct statements regarding confidence intervals (CI) include all the following EXCEPT A. CI are derived from the standard error (of the mean). B. CI can be used to assess the precision of population parameter estimates. C. The width of the CI depends on the degree of confidence required. D. The width of the CI depends on the sample size. E. The width of the CI depends on the mean value of the sample |
E. The width of the CI depends on the mean value of the sample
Correct statements regarding confidence intervals (CI) include all the following EXCEPT A. CI are derived from the standard error (of the mean). True. (Actually true in most but not absolutely all cases.) B. CI can be used to assess the precision of population parameter estimates. True, that’s exactly what they do. C. The width of the CI depends on the degree of confidence required. Among other things, yes. 99% CIs are wider than 95%. D. The width of the CI depends on the sample size. Yes. Other things being equal, large N >> narrower CI. E. The width of the CI depends on the mean value of the sample Not at all. The position of the CI, but not its width, depends on sample mean. ----------- |
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ST28
One hundred vomiting patients receive ondansetron. If 25 patients, who would not have stopped vomiting had they received a placebo, stop vomiting, then the number needed to treat (NNT) for ondansetron to stop vomiting is A. 1.3 B. 4 C. 25 D. 100 E. can't be calculated without information on placebo success rate |
B. 4
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ST31
Which of the following statements about "bias" in scientific studies is FALSE? A. bias is a systematic deviation from the truth B. observer bias can be eliminated by blinding C. randomisation is one of the most important ways to reduce bias D. the Hawthorne effect may be reduced by masking the actual intent of a study E. triple-blinding refers to the blinding of patient, observer and investigator |
B. observer bias can be eliminated by blinding
- Can not be fully eliminated. It may be reduced by blinding. see wiki ----------- |
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ST32
If a new test is developed for a particular disease, the best way to determine its SPECIFICITY is to: A. find a sample of people, some of whom have the disease and some who do not B. find a sample of people, all of whom do not have the disease C. find a sample of people, all of whom do not have the disease, and compare to the estimate of population prevalence D. find a sample of people, all of whom have the disease E. find a sample of people, all of whom have the disease, and compare to the estimate of population prevalence |
B. find a sample of people, all of whom do not have the disease
Specificity is looking for the rate of false positive for a new test, ie. true negative. Therefore, if we have a population who do not have the disease, as far as we can tell, then if any "positives" come up, then they will be false positives, so we can calculate specificity as TN/FP+TN, since we know TN from the sample. ----------- |
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ST33 [Sep04] Q129
Six patients with an uncommon but newly described complication are selected, and 7 matching patients without this complication are selected, to determine their exposure to a new drug. The results are shown in this table. Complication Present absent Exposed to new drug 5 2 Not exposed to new drug 1 5 The odds ratio of having the complication with exposure, compared with non-exposure, to the new drug is A. unable to be accurately calculated B. 0.08 C. 0.5 D. 2 E. 12.5 |
E. 12.5
Odds ratio - the ratio of the odds of an event (complication) in one group (drug exposed) to the odds of the event in another group (unexposed). ((a/b)/(c/d) = ad/bc) In this case a=5, b=2, c=1, d=2 ----------- |
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ST34
When a new diagnostic test is evaluated in a population of subjects in whom the diagnosis is known, the following results are obtained Disease known Disease known To be present to be absent New test result positive 80 40 New test result negative 20 180 In this population the POSITIVE predictive value of this test is closest to A. 10% B. 33% C. 67% D. 80% E. 90% |
C. 67%
PPV = proportion of people who have a positive test who have the disease ----------- |
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ST35
Test positive Test Negative Disease present 80 20 Disease absent 20 180 Negative predictive value is: A. 10% B. 80% C. 90% D. E. |
C. 90%
NPV = TN / TN + FN ie. 180 / 180 + 20 = 90% NPV = the proportion of those who test negative who don't have the disease ----------- |
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ST36 ANZCA version [Jul07]
Publication Bias is that A. researchers with a strong track record are more likely to get research published B. studies with positive results are more likely to be published C. studies with negative results are more likely to be published D. studies on important clinical questions are more likely to be published E. the prestige of the journal will affect readers' perception of the quality of the study |
B. studies with positive results are more likely to be published
----------- "Publication bias occurs when the publication of research results depends on their nature and direction." 1: JAMA. 1990 Mar 9;263(10):1385-9.Links The existence of publication bias and risk factors for its occurrence. Ann N Y Acad Sci. 1993 Dec 31;703:135-46; discussion 146-8.Links Publication bias: the problem that won't go away. "The validity of these conclusions is threatened if publication bias results from investigators or editors making decisions about publishing study results on the basis of the direction or strength of the study findings. " |
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ST37 [Aug10]
Positive predictive value is: A. The proportion of people without disease who are correctly identified as not having the disease. B. The proportion of people with disease who are correctly identified as having the disease. C. The proportion of people with disease who have a positive test result D. The proportion of people without disease who have a positive test result E. etc |
C. The proportion of people with disease who have a positive test result
- PPV = TP/(TP+FP) ----------- |
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ST37 [Mar10][Aug10]
Negative predictive value most closely means A. chance of a positive test in people with the disease B. chance of a negative test in people without disease C. chance of... |
Chance of a person who does not have the disease returning negative result
- A. false. chance of positive test in people with disease = TP/(TP+FN) = sensitivity B. false. chance of negative test in people without disease = TN/(TN+FP)= specificity. People without disease (denominator) will either test truly negative or falsely positive. The chance that patient does not have the disease (TN) returning a negative result (either TN or FN - the denominator) is the negative predictive value. ----------- |
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ST38 [Mar10]
What term means the number of people who are correctly identified as not having the disease? A. Sensitivity B. Specificity C. PPV D. NPV |
B. Specificity
- Specificity = TN/(TN+FP) Test sensitivity is the ability of a test to correctly identify those with the disease (true positive), whereas test specificity is the ability of the test to correctly identify those without disease. ----------- |
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ST39
If a test is negative, what proportion will not have the disease? A. Sensitivity B. Specificity C. PPV D. NPV |
D. NPV
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TMP-140 [Apr08] [Oct08]
Number needed to treat (NNT) is the number of patients who need to be treated to prevent one additional bad outcome. The NNT (of a treatment) is the reciprocal of the: A. absolute odds of the bad outcome B. absolute risk of the bad outcome C. absolute risk reduction in the bad outcome (due to the treatment) D. odds ration of the bad outcome (due to the treatment) E. relative risk of the bad outcome (due to the treatment) |
C. absolute risk reduction in the bad outcome (due to the treatment)
- ie new antiemetic reduces risk of vomiting by 1/5th. Thus absolute risk reduction of bad outcome is 1/5th. Thus NNT is 5 inorder for 1 patient to not vomit. ----------- |
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TMP-Jul10-054
[Mar11][Sep11] see also ST11 Malignant hyperthermia. The number of people in the community at any given time with a predisposition is called the: A. Prevalence B. Incidence C. D. E. |
A. Prevalence
Incidence measures the rate of occurrence of new cases of a disease or condition. Prevalence measures how much of some disease or condition there is in a population at a particular point in time. These two measures are very different. A chronic incurable disease like diabetes can have a low incidence but high prevalence, because the prevalence is the cumulative sum of past year incidence rates. A short-duration curable condition such as the common cold can have a high incidence but low prevalence, because many people get a cold each year, but few people actually have a cold at any given time (so prevalence is low and is not a very useful statistic). ----------- |
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TMP-Mar10-085.
[Mar11] If a test is negative, what proportion will not have the disease: A. Sensitivity B. Specificity C. Positive Predictive Value D. Negative Predictive Value |
D. Negative Predictive Value
- NPV = TN/(FN+TN) ----------- |
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TMP-Mar12-039
What statistical test would be best to evaluate the effects of ? 2 drugs in patients at ? 3 different points in time a. ANOVA b. Mantel Hantzel c. Kruskal Wallis d. Students t test |
?a.
-- a. ANOVA - compare means or medians of three or more sample groups (unpaired data that is normally distributed) b. Mantel-Haenszel - Case control studies of dichotomous outcomes (e.g. healed or not healed) can be represented by arranging the observed counts into fourfold (2 x 2) tables. The separation of data into different tables or strata represents a sub grouping (e.g. into age bands). Stratification of this kind is sometimes used to reduce confounding. Mantel-Haenszel method provides a pooled odds ratio across the strata of fourfold tables. Meta-analysis is used to investigate the combination or interaction of a group of independent studies, for example a series of fourfold tables from similar studies. c. Kruskal Wallis - as per ANOVA but data non-normally distributed d. Students t test - for Normally distributed data; can be one sample (compare mean/median of one sample group against a known value); Unpaired or paired (two sample groups unpaired or paired data) ----------- |
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ST30 ANZCA version [2002-Mar] Q130
If two methods of measuring a physiological parameter have a Pearson correlation co-efficient of 0.99 one could conclude that 1. there is a strong linear association between the two methods of measurement 2. there is good agreement between the two methods of measurement 3. increases in the value of one measurement will usually be associated with increases in the value of the other measurement 4. the two measurements probably use a similar methodology |
Option 1 & 3 are correct.
Pearson correlation coefficient is used for linear association. Option 2 is incorrect. While there is a good correlation, this does not imply good agreement between the methods of measurement. For example : Assume test one measures the variable as X and test two always measures the variable 0.5 * X + 0.2 Pearson Correlation (r) will still be 0.99 Option 3 Is correct Option 4 Is wrong As an aside - Myles and Gin suggest that this method SHOULD NOT be used for comparing methods of measurement - the association between measurement methods is best described by the Bland-Altman approach or a Kappa statistic! [edit] ----------- |
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ST15 [Aug96] [Apr99]
Concerning the odds ratio with regard to treatment: A. Odds ratio of 1.0 means no association B. Odds ratio of 2.0 means half the effect C. Odds ratio 0.5 if increase risk 50% D. Odds ratio of 0.0 means no association |
A. Odds ratio of 1.0 means no association
- B False - twice the effect C False - decrease risk 50% D False - can't be zero unless the probability of one group is zero ----------- |
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ST14c ANZCA version [2001-Aug] Q72
When a new diagnostic test is evaluated in a group of subjects in whom the diagnosis is known, the following results are obtained. Disease known Disease known to be present to be absent Test result positive x y Test result negative 30 60 If a subject from this population tests negative, the probability of NOT having the disease is A. unable to be calculated because “x” is unknown B. unable to be calculated because “y” is unknown C. unable to be calculated because “x” and “y” are unknown D. 0.33 E. 0.67 |
E. 0.67
- NPV = TN/FN+TN = 60/(30+60) ----------- |
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ST11 [Aug94] [Aug96] [Apr97] [Jul97] [Apr99] [Jul10] [Mar11] [Aug11]
Which of the following best describes the number of cases of a disease in a given population at a specific time: A. Incidence B. Occurrence C. Frequency D. Prevalence E. Rate |
D. Prevalence
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