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38 Cards in this Set
 Front
 Back
ST01 Mode is a statistical term relating to:
A. Most common number in a group B. Average value C. Middle value ie. equal number of values on either side of it D. ? 
ANSWER A
MODE is the value that occurs most frequently in a data set or a probability distribution. In a discrete probability distribution, it is the value which takes the maximum value, ei most likely to be sampled In a continuous probability distribution, it is the value which attains the maximum value, ei the peak. MEAN is the average value. MEDIAN is the middle value. 

ST02 [1985] For purposes of statistical analysis:
A. The correlation coefficient relates association but not causation B. The unpaired ttest relates sample means C. The chisquared test uses nominal values only D. Statistical significance does not mean clinical significance E. All of the above 
ANSWER E
A. TRUE The correlation coefficient relates association but not causation  "correlation is a measure of association" (Myles and Gin p.78) B. TRUE The unpaired ttest relates sample means  "Student's ttest is a parametric test used to compare the means of two groups" (Myles and Gin p.51) and it is used as an unpaired ttest "to test if the population means estimated by two independent samples differs significantly" (p.52) C. TRUE The chisquared test uses nominal values only  "the Chisquare test is used to compare independent groups of categorical data" (Myles and Gin p.68). "Categorical data are nominal and can be counted" (p.1) D. TRUE Statistical significance does not mean clinical significance E. All of the above  best answer 

ST03a The Standard Error of the Mean is best described as:
A. The Standard Deviation of sample means B. An indication of the likelihood of making a type II error C. Inversely proportional to statistical power D. Directly proportional to sample size E. Dependent for its validity on a normal distribution in the population 
ANSWER A
A. TRUE The Standard deviation of sample means  Standard error is also known as the standard error of the mean. If one takes a number of samples from a population, we will have a mean for each sample. The SD of the sample means is the standard error Standard deviation is a measure of spread about a central value. B. FALSE An indication of the likelihood of making a type II error  standard error is used to calculate confidence intervals, and so is a measure of precision Beta Error/Type II error occurs when one accepts the H0 incorrectly and the probability of this occurring is termed beta. Alpha Error/Type I Error : occurs when one incorrectly rejects the null hypothesis. The beta and alpha are normally predetermined before a trial to calulate the sample size required to obtain a statistically significant outcome c.f. the confidence interval which is a measure of the precision of the sampling in relation to the population. C. Inversely proportional to statistical power  false D. Directly proportional to sample size  false: SE=SD/SQRT(n) E. FALSE Dependent for its validity on a normal distribution in the population 

ST03b
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 
ANSWER C
A. FALSE : The formula for SEM (=SD/ square root n) does not assume a normal distribution. However, many of the uses of the formula do assume a normal distribution. SEM will be higher in nonnormal distributions, but just as valid. B. FALSE C. TRUE : SEM=SD/SQRT N for large samples, for small samples use the tdistribution D. FALSE : 95% CI =mean +/ 1.96 SEM E. FALSE :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 

ST04a
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 1 error D. the variability of the population E. none of the above 
ANSWER D
Power is the likelihood of detecting a specificed difference if it exists. It is equal to 1beta error. Factors influencing power 1. Significance criterion : increasing alpha error will increase the chance of obtaining a statistically significant result (reject the null hypothesis) 2. Magnitude of effect : larger the effect, the greater the chance of obtaining a significant result 3. Sample size : easiest way to boost the power of a test 4. Precision (ie SD) this is influenced by the sample size 5. Design of the test , for example RCTs have the highest power Sample size = 

ST04b The power of a statistical test used in a clinical trial is influenced by
A. the sample size B. sample variability C. Type Two error D. level of significance 
ALL TRUE


ST05 [Mar94]
The power of a statistical test is highest when: A. The expected difference between the 2 groups being compared is large B. The beta error is less than 5% C. The study groups are cohort & case controlled D. Depends on the alpha error & the number of subjects in each study group 
ANSWER A


ST07 The following data on Apgar scores are best analysed by which test?
Apgar: <4 47 >7 Group A 67 55 22 Group B 58 53 8 A. Fisher's exact test B. Wilcoxon sign rank test C. ANOVA D. Chisquared 
Fisher's Exact test
Analysis of contingency tables were sample sizees are small, categorical data, 

ST07b
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. Chisquare Test B. Chisquare Test with Yates correction C. Student's ttest D. Fisher’s Exact Test E. MannWhitney Test 
ANSWER D


ST (Q23 Aug 2008) NNT is the number of patient who need to be treated to prevent 1 additional bad outcome. The NNT is the reciprocal of the
A. absolute odds of a bad outcome B. absolute risk of a bad outcome C. absolute risk reduction in the bad outcome (due to the treatment) D. odds ratio of the bad outcome (due to the treatment) E. relative risk of the bad outcome (due to the treatment) 
ANSWER C
Number needed to treat is the number of patients that need to be treated for one to benefit compared with a control. It is the reciprocal of the absolute risk reduction. NNT = 1/(CEREER) Absolute risk reduction ARR = CER EER the decrease in risk of a given treatment in relation to a control treatment. Relative risk reduction RRR =(CEREER)/CER Absolute risk reduction over control risk, it describes how much the risk is reduced as a percentage. Relative Risk RR= EER / CER is the risk of an event occuring relative to exposure Odds Ratio OR = (EE/EN )/ (CE/CN) is the ratio of the odds of an event occuring in a treatment group to the odds of it occuring in the control EE / (EE + CE)  EN / (EN + CN) 

ST26 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 
ANSWER E
Confidence interval is a range with a particular assurance that the population parameter will fall within that range. 95%CI = mean +/ 1.96 SEM SEM = SD/SQRT(n) It is a measure of precision. 1. A narrow CI implies high precision. 2. CI that include zero are non significant ei no effect 3. Comparing CI that overlap imply non significant difference (ei clinical indifference) 

ST19 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 
ANSWER A
Sensitivity = TP / (TP + FN) measure the proportion that test positive with the disease. Specificity = TN / (TN + FP) measure of the proportion that test negative who are healthy Positive Predictive Value = TP / (TP + FP) Probability that the patient has disease in those with a positive test Negative Predictive Value = TN / (TN + FN) Probability of health in those that test negative to test Negative test in health False Positive Rate = 1  spec False Negative Rate = 1  sens False positive rate = FP / (TN + FP) = 1 – spec False negative rate = FN / (TP + FN) = 1  sens 

ST Numbers needed to treat is the inverse of:
A. ? B. reduction of absolute risk C. absolute decrease in relative risk D. relative risk E. odds ratio 
ANSWER B
NNT = 1/ARR ARR = CER  EER 

ST18 ANZCA version [2001Apr] Q70, [2001Aug] Q43, [2003Apr] Q86, [2003Aug] Q60, [2005Sep] Q41
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) 
ANSWER A
* Both relative risk and odds ratio are used to measure effect size * Both have a range of 0 to positive infinity with value of 1 meaning both groups have same probability * Relative risk = pexposed / pcontrol o Used in randomised trials o More intuitive * Odds ratio = p1(1p2) / p2(1p1) o Used in casecontrol studies and retrospective studies Easy to remember  no of times something occurs can never be negative! ie can't have negative MI's in a trial! By the by, odds ratio is an inferior tool cf risk ratios. They are used when the denominator is not known. This means that the experimenter has looked for two groups (casecontrol) to compare, rather than sampling a COHORT of the population. Risk ratios should always be used where possible; odds ratios overestimate the "risk" ie OR of 3 is actually less than three times more likely to happen. 

ST24 ANZCA version [2003Aug] Q108, [2004Apr] Q62, [Mar06]
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 
ANSWER C


ST34 ANZCA Version [Mar06] Q112
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% 
ANSWER C


ST28 ANZCA version [2004Apr] Q115, [2004Aug] Q50, [Jul06]
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 
ANSWER B
[ARR = (100  75)/100 = 0.25; NTT = 1/ARR = 4] 

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 
ANSWER B


ST33 ANZCA version [2004Aug] Q129 NB this is the same as ST 18
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 nonexposure, to the new drug is A. unable to be accurately calculated B. 0.08 C. 0.5 D. 2 E. 12.5 
ANSWER E


ST32 ANZCA Version [Jul06] Q113, [Apr07]
A new test has been developed to diagnose a disease. To determine the SPECIFICITY of this new test it should be administered to A. a mixed series of patients i.e. some known to be suffering from the disease and some known to NOT be suffering from it B. a series of patients known to NOT be suffereing from the disease C. a series of patients known to NOT be suffereing from the disease and an estimate of the prevalence of the disease in the population obtained D. a series of patients known to be suffereing from the disease E. a series of patients known to be suffereing from the disease and an estimate of the prevalence of the disease in the population obtained 
ANSWER B


ST31 ANZCA version [2005Apr] Q130
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. tripleblinding refers to the blinding of patient, observer and investigator 
ANSWER B


ST30 ANZCA version [2002Mar] Q130
If two methods of measuring a physiological parameter have a Pearson correlation coefficient 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 
ANSWER 1 and 3


ST27 [x]:
Side effect No side effect Drug given 5 2 Drug not given 1 5 Odds ratio is A. 1 B. 4 C. 8 D. 12.5 E. cannot calculate 
ANSWER D


ST22 ANZCA version [2002Aug] Q81, [2004Apr] Q88, [2004Aug] Q78
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 
ANSWER E
A, B, C and D are word for word out of Myles and Gin 115116. 

ST21 ANZCA version [2002Aug] Q122, [2003Aug] Q87, [2004Apr] Q67, [Apr07] Q87, [Jul07]
Forty patients are randomly dived 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 ttest B. MannWhitney test C. Chisquare test D. KruskalWallis test E. paired ttest 
ANSWER B
* A. unpaired ttest  false: Whilst the two groups are independent, the data is nonparametric. The unpaired ttest test "if the population means estimated by two independent samples differ significantly" (Myles and Gin p.52) * B. MannWhitney test  true: The scale is ordinal. "If there is a natural order among categories, so that there is a relative value among them... then the data can be considered ordinal data... Ordinal data are... a type of categorical data." (Myles and Gin p 23). There are two independent groups and the equivalent of the unpaired ttest for nonparametric data is the MannWhitney U test. "MannWhitney U Test (identical to the Wilcoxon rank sum) is a nonparametric equivalent to the unpaired Student's ttest" (Myles and Gin p.63) "The MannWhitney U test is the recommended test to use when comparing two groups that have data measured on an ordinal scale. However, if the data represent a variable that is, in effect, a continuous quantity, then a ttest may be used if the data are normally distributed. This is more likely with large samples (say n>100)." (p. 64) * C. Chisquare test * D. KruskalWallis test * E. paired ttest  false. The two groups are independent and data nonparametric. A paired Ttest is used "to test if the population means estimated by two dependent samples differ significantly." (p.52) "Paired tests are used when the two samples are matched or paired ("dependent"). The usual setting is when measurements are made on the same subjects before and after a treatment." (p.55) 

ST20 ANZCA version [2001Aug] Q83, [2002Aug] Q78, [Apr07] [Jul07]
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 
ANSWER A
From these data * A. the relative risk of this complication with drug exposure CANNOT be determined  true: o "Because accurate information concerning all patients at risk in a retrospective casecontrol study is not available (because sample size is set by the researcher), incidence rate and risk cannot be accurately determined, and the odds ratio is used as the estimate of the risk ratio" (Myles and gin p.74) * 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 

ST19 ANZCA version [2002Mar] Q62, [2002Aug] Q59, [2005Apr] Q58, [2005Sep] Q50
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 
ANSWER A
Sensitivity = TP / (TP + FN) Specificity = TN / (TN + FP) Positive Predictive Value = TP / (TP + FP) Negative Predictive Value = TN / (TN + FN) False Positive Rate = 1  spec False Negative Rate = 1  sens False positive rate = FP / (TN + FP) = 1 – spec False negative rate = FN / (TP + FN) = 1  sens Therefore A is correct! 

ST16 ANZCA version [2005Sep] Q42 (Similar version reported in [Apr98] [Apr99] [Mar00] [Jul00])
A placebo should be used in a clinical trial when A. observer bias is possible B. a type one error is probable C. an acceptable standard treatment is not known to exist D. a placebo effect is anticipated E. human patients are used as subjects 
ANSWER C
Placebo use and risk Article II. 3 of the current Declaration states that "in any medical study, every patient  including those of a control group, if any  should be assured of the best proven diagnostic and therapeutic method. This does not exclude the use of inert placebo in studies where no proven diagnostic or therapeutic method exists. Article 19 of the Proposed Revision states that "when the outcome measures are neither death nor disability, placebo or other notreatment controls may be justified on the basis of their efficiency".6 Helsinki declaration. 

ST14c ANZCA version [2001Aug] 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 
ANSWER E
refers to NPV which = TN/(TN + FN) = 0.67 thus E correct. how to calculate the sensitivity, specificity, PPV, NPV, FPR, FNR. Sensitivity = TP / TP + FN = TP rate Specificity = TN / TN + FP = TN rate PPV = TP / TP + FP = probability of true positive given positive test result NPV = TN / TN + FN = probability of true negative given negative test result False positive rate = FP / TN + FP = 1 – spec False negative rate = FN / TP + FN = 1 – sens 

ST14b [Jul99] [Mar00]
What is the chance of having the disease if the test is positive? Disease Yes No Test Positive 0.5 0.2 Negative X 0.8 A. 0.89 B. 0.71 C. ? D. ? 
ANSWER B
refers to the PPV which is 0.71 

ST14a [Apr96] [Apr98] [Jul98] [Apr99]
Based on the information in the 2x2 contingency table, what is the chance of having the disease if the test is positive? Disease Yes No Test Positive 80% 10% Negative 20% 90% A. 0.89 B. 0.8 C. 0.67 D. 0.5 E. <0.3 D. ? 
ANSWER A
PPV, which = TP/(TP+FP) which in this case = 0.89 

ST13 [Apr96] [Apr97] [Apr98] [Jul98] (type A)
Comparing two antihypertensive agents (or test on a new antihypertensive agent). Study reports 95% confidence interval for the difference between the two drugs is 2 to 10 mmHg. The conclusion is: A. There is no statistically significant difference because the confidence interval is small B. There is a statistically significant difference (p< 0.05) because the confidence interval does not include zero. C. There is a statistically significant difference (p< 0.01) because the confidence interval does not include zero D. There is a clinically significant difference 
ANSWER B
C would be true for a 99% CI 

ST12 [Mar93] [Jul97]
During interpretation of a report of a clinical trial the: A. Most important factor in determining the statistical power of the study is the size of the expected difference between the groups being compared B. Statistical power of the study is the probability of not making a type II error C. 99% confidence interval for the population mean is given by the sample mean +/ 1.96 x standard error of the mean D. Designs least subject to bias are cohort and case control studies E. Validity of the results is questionable if serial comparisons were made between the two samples using ANOVA (analysis of variance). 
ANSWER B
Power = 1  β where β = probability of type II error 

ST11 [Aug94] [Aug96] [Apr97] [Jul97] [Apr99]
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 
ANSWER D


Black Bank April 2008
ST Numbers needed to treat is the inverse of: A. ? B. reduction of absolute risk C. absolute decrease in relative risk D. relative risk E. odds ratio 
ANSWER B
NNT = 1/Absolute Risk Reduction 

AZ (Q120 Aug 2008) Preoperative assessment shows a malampati (ML) score of III and thyromental distance (TMD) of < 6cm. A grade 3 to 4 on Cormark and Lehanes is predicted. Compared to the ML score, the TMD is:
A less sensitive, less specific B less sensitive, more specific C more sensitive, less specific D more sensitive, more specific E equal sensitivity an specificity 
ANSWER B


ST (Q23 Aug 2008) NNT is the number of patient who need to be treated to prevent 1 additional bad outcome. The NNT is the reciprocal of the
A. absolute odds of a bad outcome B. absolute risk of a bad outcome C. absolute risk reduction in the bad outcome (due to the treatment) D. odds ratio of the bad outcome (due to the treatment) E. relative risk of the bad outcome (due to the treatment) 
ANSWER C
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. 

Black bank April 2009
Which of the following can be used to describe the spread of nonparametric data? A. standard deviation B. interquartile range C. confidence interval D. standard error E. variance coefficient 
ANSWER B
