Research Methods and Biostatistics

Research Methods and to y Biostatistics sample Possible study PLUMP studentsp Us west Lpga Texas RCTs and observational studies (cohort and case-control studies) Measures of central tendencies and dispersion stats Population vs. samples or Rongeera Type 1 and Type 2 errors Diagnostics tests Measures of risk/benefits of clinical interventions itsPgpatiqladernanorld impossible tostudyFronte Pgpalation § § § § § § Soon Goldstandard Diagnostic Biopsy test MRI pisconsheedtosechowagrattheyore 1 ifb Hierarchy of published studies 1. Hierarchy of evidence from strongest (best) to weakest (most limitations) published research Meta-analyses of RCTs, Systematic reviews y Controlled 2 experiment (RCTs) Combing results Strongest ofseveralindividual study'semeanloagel 3 Prospective cohort Retrospective cohort Case-control s Cross-sectional 7 Case series, case report, expert opinion Weakest 3 Types of published studies • There are two broad types of published studies: BE1. Descriptive I 2. Explanatory 1. Descriptive ―Here, information are simply recorded from observing patients ―Outcomes are observed but not compared between any group of patients ―Descriptive studies are generally considered as “reports” and not necessarily “studies” ―Examples 1. 2. Case reports: description of individual patients e.g., detailed report of the treatment and outcome of an individual patient Case series: description of a series of patients e.g., description of clinical outcomes in a series of patients who have received similar drug treatment 4 Types of published studies (cont’d) 2. Explanatory/analytical ―Here, there is group comparison and this is used as the basis for determining whether an exposure or treatment allocation might cause or affect an outcome ―Explanatory studies are considered “studies” ―Examples 1. 2. Experimental studies (e.g., randomized control trials, RCTs) so Observational studies (e.g., cohort studies, case-control studies, and surveys) prespectin caseropoat series Case explanatory Eton RG survey studies cohort 5 O Experimental studies Has acontestedgroup me 1. Controlled (e.g., RCTs) Is 2. Uncontrolled 6 Experimental studies • These studies involve actual intervention by the investigators ―Intervention means subjects are assigned into treatment groups by the investigator • There are two types of experimental studies 1. Controlled (has a control group) 2. Uncontrolled (lacks a comparison or a control group) observational Experimental independent variable ge 19,11ft Assignpeopleinto gaps there's givinsoftraetatientmention peoplealreadycometothestudytakingtve anglinterention 7 Controlled experimental studies • They are the gold standard and the best • They use a treatment group (intervention) and a control group (comparison) • They are used to determine therapeutic efficacy for FDA drug approval ―Efficacy means can the drug work in ideal/controlled conditions?RET ―Effectiveness means does the drug work in real-world clinical environment? clinical AThome ans 8 Types of control groups • There are different types of control groups 1. PlaceboSB Ambdipireplacebo →A substance that has no therapeutic effect →Used in placebo-controlled trials 2. Active control Andipinentfidipine →Use of another treatment with established efficacy for the condition studied 3. No treatment AmodipineNotraitaat →The treatment or intervention is withheld from the control group →Difficult to blind • Advantages of using a control group ―Helps account for factors (other than the treatment) that might affect the study results at W only →In other words, any effect (or outcome) seen among the treatment group can be attributed to the treatment or drug given and not to any other factor(s) 9 Types of RCTs psport fohunppend stout folinuppend B Mta Parallel design Cross-over design to A B t B dash ou agashat Bt a 10 35 Features of Parallel design • Features of parallel design: ―Patients are randomly assigned to receive either a study treatment or control (usually placebo) ―Patients receive ONLY one of the intervention (treatment or control) ―Results are then compared between the treatment and the control groups →Randomization ensures patients in the treatment and control groups have similar characteristics as much as possible ―It is the most straightforward design to analyze statistically ―There is absence of “carry-over” effects →Effects from one study intervention (e.g., Treatment A) cannot carry over and affect the results from another study intervention (e.g., Treatment B) as can happen in cross-over designs ―Requires less time to conduct than cross-over designs 13 Features of Cross-over design • Features of cross-over design ―Initially, subjects are assigned to receive either the treatment or the control ―After completing the treatment, and after the wash-out period, the patients are then assigned to the second treatment they have not taken →In essence, each patient serve as their own control ―Wash-out period is necessary in cross-over design →This is the amount of time during which no therapy is given in-between each intervention →In theory, this allows the effects of one treatment to be eliminated or “washed out” from the body prior to the beginning of the next intervention 14 Features of cross-over design • Advantage(s) ―Easier at eliminating patient differences (if there is one) as the cause of treatment effect than the parallel design →This is because each patients receive both interventions (treatment or control) being investigated • Disadvantages ―Requires more time to complete than parallel design (additional time to take the cross-over drug + the wash-out period) ―Possibility of treatment carry-over (if two treatments are being evaluated) →Wash out period is used to mitigate this problem ―Requires more complex statistical analyses to account for the carry-over effects ―Have a higher drop-out rate than parallel design 15 Case 1 • Investigators conducted a randomized, single-blind, placebo-controlled study of galantamine given for 12 weeks in 60 patients aged 65-85 years with mild Alzheimer’s disease. The investigators state that this study is being conducted “to determine the efficacy of galantamine on quality of life in Alzheimer’s disease patients” • Question ―How would this study be conducted using a parallel study design? ―How would it be conducted using a cross-over design? 16 Case 1-answer • How would this study be conducted using a parallel study design? trt ―The 60 patients would be randomly assigned to receive either galantamine (treatment=30 patients) or placebo (control=30 patients) for 12 weeks ―Patients can only receive just one therapy (treatment or placebo) • How would it be conducted using a cross-over design? ―Initially, the 60 patients would be randomly assigned to receive either galantamine (treatment=30 patients) or placebo (control=30 patients) for 12 weeks __ ―A wash-out period of no drug administration should then occur →Might depend on drug’s pharmacokinetics and elimination half-life ―Patients will then receive the drug (treatment or placebo) they did not receive initially and use for another 12 weeks →Notice that the total study duration is longer than parallel design 17 Random Assignment • Patient assignment to the treatment or control group is best done through randomization ―Bias could be present if an investigator simply chooses who receives what →Sicker patients might be assigned to one group over another ―Randomization is often accomplished through the use of computer-generated random numbers or using a random number table →Thus, each patients has an equal chance of being assigned to a particular group • Reasons for random assignment ―To reduce bias and ensure study quality ―Reduces the chance that any extraneous factors might affect the study results since differences in patient characteristics would be “balanced” between groups 19 groupsaresimilar P-values >0.05 are not significant =evidence that randomization worked 21 know Resrcherer fBothdont Fudlebynd the knowwhatshouldn't subjectsoretoting subjatsdonknowwhatthayere Blindingya takings • Blinding (or masking) occurs when the patients and/or investigators do not know the intervention group patients were assigned to • Reasons for blinding 1. Outcomes that are subjective in nature can be affected by a person’s belief that a therapy will work or have a certain effect →e.g., pain relief, changes in mood, development of side effects 2. An investigator might be hopeful that a new drug will be efficacious and unintentionally skew the results in that direction →By asking patients more questions about the drug →Repeating tests if findings are unexpected 3. If patients believe the drug they are taking will be beneficial →They might report changes they would otherwise overlook →Fail to mention potential adverse events they experienced or is innestiypeserch SinghDouble V X triple X x 22 x Three types of blinding 1. Single-blind ―The patients are not aware of the therapy they are receiving but the investigators know 2. Double-blind ―Neither the patients nor the investigators (assuming they perform the analyses) know which therapy each patient is receiving →This is the preferred “gold standard” in RCTs 3. Triple-blind ―If any noninvestigator perform the analyses, neither they, the patients, nor the investigators know which therapy each patient is receiving • What is an open-label study? ―These are clinical studies that are unblinded and where both the patients and the investigators know which therapy each patient is receiving →Less quality compared to blinded studies 23 cohort whatarewe measuring whostarts theprocess casecontrol werepeople DiseaseIncidence Exposure what taking exposedfrontman peoplewithdisease identifylases unexposed Iggy identifycontrolipeophenotoeditose Observational studies host looksexaytlylikeRCI prospective co There's randomization pocalarateRRbetweenexposedus.org 1. Cohort 2. Case-control GrOddsratiooftheexposmebetween congers ODI 24 Should you use experimental or observational study? • Smoking and lung cancer ―Investigators wanted to know if cigarette smoking is associated with an increased risk of developing lung cancer • Would it be appropriate for the investigators to use an experimental design to test their hypothesis? ―No →It will be unethical to administer cigarette smoke to subjects for the sole purpose of investigating a detrimental outcome (lung cancer) without possible offsetting advantages ―An observational study will be more appropriate →In addition, an experimental study will not be practical since it takes a very long time for cancer to develop →An observational study (case-control) that compares patients already diagnosed with lung cancer (cases) and those without cancer (control), will be more appropriate 26 Cohort studies • These follow a study “cohort” over time to determine if a drug or other exposure will lead to the development of an outcome of interest ―Cohort: a group of subjects who share a common characteristics →e.g., men and women taking statin therapy (statin therapy=shared common characteristic) who are followed over time for an outcome such as reduction in the occurrence of myocardial infarction (MI) • There are two types of cohort study design It out 1. Prospective cohort 2. Retrospective cohort at 27 Advantages and disadvantages of cohort study • Advantages ―Good for studying possible causes of adverse events or outcomes especially if they occur relatively commonly • Disadvantages ―Not ideal for rare diseases or diseases that take a long time to develop (especially prospective design) ―There is possibility of selection bias →Are the exposed truly comparable to the unexposed? ―With prospective design: →Can take a long time to complete →Loss to follow-up (subjects dropping out of the study) →Could be costly to implement ―With retrospective design →Records or database may be inaccurate or incomplete 29 RR risk aintervention't riskcontrolled Worked example (cohort study) • A cohort study wanted to evaluate the effectiveness of using aspirin to prevent myocardial infarction (MI) among 610 patients. The study identified 154 patients who are already taking aspirin (the exposed) and another 456 patients who are not taking aspirin (the unexposed). Both groups were followed over time for 3 years. After 3 years of follow-up, 7 patients who took aspirin developed MI, while 61 patients who did not take aspirin developed MI. Calculate the relative risk (RR) of MI between the exposed and the unexposed groups and explain the meaning of the calculated RR. Transforming question into a 2 by 2 table: numbers in red are from the question MI (outcome=yes) No MI (outcome=no) Total Aspirin (exposed) 7 (a) 147 (b) 154 (e) No aspirin (unexposed) 61 (c) 395 (d) 456 (f) e+f=610 30 Worked example (cohort study=relative risk) MI (outcome=yes) intention No MI (outcome=no) riskinterationi RR riskcontrol Total Aspirin (exposed) 7 (a) 147 (b) 154 (e) No aspirin contre 61 (c) (unexposed) 395 (d) 456 (f) BR 7 154 611456 e+f=610 !"#$%&'" )&*+ = !"#$ &! '!&()("*"+, &- &.+/&01 ()0&34 15') !"#$ &! '!&()("*"+, &- &.+/&01 ()0&34 .315') = # #$ %&'#(%) *+,- #.,/#0% ,#,12 # #$ %&'#(%) # #$ .3%&'#(%) *+,- #.,/#0% ,#,12 # #$ .3%&'#(%) 7% #⁄ $ 154 = 0.04545 = 0.34 =& = 61 %' %456 0.13377 0339 0.34 RRC Lower Omorginterentio proterthe Since RR<1, this is a protective effect and means that the risk of the outcome is lower among the exposed than among the unexposed (i.e., 0.045<0.134). The exposure (aspirin) is protective against the outcome (myocardial infarction) 31 noo Odds PNbabilityexposed being probabilitynot Worked example (case-control) exposed • A case-control study wanted to evaluate whether smoking was associated with the development of lung cancer. The study identified 110 patients who already had lung cancer (the cases) and another 100 patients who do not have cancer (the control). Both groups were then asked about their exposure to smoking for the past 70 years. It turned out that 65 patients among the cases were smokers (exposed cases) compared to 40 patients among the controls who were smokers (exposed controls). What is the odds ratio (OR) of smoking between cases and controls? Transforming question into a 2 by 2 table: numbers in red are from the question Exposed Cases (lung cancer) Control (no lung cancer) Yes (smokers) a (65) b (40) No (non-smokers) c (45) d (60) Total e=110 f=100 32 Odds ratio (using probability of exposure) • Data given: e, a (cases); b, f (control) cases e=total number of cases (people with lung cancer)=110 a=number of cases who were smokers (or exposed)=65 → c=number of cases who were non-smokers (not exposed)= 110-65=45 6 00 To probability of being exposed !/# ! ―Odds of exposure among cases = probability of not being exposed = $/# = $ !"⁄ &' ―Odds of exposure (among cases)= %"⁄##$ = (() × ##$ (() *' t.gg Cases &' = *' f=total number of controls (people without lung cancer)=100 b=number of controls who were smokers (or exposed)=40 iontrol 901000 o → d=number of controls who were non-smokers (not exposed)=100-40=60 ―Odds of exposure among controls = probability of being exposed +/, + = = probability of not being exposed -/, %$⁄ ―Odds of exposure (among controls)= !$⁄#$$ = #$$ *) ()) × ()) &) = 66 O Control *) &) 33 Odds ratio (using probability of exposure) • Odds ratio = = ./⁄ 0/ 01⁄ .1 CD 0112 04 56702895 (;<0=> ?;252) 0112 04 56702895 (;<0=> ?0=A90B2) CF GHFF = ED × EF = IJFF =2.17 • Alternatively, using the formula in the 2*2 table: • Odds ratio = !" #$ = %&×%( )(×)& = *+(( ,-(( =2.17 Exposed Cases (lung cancer) Control (no lung cancer) Yes (smokers) a (65) b (40) No (nonsmokers) c (45) d (60) Total e=110 f=100 • Interpretation: Cases (people with lung cancer) have a higher odds (2.17 to 1) of being exposed to smoking than controls (people without lung cancer) • This implies that smoking “causes” (or is associated with) a higher risk of lung cancer than not smoking 34 Odds ratio (using number with exposure) Exposed Cases (lung cancer) Control (no lung cancer) Yes (smokers) a (65) b (40) No (non-smokers) c (45) d (60) Total e=110 f=100 # !* 3&#+# 'ℎ&' &%+ +,-!#+" (&) !""# !* +,-!#.%+ (&0!12 3&#+#) # !* 3&#+# 'ℎ&' &%+ .1+,-!#+" (3) !""# %&'(! = = # !* 3!1'%!5# 'ℎ&' &%+ +,-!#+" (8) !""# !* +,-!#.%+ (&0!12 3!1'%!5#) # !* 3!1'%!5# 'ℎ&' &%+ .1+,-!#+" (") 65: &⁄ 3 45 = 65×60 = 3,900 = 2.17 = = 40: 8: 40×45 1,800 60 " Interpretation: Cases (people with lung cancer) have a higher odds (3,900 vs. 1,800) of being exposed to smoking than controls (people without lung cancer). This implies that smoking “causes” (or is associated with) a higher risk of lung cancer than not smoking 35 Calculation of odds ratio 89 Odds Ratio Oatcom 236689 98 237998 1.268 36 Interpretation of odds ratio • OR=1 (or close to 1) means that the odds of exposure among cases is the same as the odds of exposure among controls ―The exposure is not associated with the disease • OR >1 means that the odds of exposure among cases is greater than the odds of exposure among controls ―The exposure may be a risk factor for the disease • OR<1 means that the odds of exposure among cases is lower than the odds of exposure among controls ―The exposure may be protective against the disease 37 Advantages and disadvantages of case-control study • Advantages 1. Good for studying possible causes of adverse events or negative outcomes especially if the outcomes are →Rare or infrequent →Take a long time to develop oft 2. Faster to execute than prospective cohort design 3. Less expensive than prospective cohort design presentin allobservationalstudies BcltheresNorondom ind • Disadvantages dramatic and Thetwogroupsmaynotbalance on 1. Possibility of selection bias to a →Are cases truly comparable to controls? dinicathonacterstis 2. Possibility of recall bias →patients’ memories of an exposure may be inaccurate 3. Records or database may be inaccurate or incomplete 38 Case 2 • Study purpose ―Investigator wish to study if aspirin might lead to Reye’s syndrome development in small children who take aspirin for a viral illness • Method ―Children with a viral illness who later developed Reye’s syndrome were identified, along with other children with a viral illness who did not develop Reye’s syndrome ―The use of aspirin during the viral illness, including dose and duration of therapy, was analyzed and compared in both groups through interviews with parents • Questions ―What study design was used ―What was compared and analyzed? 39 Case 2-answer • What study design was used? ―Case-control design was used (because cases and controls were first identified, exposures were then determined) →Cases = children with Reye’s syndrome (condition/outcome) →Controls = children without Reye’s syndrome • What was compared/analyzed? ―Exposure was compared →Exposure (aspirin use) was compared between cases and controls →Since the exposure (aspirin) must precede development of the condition (Reye’s syndrome) • Information about exposure is obtained by looking in the past (retrospectively) • Greater aspirin use among the cases might indicate a link between aspirin use and development of Reye’s syndrome • However, even if aspirin use was greater among the cases, this study design cannot ‘prove’ that aspirin ‘causes’ Reye’s syndrome • An association could be established but not causality • Not a randomized control trial • Other factors not accounted for could be causing the disease 40 Case 3 • Study background ―Previous studies in adults have indicated a possible link between low vitamin D levels and depression symptoms • Study purpose ―Since children can also develop depression, investigators wanted to know whether low vitamin D levels might be associated with depression in children • Method ―Several hundred children were identified and their vitamin D concentrations were measured ―The children were then divided into three groups based on their vitamin D levels (low, normal, and high) ―The children were then followed up over the next 3-4 years and were periodically tested to determine if they developed symptoms of depression • Questions ―What study design was used? ―What was compared/analyzed? 41 Case 3-answer • What study design was used? ―This is a prospective cohort design →Prospective because they were followed into the future for 3-4 years →Cohort because exposure (vitamin D intake) was determined before outcome/disease occurrence (depression) ―The children were divided into three groups based on the extent to which they had the exposure of interest (Vitamin D) →Group of interest (low vitamin D exposure) →Comparison groups (normal and high vitamin D exposures) • These are active “controls”. They could have also used a group not exposed to vitamin D at all ―There was no intervention →Investigators did not administer the vitamin D to the subjects but rather measured existing concentrations of children already taking vitamin D →Therefore, the design was not experimental but observational 42 Case 3-answer (cont’d) • What was compared/analyzed? ―Outcome of interest (development of depressive symptoms) was compared between the exposure groups (low vs. normal vs. high vitamin D) ―Children were followed prospectively to determine the extent to which depression occurred ―Greater depressive symptoms in children with low vitamin D levels compared to normal or high levels could indicate a link between vitamin D deficiency and depression →However, even if depression developed to a much greater degree among children with low vitamin D concentrations, this study design cannot ‘prove’ that vitamin D deficiency ‘causes’ depression • An association could be established but not causality • Not a randomized control trial • Other factors not accounted for could be causing the disease 43 Measures of central I tendency avenge Mean avengefirstinorder Medianso Ten sp twomiddlevote09 Mode 44 0 Median • The median is the score that divides a distribution exactly in half • The median is calculated by listing the values of a variable in ascending or descending order and reporting the value that lies in the middle of this list (for odd values) ―For example, calculate the median of 32, 76, 36, 45, 23, 52, 49 →First arrange in ascending order: 23, 32, 36, 45, 49, 52, 76 →Median is the middle value=45 • If a list has an even number of values, then the arithmetic mean between the two middle values is calculated and reported as the median ―For example, calculate the median of 32, 76, 36, i 45, 23, 52, 49, 62 →First arrange in ascending order: 23, 32, 36, 45, 49, 52, 62, 76 →Find the average of the two middle numbers: *'3*6 2 = 6* 2 =47 47 When is it more appropriate to use the median? • The median is most appropriately used in situations where a few values of a variable are noticeably larger or smaller than most of the rest of the values ―These values are known as outliers ―In these situations, a median is preferable to the mean because it is not easily affected by extreme scores →For example, imagine trying to accurately describe the typical annual income of a group of people who made $25K, $30K, $27K, $31K, $28K, and $60 million →The person making a substantially higher income than the rest of the group would cause the “average” annual income of the group (=$10,023,500 million) to appear much higher, thus distorting the measure to reflect disproportionally more of the outlier’s income 48 Mode • The mode is calculated by counting the occurrence of each value of the variable and is assigned as the value that appears most often ―For example, evaluate the mode of 32, 76, 45, 36, 32, 45, 23, 57, 49, 45 ―32 appears 2 times, 45 appears 3 times, and 76, 36, 23, 57, and 49 each appear once ―Thus, 45 appears the most often (i.e., 3 times) and is therefore the mode • The mode is the only measure of central tendency that can be used with data from a nominal scale of measurement ―Although a distribution will have only one mean and only one median, it is possible to have more than one mode ―Specifically, it is possible to have two or more scores that have the same highest frequency →A distribution with two modes is called bimodal, while a distribution with more than two modes is called multimodal • The mode is not reported as often in clinical research as other measures of central tendency such as the mean or median 49 Exe I 96 8 8 9 9 102560 60 1 59 Measures of dispersion what 1 4 68 462 5 median a g gg 2g 93 GE f wave Highestwave lover Range noInterquartile range art Variance Standard deviation Skewness AGR 93 69 7 2.5 9.5 50 Measures of dispersion or variability • Measures of dispersion are used to describe how data are spread ―They provide information about the variability in a distribution of observations • If the scores in a distribution are all the same, then there is no variability ―If there are small differences between scores, then the variability is small ―If there are large differences between scores, then the variability is large • Variability provides a quantitative measure of the degree to which scores in a distribution are spread out or clustered together ―Variability tells how much distance to expect between two scores or between an individual score and the mean →Variability provides information about how much error to expect if you are using a sample to represent a population 51 Measures of dispersion or variability (cont’d) • Common examples of measures of dispersion or variability include: ―Range ―Interquartile range ―Variance ―Standard deviation ―Skewness →Positive skew →Negative skew →No skew (normally distributed) • Combining measures of central tendency and dispersion or variability yield a detailed illustration of what the data represent 52 53 Interquartile range (IQR) • One way to avoid the excessive influence of one or two extreme scores is to measure variability with the !"#$%&'(%#!)$ %("+$ (IQR) • The IQR ignores the extreme scores and instead measures the range covered in the middle 50% of the distribution • ,"#$%&'(%#!)$ %("+$ ,-. = -3 − -1 • To find the interquartile range (IQR) ―Order the data from lowest to highest ―Find the median of the whole data ―Find the median (middle value) of the lower and upper half of the data →These values are quartile 1 (Q1) and quartile 3 (Q3), respectively ―The IQR is the difference between Q3 and Q1 55 IQR (odd data) • Assuming the even data in Table 8-2 is made odd by adding an extra number (60) ―Original even data arranged in order: 20, 20, 20, 38, 42, 42, 51, 68 ―New odd data arranged in order: 20, 20, 20, 38, 42, 42, 51, 60, 68 ―Calculate the median →20, 20, 20, 38, 42, 42, 51, 60, 68 = 42 ―Estimate the median of the lower and upper half of the data →20, 20, 20, 38, 42, 42, 51, 60, 68 →$1: Lower half median=(20+20)/2=20 →$3: Upper half median=(51+60)/2=55.5 ,-. = -3 − -1 = 55.5 − 20 = 35.5 56 IQR (even data) • Calculate the IQR of the even data in Table 8-2 ―Arrange the data from lowest to highest →20, 20, 20, 38, 42, 42, 51, 68 ―Calculate the median →20, 20, 20, 38, 42, 42, 51, 68 = (38+42)/2= 40 ―Insert the median into the data (since there was no true median before) →20, 20, 20, 38, 40, 42, 42, 51, 68 ―Estimate the median of the lower and upper halves of the data →20, 20, 20, 38, 40, 42, 42, 51, 68 →Q1: Lower half median=(20+20)/2=20 →Q3: Upper half median=(42+51)/2=46.5 ―'$( = $3 − $1 = 46.5 − 20 = 26.5 57 Standard deviation • The standard deviation (7 8% 9) is the most commonly used and the most important measure of dispersion or variability • Standard deviation uses the mean of the distribution as a reference point and measures variability by considering the distance between each score and the mean ―That is, are the scores clustered together or scattered? • In simple terms, the standard deviation approximates the average distance of all the scores from the mean 2 5 4312 9 4 5 6012 9 6 5291229 8 5 Ex 2 I mean Shinn 4 6 8 6 89942 2 5 58 9949 mean 9 9 Standard deviation T 2.29 41 • Standard deviation (7 8% 9) is the square root of the variance ;#("<(%< <$=!(#!8" = =(%!(">$ • The standard deviation tends to be reported more often than variance because it is expressed using the same units as the original data 0123456718 9658:5;: :<=756718, ? = F5C24< 9658:5;: :<=756718, 9 = ∑(BK −C<58)L = E ∑(BK −C<58)L = 8−1 FF E FF 8−1 59 Standard deviationusing the definitional formula Age (M! ) (M! − OPQR)" (M! − OPQR) 20 20 − 37.63 = −17.63 310.82 38 38 − 37.63 = 0.37 0.14 42 42 − 37.63 = 4.37 19.10 20 20 − 37.63 = −17.63 310.82 20 20 − 37.63 = −17.63 310.82 42 42 − 37.63 = 4.37 19.10 68 68 − 37.63 = 30.37 922.34 51 51 − 37.63 = 13.37 178.76 Mean=37.63 Sample variance, Z # = V (M! − OPQR) = 0.04 ≅ 0 $$ %&' = #,)*'.,) -&' = #,)*'.,) * =295.99 Sample standard deviation, s = \#]^#_&$ = 295.99 = 17.20 O XX = V(M! − OPQR)" = 2,071.90 60 Standard deviation- SE using the computational formula FA M" Age, M 20 400 38 1,444 42 1,764 20 400 20 400 42 1,764 68 4,624 51 ∑ / = 301 ; Standard deviation= 2,601 4 (∑ /) = ∑ / !& ! (∑ $) & %&' (301)4= = 90,601 '0,0,*& '(,*(+ , -&' V ` # = 13,397 = '0,0,*&'',0#1.'#1 * = #,)*'.-*1 * = 295.982 = 17.20 61 Confidence interval for a mean-Example N • Assuming a two-tailed test, what is the 95% confidence interval for a sample of 63 patients on statin therapy with a mean age of 71 years and a standard deviation of 25.5 year? (z-critical= 1.96) "# = $%&'( *+(&,-(* ± /0&(&/-1 2-13* +(-'4-04 *00%0 %5 (ℎ* ,*-' 95% "# = ,*-' ± 1.96 =>9:!; =>9:!; = =? 15.5 25.5 = = = 3.23 7.9 ' 63 95% "# = ,*-' ± 1.96 =>9:!; = 71 ± 1.96 3.23 3$$*0 C%3'4 = 71 + 1.96 3.23 = 71 + 6.33 = 77.3 1%E*0 C%3'4 = 71 − 1.96 3.23 = 71 − 6.33 = 64.7 95% "# = 64.7 H*-0+ − 77.3 H*-0+ 62 Errors in hypothesis testing • Hypothesis testing is an inferential process, which means that it uses limited information as the basis for reaching a general conclusion ―A sample provides only limited or incomplete information about the whole population ―Thus, there is always the possibility that an incorrect conclusion will be made • Although sample data are usually representative of the population ―There is always a chance that the sample is misleading and will cause a researcher to make a wrong decision about the research result • In hypothesis testing, there are two kinds of errors that can be made ―Type I error ―Type II error 63 Type I error • A Type I error occurs when a researcher incorrectly rejects a null hypothesis ―Here, the researcher concludes there is a treatment effect when in fact there was no effect →Some extreme samples can be very different from the population they are supposed to represent →If a researcher selects one of these extreme samples by chance, then the data from the sample may give the appearance of a strong treatment effect, even though there is no real effect →A Type I error can occurs when a researcher unknowingly obtains an extreme, nonrepresentative sample 64 Type I error -example • A research study examined how a food supplement (blueberry) that is high in antioxidants affects the cognitive functioning of elderly patients ―Suppose the researcher selected a sample of n=25 people who already had cognitive functioning that was well above average ―When these people are tested after 6 months of taking the supplement, they will score higher than average on the neuropsychological test even though the blueberry supplement (the treatment) may have had no effect →In this case, the researcher is likely to conclude that the treatment has had an effect, when in fact it really did not →This is an example of what is called a Type I error 65 split Type I error-consequences • In most research situations, the consequences of a Type I error can be very serious • Because the researcher has rejected the null hypothesis and believes that the treatment has real effect, it is likely that the researcher will report or even publish the research results ―A Type I error means that this is a false report ―Other researchers may try to build theories or develop other experiments based on the false results →A lot of precious time and resources may be wasted 66 Type II error • Type II error is also known as beta (!) $%%&% • This occurs when you incorrectly fail to reject the null hypothesis ―This means you are saying there is no treatment effect when there might have been a treatment effect →E.g., if a medication is known to be effective (from several studies) but a new study concluded that the medication is NOT effective • A Type II error would have been committed 67 Type II error (cont’d) • Typical values for Type II error are 20% (Power=80%) and 10% (Power=90%) • Statistical power is related to Type II error ―Power= 100 − GH2< 2 <;;1; = 100 − 20% = 80% ―Thus, a study with low power will have high Type II error rate →This means that there is a higher probability that low-powered studies will falsely claim there is no treatment effect when in fact there might have been, but the study has low sample size and is not powered enough to detect the treatment effect 68 Type II error-consequences • The consequences of a Type II error are usually not as serious as those of a Type I error • In general terms, a Type II error means that the research data do not show the results that the researchers had hoped to obtain ―The researcher can accept this outcome and conclude that the treatment either has no effect or has only a small effect that is not worth pursuing ―Or the researcher can repeat the experiment (usually with some improvements) and try to demonstrate that the treatment really does work 69 Diagnostic Tests Sensitivity Specificity Predictive value Diagnostic tests • In order to estimate disease incidence or prevalence, it is necessary to correctly classify individuals as either diseased or not diseased ―This is achieved through the use of diagnostic tests • However, diagnostic tests are not perfect and misclassification can occur ―Gold standard diagnostic tests which provide a definitive diagnosis are generally impractical due to cost or complexity t I ―There is a trade-off between identifying disease in individuals who really have the disease (true positives) and avoiding detection in those who do not have the disease (false positives) Performance measures of diagnostic tests • These include sensitivity, specificity, and predictive value ―These performance measures provide a standard mechanism for quantifying the ability of a test to correctly classify individuals as either ‘diseased’ or ‘not-diseased’ • The hypothetical ‘gold standard’ diagnostic test has a sensitivity of 100% and a specificity of 100% ―In practice, there are sometimes no true gold standard tests Sensitivity • Sensitivity is the proportion of individuals with the disease who are correctly identified by the test as having the disease ―Also called a true positive rate, indicating that the diagnostic test is positive for a patient who truly has the disease ―Tests with high sensitivity have low false negative rates →Highly sensitive tests do not miss identifying many patients who actually have the disease 66 Specificity • Specificity is the proportion of individuals without the disease who are correctly identified by the test as not having the disease ―Also called a true negative rate, indicating the probability that the diagnostic test is negative for a patient who does not have the disease ―Tests with high specificity will have low false positives →Highly specific tests do not identify many patients as having a disease when they are actually disease-free Predictive value • Predictive values provide information about how likely it is that the individual does, or does not, have the disease given their test result • In clinical practice, test results is all that is known, so we want to know how good the test is at predicting abnormality ―Positive predictive value: the probability that a patient has the disease given that a positive result was obtained ―Negative predictive value: the probability that a patient does not have the disease given that a negative test result was obtained 2-by-2 table for evaluating diagnostic tests True condition of disease (biopsy) Diagnostic test result Disease present (+) Disease absent (-) Positive (+) w x w + x = total testing positive Negative (-) y z y + z = total testing negative Total (column) w + y = total patients with the disease x + z = total patients without the disease Total (row) w + x + y + z = n (total number of patients) w: patients who truly have the disease and have positive test results; true positives x: patients who DO NOT truly have the disease but have positive test result; false positives y: patients who truly have the disease but have negative test result; false negatives z: patients who DO NOT truly have the disease and have negative test result; true negatives 2-by-2 table for evaluating diagnostic tests • Prevalence = total number of patients with the disease divided by overall ("#$) total number of patients = & • Sensitivity = proportion of individuals with the disease who are correctly " identified by the test as having the disease (true positive rate); ("#$) • Specificity= proportion of individuals without the disease who are correctly' identified by the test as not having the disease (true negative rate); ((#') • Positive predictive value (PPV)" = proportion of individuals with a positive test who have the disease; ("#() • Negative predictive value (NPV) = proportion of individuals with a negative ' test who DO NOT have the disease; ($#') Ideal diagnostic test • An ideal diagnostic test would be 100% sensitive and 100% specific ―This means that: (1) all patients with the disease would be identified (100% true positive rate); (2) no patients with the disease would be missed (0% false negative rate), and (3) patients without the disease would never be misclassified as having the disease (0% false positive rate) • However, there is no ideal diagnostic test and sensitivity is gained at the expense of specificity and vice-versa ― In practice, tests used to diagnose disease commonly have a sensitivity of 80% and a specificity of 90% Calculations for specificity, sensitivity, PPV, NPV, and prevalence • The accuracy of a liver scan at correctly diagnosing a liver disease (cirrhosis) was being evaluated among 344Npatients with or without the disease. Among 258 patients who actually had the disease (through biopsy), 231 patients had a true positive diagnostic test result. However, among 86 patients who do not actually have the disease (again through biopsy), 54 patients had a true negative diagnostic test result. Calculate the sensitivity, specificity, positive predictive value, and negative predictive value of the liver scan. Also, calculate the prevalence of the disease among the 344 patients. Best to first create a 2 by 2 table to visualize. Provided numbers in the question are in red. You can calculate others from these. Disease present (through biopsy) Disease absent (through biopsy) Total 231 (true positive) 32 (false positive) 263 Negative diagnostic test 27 (false negative) 54 (true negative) 81 Total 86 344 Positive diagnostic test 258 79 How good is a liver scan at diagnosis of abnormal liver disease? True condition of disease (through biopsy) Diagnostic test (liver scan) Disease present (+) Disease absent (-) Positive (+) True positive 231 False positive 32 PPV = <*, (<*,=*<) Negative (-) False negative 27 True negative 54 NPV = &) (<>=&)) Sensitivity = 89.5% <*, = Specificity = (<*,=<>) 62.8% &) (*<=&)) = Prevalence = <*,=<> <*,=*<=<>=&) =75% = 87.8% = 66.7% AI risklintenention riskcontay NII Ipp NNH Ap Roundup dearest rounddough Measures of risk/benefits of clinical interventions Risksintervention's risksontrow Risk, ARR, RR, RRR, NNT, NNH, OR We Riskcontgroy riskeintergntion Be Absolaterigeduction risk control too 81 82 83 Questions 11.21 1. What is the risk of fatal or nonfatal stroke among the atorvastatin group? 2. What is the risk of fatal or nonfatal stroke among the placebo group? BoIto 3. What is the absolute risk reduction of fatal or nonfatal stroke among the atorvastatin group relative to the placebo group ARE13 I 11.2 9.9 placebo intention 4. What is the relative risk (or hazard ratio) of fatal or nonfatal stroke among the atorvastatin group compared to the placebo group? 5. What is the relative risk reduction of fatal or nonfatal stroke among the atorvastatin group compared to the placebo group? 6. What is the NNT to prevent one fatal or nonfatal stroke? 7. What is the NNH if muscle pain was higher among atorvastatin group? 84 Risk 1. What is the risk of fatal or nonfatal stroke among the atorvastatin group? # +, -.-&/0 12+&3 1/+4.10/1/5& Risk (atorvastatin) = ×100% = # +, 1/+4.10/1/5& 61/5-&/0 ? (("9C$% D%8=!<$< !" %$9')#9 9$>#!8" = 11.2%) 789 ? ×100% = 2. What is the risk of fatal or nonfatal stroke among the placebo group? # +, -.-&/0 12+&3 6;1<-=+ Risk (placebo) = ×100% = # +, 6;1<-=+ 61/5-&/0 ? (("9C$% D%8=!<$< !" %$9')#9 9$>#!8" = 13.1%) >?? ? ×100% = 85 ARR 3. What is the absolute risk reduction of fatal or nonfatal stroke among the atorvastatin group relative to the placebo group? ARR= .!9E >8"#%8) 8% D)(>$F8 − .!9E (#8%=(9#(#!" = 13.1% − 11.2% = 1.9% 86 Relative risk (RR) or Hazard ratio (HR) 4. What is the relative risk (or hazard ratio) of fatal or nonfatal stroke among the atorvastatin group compared to the placebo group? Round final answer to 2 decimal places, no units or commas RR or HR = 1234(2&67187&629&) 1234 (;9&619<) = 1234(=6918=36=62&) 1234(><=;7?9) @@.A% = @C.@% = 0.8549 = 0.85 • Interpretation: ―Since RR<1, this is a protective effect and risk of the event is lower among atorvastatin group relative to the placebo group (this is a good thing. Means the drug is working) ―If RR>1, risk of the event is higher among the intervention group (this is a bad thing) ―If RR=1, risk of the event is the same between the two groups (not a good thing. Drug is not better than placebo) 87 Relative risk reduction (RRR) 5. What is the relative risk reduction of fatal or nonfatal stroke among the atorvastatin group compared to the placebo group? Round final answer to 1 decimal places, no units or commas RRR= !#?@ABC: EF?G E:"B$CF@; (HII) EF?G (JA!$:#@) ×100% = ,*.,%M,,.<% ,,.<% ×100% = 16.964% =16.9 Interpretation: ―Risk of fatal or nonfatal stroke was 16.9% lower among the atorvastatin group compared to the placebo group To verify: RRR= 1 − KK = 1 − 0.85 = 0.15 = 15% (15% is close to 16.9%, but just a rough estimate) 88 NNT 6. What is the NNT to prevent one fatal or nonfatal stroke? OR 6. How many patients would need to be treated with atorvastatin in order to avoid one additional fatal or nonfatal stroke? ? ? NNT= 1=0+;@/- 450A 4-B@.>%J?.8% = ? ?.K% = ? !.% !$$ = ? H.H?K = 58.82 =58 (NNH should be rounded down to the nearest whole number) Scenarios: 1. If NNH=58.03, rounding down = 58 2. If NNH=57.96, rounding down=57 90 Calculation of odds ratio 91 8. The study reported that the overall mortality rate was similar, with 216 deaths in the atorvastatin group and 211 deaths in the placebo group (p=0.98). Evaluate the following sentences a. b. c. d. e. f. g. The null hypothesis should be rejected We should fail to reject the null hypothesis There was a significant difference in overall mortality rate between the two groups since p>0.05 If the level of significance was set at 0.05, we should reject the null hypothesis At p=0.98, the null hypothesis should be rejected The ARR is 216-211=5 For the results section as a whole a. b. What are the independent variables? What are the dependent variables? 92 • Which of the outcomes are statistically significant at alpha of 5%? • How do you interpret the HR for the following outcomes? • Nonfatal stroke • Acute coronary event • Death from cancer 93 Practice Questions 94 95 1.Independent and dependent variables In the study comparing rivaroxaban to placebo, which of the following correctly identifies the independent and dependent variables? A)Independent variable: Death from any cause, myocardial infarction, or stroke Dependent variable: Rivaroxaban vs. placebo B) Independent variable: Rivaroxaban vs. placebo Dependent variable: Composite of death from any cause, myocardial infarction, or stroke C)Independent variable: Left ventricular ejection fraction Dependent variable: Rivaroxaban vs. placebo D)Independent variable: Natriuretic peptide levels Dependent variable: Treatment with rivaroxaban 96 1. Independent and dependent variables Answer: B) Independent variable: Rivaroxaban vs. placebo Dependent variable: Composite of death from any cause, myocardial infarction, or stroke. Rationale: • The independent variable is the factor that the researchers manipulate—in this case, whether the patient receives rivaroxaban or placebo. • The dependent variable is the outcome being measured—in this case, the composite outcome of death, myocardial infarction, or stroke. Why the other answers are wrong? A: Death, MI, and stroke are outcomes (dependent variables), not the independent variable. C: Left ventricular ejection fraction was a baseline characteristic, not an independent variable. D: Natriuretic peptide levels were used as an inclusion criterion, not the independent variable. 97 2. Hypothesis Testing Based on the study results, if the p-value for the primary end point is 0.27, which of the following is the correct conclusion? A) Reject the null hypothesis; rivaroxaban significantly reduces the primary end point B) Fail to reject the null hypothesis; rivaroxaban does not significantly reduce the primary end point C) Accept the alternative hypothesis; rivaroxaban is beneficial D) Reject both the null and alternative hypotheses because the p-value is above 0.05 98 2. Hypothesis Testing Answer: B Fail to reject the null hypothesis; rivaroxaban does not significantly reduce the primary endpoint. Rationale: • A p-value of 0.27 is greater than 0.05, meaning the results do not provide enough evidence to reject the null hypothesis • The null hypothesis states that there is no difference between rivaroxaban and placebo • Since the p-value is high, we cannot conclude a statistically significant benefit of rivaroxaban Why the other answers are wrong: A: A p-value of 0.27 does not allow us to reject the null hypothesis C: Accepting the alternative hypothesis is incorrect when the p-value is >0.05 D: We do not reject both hypotheses; we only fail to reject the null hypothesis 99 3. Absolute Risk Reduction (ARR) Calculate the absolute risk reduction (ARR) for the primary end point A) 1.2% B) 0.3% C) 25% D) 26.2% 100 3. Absolute Risk Reduction (ARR) Answer: A) 1.20% Rationale: • ARR = Event rate in placebo - Event rate in treatment • ARR = 26.2% - 25.0% • ARR = 1.20% Why the other answers are wrong: B (0.3%): Incorrect; this is the ARR for “all-cause mortality” (22.1% - 21.8%) C (25%): This is the event rate in the rivaroxaban group, not ARR D (26.2%): This is the event rate in the placebo group, not ARR 101 4. Relative Risk What is the relative risk (RR) of the primary end point for the rivaroxaban group compared to the placebo? A) 0.80 B) 1.05 C) 0.95 D) 1.20 102 4. Relative Risk Answer: C) 0.95 Rationale: • RR = Event rate in treatment group / Event rate in placebo group • RR = 25.0% / 26.2% • RR = 0.95 Interpretation: • Since RR < 1, rivaroxaban reduces the risk of the primary end point compared to placebo • However, the reduction is small (5% lower risk compared to placebo) Why the other answers are wrong: A) 0.80 → Would mean a 20% lower risk, which is a much larger effect than what the study found B) 1.05 → Would indicate higher risk in the rivaroxaban group (i.e., rivaroxaban increases events), which is incorrect D) 1.20 → Would indicate rivaroxaban increased risk by 20%, which is incorrect 103 5. Relative Risk Reduction What is the relative risk reduction (RRR) of the primary end point for the rivaroxaban group compared to placebo? A) 4.58% B) 4.80% C) 20% D) 50% 104 5. Relative Risk Reduction Answer: A) 4.58% Rationale: • RRR = (ARR / Event rate in placebo) × 100 • RRR = (26.2% - 25.0%) / 26.2% × 100 • RRR = 1.2% / 26.2% × 100 • RRR = 4.58% Interpretation: • Rivaroxaban reduces the risk of the primary endpoint by 4.58% relative to placebo. • This means that compared to placebo, rivaroxaban has a small but measurable effect in lowering the composite risk of death, MI, or stroke. Why the other answers are wrong: B) 4.80% → This divides the ARR by event rate in treatment (25%) rather than by 26.2% C) 20.0% → Would mean a much stronger risk reduction than observed. D) 50.0% → Would imply rivaroxaban halves the risk, which is far from the true effect. 105 6. Number needed to treat What is the number needed to treat (NNT)? A) 84 B) 83 C) 0.83 D) 1 106 6. Number needed to treat Answer: A) 83 Rationale: • NNT = 1 / ARR • NNT = 1 / (26.2% - 25.0%)= 1/1.2% = 1/(1.2/100) • NNT = 1 / 0.012 • NNT = 83.33= 84 (you should always round up NNT) Interpretation: • 84 patients need to be treated with rivaroxaban to prevent one additional primary endpoint event compared to placebo • A higher NNT (such as 84) indicates that the treatment effect is small, meaning most patients will not benefit significantly (1 is the ideal number. The smaller the NNT, the better the treatment effect) Why the other answers are wrong: • B) 83 → Should round-up NNT, not down • C) 0.83 → You should divide the 1.2 by 100 (i.e., 1.2%=1.2/100) • D) 1 → rounding up 0.83 to 1 is wrong since 0.83 itself was obtained from a wrong calculation 107 7. Number needed to harm If the major bleeding rate in the rivaroxaban group is 4.2% and in the placebo group is 2.3%, what is the number needed to harm (NNH)? A) 53 B) 52 C) 0.53 D) 0 108 7. Number needed to harm Answer: B) 52 Rationale: • NNH = 1 / Absolute Risk Increase (ARI) • NNH = 1 / (4.2% - 2.3%) = 1 / 1.9% = 1/ (1.9/100) • NNH = 1 / 0.019 = 52.63 = 52 (always round down NNH) Interpretation: • For every 52 patients treated with rivaroxaban, one patient is harmed compared to placebo • The ideal number is a number that is as high as possible. The higher the NNH, the less the harm caused by the treatment) Why the other answers are wrong: • A) 53 → Should round down NNH, not up • C) 0.53 → You should divide the 1.9 by 100 (i.e., 1.9%=1.9/100) • D) 0 → rounding down 0.53 to 0 is wrong since 0.53 itself was obtained from a wrong calculation 109 8. Odds ratio What is the odds ratio (OR) of the primary end point for the rivaroxaban group compared to the placebo group? Round answer to two decimal places A) 1.20 B) 1.06 C) 0.95 D) 0.94 110 8. Odds ratio Answer: D) 0.94 Rationale: • Odds = Number of events / Number of non-events • Odds in rivaroxaban group = 626 / (2507 - 626) = 626 / 1881 = 0.3329 • Odds in placebo group = 658 / (2515 - 658) = 658 / 1857 = 0.3544 • Odds Ratio (OR) = Odds in treatment / Odds in placebo • OR = 0.3329 / 0.3544 = 0.9393=0.94 Interpretation: • Since OR < 1, rivaroxaban is associated with lower odds of the primary endpoint compared to placebo • The reduction in odds is modest (~6%), indicating a small benefit of rivaroxaban Why the other answers are wrong: A) 1.20 → Would indicate rivaroxaban increases the odds of the primary endpoint, which is incorrect B) 1.06 → Incorrect formula: odds in placebo/odds in treatment (0.3544/0.3329), which is incorrect C) 0.95 → Not doing due diligence to calculate the number of non-events (626/2507)/(658/2515) 111 9. Type I and Type II Errors A researcher conducts a study evaluating the effectiveness of a new anticoagulant compared to standard therapy. The study finds no significant difference between the two treatments, but in reality, the new anticoagulant is superior. What type of error has occurred? A) Type I error B) Type II error C) Selection bias D) Confounding 112 9. Type I and Type II Errors Answer: B) Type II error Rationale: • A Type II error (β error) occurs when the null hypothesis is not rejected even though a true difference exists (false negative) • In this case, the new anticoagulant is truly more effective, but the study failed to detect it as significant Why the other answers are wrong: A (Type I error): Occurs when the null hypothesis is incorrectly rejected (false positive). That would mean concluding the drug is effective when it is not C (Selection bias): Happens when the study population is not representative of the target population, which is unrelated to hypothesis testing D (Confounding): Occurs when an external factor distorts the association between the intervention and the outcome 113 10. Confidence Intervals (CI) and Significance A clinical trial evaluating a new heart failure medication reports a relative risk (RR) of 0.85 with a 95% confidence interval (CI) of 0.72–1.04. Based on this, which of the following is the best interpretation? A) The new medication significantly reduces heart failure risk B) The new medication has no significant effect on heart failure risk C) The new medication increases heart failure risk D) The study was underpowered 114 10. Confidence Intervals (CI) and Significance Answer: B) The new medication has no significant effect on heart failure risk Rationale: • A confidence interval that crosses 1 (0.72–1.04) means that the effect could include no difference, so the result is not statistically significant Why the other answers are wrong: A: If the CI were entirely below 1, the drug would significantly reduce risk C: If the CI were entirely above 1, the drug would increase risk D: While underpowering could contribute, it is not confirmed solely by the CI crossing 1 115 11. Study design identification A researcher wants to evaluate the long-term effects of a new osteoporosis medication. They recruit 5,000 women, randomly assign them to treatment or placebo, and follow them for 10 years, recording hip fracture rates. What type of study is this? A) Case-control study B) Cross-sectional study C) Randomized controlled trial (RCT) D) Cohort study 116 11. Study design identification Answer: C) Randomized controlled trial (RCT) Rationale: • An RCT is a prospective study where participants are randomly assigned to groups and followed over time. • The randomization helps control for confounders Why the other answers are wrong: A (Case-control study): Retrospective, compares the odds of exposure in individuals with (cases) and without (control) an outcome B (Cross-sectional study): Observational, measures exposure and outcome at the same time D (Cohort study): Prospective like RCT and compares the risk of outcomes between the exposed (treatment) and unexposed (no treatment) groups. Unlike RCT, it has no randomization 117 12. Incidence vs. Prevalence A hospital study finds that 500 patients have active tuberculosis. Over the next year, 100 new cases are diagnosed. What is the incidence rate of tuberculosis? A) 100 new cases per year B) 500 cases in total C) 100 / 500 = 20% D) 100 / total population at risk 118 12. Incidence vs. Prevalence Answer: D) 100 / total population at risk Rationale: • Incidence = New cases / total population at risk during a specific period • The denominator should include those at risk but without disease at baseline Why the other answers are wrong: A: Just stating "100 new cases per year" is incomplete; we need the denominator B: 500 represents prevalence (total cases), not incidence C: 100/500 is incorrect unless 500 represents the population at risk 119 13. Intention to treat vs. Per-protocol analysis A clinical trial includes 1,000 patients, but 50 patients in the treatment group stop taking the medication. The researchers analyze all patients as originally assigned, regardless of medication adherence. What type of analysis is this? A) Per-protocol analysis B) Intention-to-treat (ITT) analysis C) Observational analysis D) Post-hoc analysis 120 13. Intention to treat vs. Per-protocol analysis Answer: B) Intention-to-treat (ITT) analysis Rationale: • ITT analysis includes all randomized participants, regardless of whether they followed the protocol • Preserves randomization, making it more real-world applicable Why the other answers are wrong: A (Per-protocol analysis): Includes only those who followed the study protocol C (Observational analysis): The study is an RCT, not observational D (Post-hoc analysis): Refers to unplanned analysis, not ITT vs. per-protocol 121 14. Hazard ratio interpretation A lung cancer drug trial reports a hazard ratio (HR) of 0.70 (95% CI: 0.55–0.88) for mortality. What does this indicate? A) The drug reduces the risk of death by 30% B) The drug increases the risk of death by 70% C) The drug has no significant effect D) The drug reduces lung cancer incidence 122 14. Hazard ratio interpretation Answer: A) The drug reduces the risk of death by 30% Rationale: • HR <1 means the treatment group has a lower hazard (risk) than the control • HR of 0.70 means a 30% reduction in the risk of death (1 - 0.70 = 30%) • CI (0.55–0.88) does not include 1, so it's statistically significant Why the other answers are wrong: B: A hazard ratio above 1 would indicate increased risk. HR=0.7 is not above 1 C: The CI does not include 1, so the result is significant D: The HR applies to mortality, not incidence 123 15. Power and sample size A researcher designs a study comparing two antihypertensive drugs. If the sample size is too small, what is the most likely consequence? A) Increased Type I error B) Increased Type II error C) Increased statistical power D) Increased external validity 124 15. Power and sample size Answer: B) Increased Type II error Rationale: • Underpowered studies (small sample size) have a higher risk of failing to detect a true difference, leading to a Type II error (false negative) • This means a real effect may be missed due to insufficient statistical power Why the other answers are wrong: A: Type I error (false positive) occurs when a difference is detected when none exists, not from low power C: Power is reduced in small sample sizes, not increased D: External validity is not directly impacted by sample size 125