SMHS AssociationTests - Revision history

Dinov at 14:15, 6 November 2014

2014-11-06T14:15:33Z

Dinov: /* Fisher’s exact test */

2014-09-15T17:03:47Z

‎Fisher’s exact test

Dinov: /* Fisher’s exact test */

2014-09-15T17:03:29Z

‎Fisher’s exact test

Dinov: /* Software */

2014-09-06T19:48:41Z

‎Software

Dinov: /* Software */

2014-09-06T19:48:26Z

‎Software

Dinov: /* [AP_Statistics_Curriculum_2007_Contingency_Indep|Chi-square test]] */

2014-09-05T22:41:58Z

‎[AP_Statistics_Curriculum_2007_Contingency_Indep|Chi-square test]]

Dinov: /* Theory */

2014-09-05T22:41:44Z

‎Theory

Dinov: /* Theory */

2014-09-05T22:29:45Z

‎Theory

Dinov: /* Theory */

2014-09-05T22:16:45Z

‎Theory

Clgalla at 18:16, 2 September 2014

2014-09-02T18:16:12Z

← Older revision		Revision as of 14:15, 6 November 2014
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	The case frequencies were the same between the two studies but control frequencies were different. We fail to reject the null hypothesis of no association and failed to find association suggesting that the first study was false positive.		The case frequencies were the same between the two studies but control frequencies were different. We fail to reject the null hypothesis of no association and failed to find association suggesting that the first study was false positive.
		+
		+	* [[AP_Statistics_Curriculum_2007#Chapter_XIII:_Multinomial_Experiments_and_Contingency_Tables\|See also the Chi-Square tests for association and model-fitting quiality]].

	====[[SOCR_EduMaterials_AnalysisActivities_Fisher_Exact\|Fisher’s exact test]]====		====[[SOCR_EduMaterials_AnalysisActivities_Fisher_Exact\|Fisher’s exact test]]====

@@ Line 87: / Line 87: @@
 Fisher showed the probability of obtaining any such set of values was given by the hyper-geometric distribution:
-$$p=\frac{{a+b\choose a}{c+d\choose c}}{n\choose a+c}$ =$\frac{(a+b)!(c+d)!(a+c)!(b+d)!} {a! b! c! d! n!}$=$\frac{{10\choose1} {14\choose11}} {24\choose12}$ = $\frac{10!\,14!\,12!\,12!} {11!\,9!\,11!\,3!\,24!}\sim 0.00135.$$
+$p=\frac{{a+b\choose a}{c+d\choose c}}{n\choose a+c}$ =$\frac{(a+b)!(c+d)!(a+c)!(b+d)!} {a! b! c! d! n!}$=$\frac{{10\choose1} {14\choose11}} {24\choose12}$ = $\frac{10!\,14!\,12!\,12!} {11!\,9!\,11!\,3!\,24!}\sim 0.00135.$
 This is the exact hyper-geometric probability of observing this particular arrangement of the data assuming the given marginal totals on the null hypothesis that men and women are equally likely to be dieters. The smaller the p value, the greater the evidence to reject the null hypothesis, so we have significant evidence to reject the null hypothesis and conclude that women and men are not equally likely to be dieters.

@@ Line 85: / Line 85: @@
 |}
 </center>
-Fisher showed the probability of obtaining any such set of values was given by the hyper-geometric distribution
+Fisher showed the probability of obtaining any such set of values was given by the hyper-geometric distribution:
-$p=\frac{{a+b\choose a}{c+d\choose c}}{n\choose a+c}$ =$\frac{(a+b)!(c+d)!(a+c)!(b+d)!} {a! b! c! d! n!}$=$\frac{{10\choose1} {14\choose11}} {24\choose12}$ = $\frac{10!\,14!\,12!\,12!} {11!\,9!\,11!\,3!\,24!}\sim 0.00135$
+$$p=\frac{{a+b\choose a}{c+d\choose c}}{n\choose a+c}$ =$\frac{(a+b)!(c+d)!(a+c)!(b+d)!} {a! b! c! d! n!}$=$\frac{{10\choose1} {14\choose11}} {24\choose12}$ = $\frac{10!\,14!\,12!\,12!} {11!\,9!\,11!\,3!\,24!}\sim 0.00135.$$
-this is the exact hyper-geometric probability of observing this particular arrangement of the data assuming the given marginal totals on the null hypothesis that men and women are equally likely to be dieters. The smaller the p value, the greater the evidence to reject the null hypothesis, so we have significant evidence to reject the null hypothesis and conclude that women and men are not equally likely to be dieters.
+This is the exact hyper-geometric probability of observing this particular arrangement of the data assuming the given marginal totals on the null hypothesis that men and women are equally likely to be dieters. The smaller the p value, the greater the evidence to reject the null hypothesis, so we have significant evidence to reject the null hypothesis and conclude that women and men are not equally likely to be dieters.
 ====Randomized Controlled Trials (RCT)====

@@ Line 214: / Line 214: @@
 ===Software===
-[http://stat.ethz.ch/R-manual/R-patched/library/stats/html/cor.test.html
+[http://stat.ethz.ch/R-manual/R-patched/library/stats/html/cor.test.html cor.test(x, y, …)]  ## test for association/correlation between paired samples
 [http://www.socr.ucla.edu/htmls/ana/FishersExactTest_Analysis.html  Fishers Exact Test]

@@ Line 215: / Line 215: @@
 ===Software===
 [http://stat.ethz.ch/R-manual/R-patched/library/stats/html/cor.test.html
-cor.test(x, y, …)  ## test for association/correlation between paired samples
+cor.test(x, y, …)]  ## test for association/correlation between paired samples
-[http://www.socr.ucla.edu/htmls/ana/FishersExactTest_Analysis.html   Fishers Exact Test]
+[http://www.socr.ucla.edu/htmls/ana/FishersExactTest_Analysis.html  Fishers Exact Test]
-[http://cran.r-project.org/web/packages/AssotesteR/AssotesteR.pdf   Package ‘AssotesteR’]
+[http://cran.r-project.org/web/packages/AssotesteR/AssotesteR.pdf  Package ‘AssotesteR’]
 [http://cran.r-project.org/web/packages/HSAUR/vignettes/Ch_conditional_inference.pdf A Handbook of Statistical Analyses   Using R]
 ===Problems===

@@ Line 18: / Line 18: @@
 We are going to interpret the measurement results and conclude about the association between variables through examples in different types of trials.
-====[AP_Statistics_Curriculum_2007_Contingency_Indep|Chi-square test]]====
+====[[AP_Statistics_Curriculum_2007_Contingency_Indep|Chi-square test]]====
 The Chi-square test is a non-parametric test of statistical significance of two variables. It tests if the measured factor is associated with the members in one of two samples with chi square. For example, the chi-square test tests whether there is statistical evidence that the measured factor is not randomly distributed in the cases compared to the controls in a case-control study. The test statistic is $X_{o}^{2}=\sum_{i=1}^{n}\frac{(O_i-E_i )^{2}}{E_{i}} \sim χ_{df}^{2}$, where $E_{i}$ is the expected frequency under the null hypothesis and $O_{i}$ is the observed frequency, n is the number of cells in the table and $ df=(\#\,rows-1)(\#\,columns-1)$, $E=\frac{row\,total*column\, total} {grand total}$. The null hypothesis is that there is no association between exposure group the disease studied.
 *Conditions for validity of the $χ^{2}$test are:

@@ Line 18: / Line 18: @@
 We are going to interpret the measurement results and conclude about the association between variables through examples in different types of trials.
-'''Chi-square test:''' a non-parametric test of statistical significance of two variables. It tests if the measured factor is associated with the members in one of two samples with chi square. For example, the chi-square test tests whether there is statistical evidence that the measured factor is not randomly distributed in the cases compared to the controls in a case-control study. The test statistic is $X_{o}^{2}=\sum_{i=1}^{n}\frac{(O_i-E_i )^{2}}{E_{i}} \sim χ_{df}^{2}$, where $E_{i}$ is the expected frequency under the null hypothesis and $O_{i}$ is the observed frequency, n is the number of cells in the table and $ df=(\#\,rows-1)(\#\,columns-1)$, $E=\frac{row\,total*column\, total} {grand total}$. The null hypothesis is that there is no association between exposure group the disease studied.
+====[AP_Statistics_Curriculum_2007_Contingency_Indep|Chi-square test]]====
 *Conditions for validity of the $χ^{2}$test are:
 **Design conditions
@@ Line 39: / Line 40: @@
 |}
 </center>
 The [[SMHS_OR_RR|Odds Ratio (OR) of a disease (D) in relative to exposure (E)]] is $OR =  \frac{\frac{P \left( D \mid E \right)}{1 - P \left( D \mid E \right)}}{\frac{P \left( D \mid E' \right)}{1 - P \left( D \mid E' \right)}}$, where $E'$ is the complement of exposure (''Reference group''). Thus,
@@ Line 69: / Line 69: @@
 The case frequencies were the same between the two studies but control frequencies were different. We fail to reject the null hypothesis of no association and failed to find association suggesting that the first study was false positive.
-'''Fisher’s exact test:''' a statistical significance test used in the analysis of contingency tables and is valid for all sample sizes especially when the sample size is small. Suppose we want to study if the proportion of dieting is higher among women than among men and the following data is collected:
+====[[SOCR_EduMaterials_AnalysisActivities_Fisher_Exact|Fisher’s exact test]]====
 <center>
@@ Line 90: / Line 91: @@
 this is the exact hyper-geometric probability of observing this particular arrangement of the data assuming the given marginal totals on the null hypothesis that men and women are equally likely to be dieters. The smaller the p value, the greater the evidence to reject the null hypothesis, so we have significant evidence to reject the null hypothesis and conclude that women and men are not equally likely to be dieters.
-'''Randomized Controlled Trials (RCT):''' Investigator assigns exposure at random to study participants, investigator then observes if there are differences in health outcomes between people who were (treatment group) and were not (comparison group) exposed to the facto. Special care is taken in ensuring that the follow-up is done in an identical way in both groups. The essence of good comparison between “treatments” is that the compared groups are the same except for the “treatment”.
+====Randomized Controlled Trials (RCT)====
 *Steps of a RCT: hypothesis formed; study participant recruited based on specific criteria and their informed consent is sought; eligible and willing participants randomly allocated to receive assignment to a particular study group; study groups are monitored for outcome under study; rates of outcome in the various groups are compared:
@@ Line 96: / Line 98: @@
 [[Image:Association_Tests_Fig1.png|500px]]
-'''External and internal validity:'''
+====External and internal validity====
 External validity: Generalization of study to larger source population. Influenced by factors like: demographic differences between eligible and ineligible subgroups; intervention mirror what will happen in the community or source population.
 Internal validity: Ability to reach correct conclusion in study. Influenced by factors like: ability of subjects to provide valid and reliable data; expected compliance with a regimen; low probability of dropping out.

@@ Line 40: / Line 40: @@
 </center>
-$Odds\,ratio=1.97,(95\%\, CI:1.29-3.00)$, $χ^{2}=\frac{(89-74.10)^{2}}{74.10}$+$\frac{(119-133.90)^{2}}{133.90}$+$\frac{(51-65.90)^{2}}{65.90}$+$\frac{(134-119.10)^{2}}{119.10}=9.89$,$(p=0.0017)$
 Hence, we have significant evidence that we can reject the null hypothesis and conclude that there is statistically significant association between the exposure group and case and control status in the sample. Cases with AD are about 2 times as likely to carry the exposure group genotype as the controls.

@@ Line 40: / Line 40: @@
 </center>
-$Oodds\,ratio=1.97,(95\%\, CI:1.29-3.00)$, $χ^{2}=\frac{(89-74.10)^{2}}{74.10}$+$\frac{(119-133.90)^{2}}{133.90}$+$\frac{(51-65.90)^{2}}{65.90}$+$\frac{(134-119.10)^{2}}{119.10}=9.89$,$(p=0.0017)$
+$Odds\,ratio=1.97,(95\%\, CI:1.29-3.00)$, $χ^{2}=\frac{(89-74.10)^{2}}{74.10}$+$\frac{(119-133.90)^{2}}{133.90}$+$\frac{(51-65.90)^{2}}{65.90}$+$\frac{(134-119.10)^{2}}{119.10}=9.89$,$(p=0.0017)$
 Hence, we have significant evidence that we can reject the null hypothesis and conclude that there is statistically significant association between the exposure group and case and control status in the sample. Cases with AD are about 2 times as likely to carry the exposure group genotype as the controls.
@@ Line 195: / Line 195: @@
 $Prevalence\,Ratio\,=\frac{Prevalence\,of\,disease\,in\,exposed} {Prevalence\,of\, disease\,in\,unexposed}=\frac{a/(a+b)}{c/(c+d)}$
 $Odds\,Ratio\,=\frac{Odds\,of\,diseases\,in\,exposed}{Odds\,of\,disease\,in\,unexposed}=\frac{a/b}{c/d}=\frac{ad}{cd}$
 ===Applications===

@@ Line 18: / Line 18: @@
 We are going to interpret the measurement results and conclude about the association between variables through examples in different types of trials.
-'''Chi-square test:''' a non-parametric test of statistical significance of two variables. It tests if the measured factor is associated with the members in one of two samples with chi square. For example, the chi-square test tests whether there is statistical evidence that the measured factor is not randomly distributed in the cases compared to the controls in a case-control study. The test statistic is $X_{o}^{2}=\sum_{i=1}^{n}\frac{(O_i-E_i )^{2}} {E_{i}} \sim χ_{df}^{2}$, where $E_{i}$ is the expected frequency under the null hypothesis and $O_{i}$ is the observed frequency, n is the number of cells in the table and $ df=(\#\,rows-1)(\#\,columns-1)$, $E=\frac{row\,total*column\, total} {grand total}$. The null hypothesis is that there is no association between exposure group the disease studied.
+'''Chi-square test:''' a non-parametric test of statistical significance of two variables. It tests if the measured factor is associated with the members in one of two samples with chi square. For example, the chi-square test tests whether there is statistical evidence that the measured factor is not randomly distributed in the cases compared to the controls in a case-control study. The test statistic is $X_{o}^{2}=\sum_{i=1}^{n}\frac{(O_i-E_i )^{2}}{E_{i}} \sim χ_{df}^{2}$, where $E_{i}$ is the expected frequency under the null hypothesis and $O_{i}$ is the observed frequency, n is the number of cells in the table and $ df=(\#\,rows-1)(\#\,columns-1)$, $E=\frac{row\,total*column\, total} {grand total}$. The null hypothesis is that there is no association between exposure group the disease studied.
 *Conditions for validity of the $χ^{2}$test are:
 **Design conditions
@@ Line 82: / Line 82: @@
 Fisher showed the probability of obtaining any such set of values was given by the hyper-geometric distribution
-$p=(■(a+b@a))(■(c+d@c))/((■(n@a+c)) )=(a+b)!(c+d)!(a+c)!(b+d)!/(a! b! c! d! n!)=(■(10@1))(■(14@11))/((■(24@12)) )=10!14!12!12!/11!9!11!3!24!≈0.00135
+$p=\frac{{a+b\choose a}{c+d\choose c}}{n\choose a+c}$ =$\frac{(a+b)!(c+d)!(a+c)!(b+d)!} {a! b! c! d! n!}$=$\frac{{10\choose1} {14\choose11}} {24\choose12}$ = $\frac{10!\,14!\,12!\,12!} {11!\,9!\,11!\,3!\,24!}\sim 0.00135$
 this is the exact hyper-geometric probability of observing this particular arrangement of the data assuming the given marginal totals on the null hypothesis that men and women are equally likely to be dieters. The smaller the p value, the greater the evidence to reject the null hypothesis, so we have significant evidence to reject the null hypothesis and conclude that women and men are not equally likely to be dieters.
@@ Line 115: / Line 114: @@
 </center>
- Relative Risk=Cumulative Incidence in exposed/Cumulative Incidence in unexposed)=ratio of risks=Risk Ratio=a/(a+b))/c/(c+d))=〖CI〗_drugA/〖CI〗_placebo
+$Relative\,Risk\,=\frac{Cumulative\,Incidence\,in\,exposed}{Cumulative\,Incidence\,in\, unexposed}=ratio\,of\,risks=Risk\, Ratio=\frac{a/(a+b)}{c/(c+d)}=\frac{CI_{drugA}}{CI_{placebo}}$
- Rate Ratio=(Incidence rate in exposed)/(Incidence rate in unexposed
+$Rate\,Ratio=\frac{Incidence\,rate\,in\,exposed} {Incidence\,rate\,in\,unexposed}$
 ''Interpretation:'' RR>1, The risk of X is RR times more likely to occur in group A than in group B; RR=1, Null value (no difference between groups); RR<1, Either calculate the reduction in risk ratios (100%-xx%) or invert (1/RR) to be interpreted as “less likely” risk.
- Efficacy=(C.I.in placebo-C.I.rate in the treatment)/(C.I.Rate in placebo group)
+$Efficacy=\frac{C.I.\,in\, placebo\,-\,C.I.\,Rate\,in\,the\,treatment} {C.I.\,Rate\,in\,placebo\,group}$
 *Situations that favor the use of RCT: (1) Exposure of interest is a modifiable factor over which individuals are willing to relinquish control; (2) Legitimate uncertainty exists regarding the effect of interventions on outcome, but reasons exist to believe that the benefits of the intervention in question overweight the risks; (3) Effect of intervention on outcome is of sufficient importance to justify a large study.
@@ Line 160: / Line 159: @@
 </center>
- Odds Ratio=(odds of a case being exposed)/(odds of a control being exposed)=(a/c)/(b/d)=ad/bc.
+$Odds\,Ratio\,=\frac{odds\,of\,a\,case\,being\,exposed} {odds\,of\,a\,control\,being\,exposed}=\frac{a/c} {b/d}=\frac {ad}{bc}.$
 ''Interpretation:'' Odds of being exposed is OR times higher (if OR > 1) in the cases than the controls (1/OR times lower (if OR < 1) in the cases than the controls; No association – odds is the same in cases and controls (if OR = 1)).
@@ Line 194: / Line 193: @@
 </center>
- Prevalence Ratio=(Prevalence of disease in exposed)/(Prevalence of disease in unexposed)=(a/(a+b))/(c/(c+d))
+$Prevalence\,Ratio\,=\frac{Prevalence\,of\,disease\,in\,exposed} {Prevalence\,of\, disease\,in\,unexposed}=\frac{a/(a+b)}{c/(c+d)}$
- Odds Ratio=(Odds of diseases in exposed)/(Odds of disease in unexposed)=(a/b)/(c/d)=ad/bc
+$Odds\,Ratio\,=\frac{Odds\,of\,diseases\,in\,exposed}{Odds\,of\,disease\,in\,unexposed}=\frac{a/b}{c/d}=\frac{ad}{cd}$
@@ Line 262: / Line 261: @@
 d. 1.19
 .5) Calculate the prevalence ratio of being fully immunized, comparing children on Medicaid to children not on Medicaid