Chi Square Tests PhD Özgür Tosun

IMPORTANCE OF EVIDENCE BASED MEDICINE

The Study Objective: To determine the quality of health recommendations and claims made on popular medical talk shows. Sources: Internationally syndicated medical television talk shows that air daily (The Dr Oz Show and The Doctors). Interventions: Investigators randomly selected 40 episodes of each of The Dr Oz Show and The Doctors from early 2013 and identified and evaluated all recommendations made on each program. A group of experienced evidence reviewers independently searched for, and evaluated as a team, evidence to support 80 randomly selected recommendations from each show. Main outcomes measures: Percentage of recommendations that are supported by evidence as determined by a team of experienced evidence reviewers.

Results On average, The Dr Oz Show had 12 recommendations per episode and The Doctors 11. At least a case study or better evidence to support 54% (95% confidence interval 47% to 62%) of the 160 recommendations (80 from each show). For recommendations in The Dr Oz Show, evidence supported 46%, contradicted 15%, and was not found for 39%. For recommendations in The Doctors, evidence supported 63%, contradicted 14%, and was not found for 24%. The most common recommendation category on The Dr Oz Show was dietary advice (39%) and on The Doctors was to consult a healthcare provider (18%). The magnitude of benefit was described for 17% of the recommendations on The Dr Oz Show and 11% on The Doctors

Conclusions Recommendations made on medical talk shows often lack adequate information on specific benefits or the magnitude of the effects of these benefits. Approximately half of the recommendations have either no evidence or are contradicted by the best available evidence. The public should be skeptical about recommendations made on medical talk shows.

A Fictional Answer for a Random Dr. Oz’s Recommendation Dr Oz: ◦ "Saturated fat is solid at room temperature, so that means it's solid inside your body." Patient: ◦ Thanks, Dr. Oz. You give the best advice. ◦ Carrots are very hard and dense, so they'll petrify (transform into stone) your body, turning you into an orange statue. Am I doing it right?

Pull down that bread, kiddo!!!

CATEGORICAL ONE SAMPLETWO SAMPLES>2 SAMPLES

CATEGORICAL ONE SAMPLETWO SAMPLES>2 SAMPLES IndependentPairedIndependent

CATEGORICAL ONE SAMPLE TWO SAMPLES>2 SAMPLES IndependentPairedIndependent 2 x 2 Chi Square Test Mc Nemar Test N x M Chi Square Test One Sample Chi Square Test Fisher’s Exact Test Nonparametric One sample difference of proportions test Parametric

Cross Table (Contingency Table) enables showing two or more variables simultaneously in table format a table of counts cross-classified according to categorical variables best way to include sub-group descriptive statistics simplest contingency table is a 2 x 2 table is good for demonstrating possible relationships among variables

Cross Table (Contingency Table) An r X c contingency table shows the observed frequencies for two variables. The observed frequencies are arranged in r rows and c columns. The intersection of a row and a column is called a cell

Misreading the Table it is important to correctly read the information given in a table although the original data do not change at all, tables can be arranged in several different views looking at the table does not necessarily show the reader about possible relationships among variables – in order to decide on the existence of relationship, «statistical hypothesis testing» is required

Observed versus Expected In a cross tabulation, the actual numbers in the cells of the table are called the observed values Observed Frequencies are obtained empirically through direct observation Theoretical, or Expected Frequencies are developed on the basis of some hypothesis

Expected Frequencies Assuming the two variables are independent, you can use the contingency table to find the expected frequency for each cell. Finding the Expected Frequency for Contingency Table Cells The expected frequency for a cell E r,c in a contingency table is

Example : Find the expected frequency for each “Male” cell in the contingency table for the sample of 321 individuals. Assume that the variables, age and gender, are independent Female and older Total Male Total51 – 6041 – 5031 – 4021 – 3016 – 20Gender Age

Expected Frequency Example continued : Female and older Total Male Total51 – 6041 – 5031 – 4021 – 3016 – 20Gender Age

Chi-Square Independence Test A chi-square independence test is used to test the independence of two variables. Using a chi-square test, you can determine whether the occurrence of one variable affects the probability of the occurrence of the other variable. For the chi-square independence test to be used, the following must be true. 1.The observed frequencies must be obtained by using a random sample. 2.Each expected frequency must be greater than or equal to 5.

Chi-Square Independence Test We are looking for significant differences between the observed frequencies in a table (f o ) and those that would be expected by random chance (f e )

2 x 2 Chi Square df = (r-1)(c-1)= O 11 2 TotalN First criteria Total Second Criteria O 12 O 21 O 22 O.1 O.2 O1.O1. O 2. E ij should be greater than or equal to 5.

Is squint more common among children with a positive family history? Is there an association between squint and family history of squint? Total Squint ( Şaşılık) Total Family History  2 (1,0.025) =5.024 > Accept H 0. There is no relation between squint and family history

Attention In 2 X 2 contingency tables, if any expected frequencies are less than 5, then alternative procedure to called Fisher’s Exact Test should be performed.

An Example A study was conducted to analyze the relation between coronary heart disease (CHD) and smoking. 40 patients with CHD and 50 control subjects were randomly selected from the records and smoking habits of these subjects were examined. Observed values are as follows:

Observed and expected frequencies + - Yes No Total 90 Smoking Total CHD

df = (r-1)(c-1)=(2-1)(2-1)=1  2 (1,0.05) =3.841 Conclusion: There is a relation between CHD and smoking.  2 =4. 95 > reject H 0

An Example for Fisher’s Exact Test Research question: does positive BRCA1 gene actually affects the occurrence of breast cancer?

Since the percentage of the cells which have expected count < 5 is 50%, Fisher’s exact test should be applied. According to Fisher’s test, p value is p>α Fail to reject H 0 BRCA1 gene has no affect on breast cancer

McNemar Test 35 patients were evaluated for arrhythmia with two different medical devices. Is there any statistically significant difference between the diagnose of two devices? Device I Device II Total Arrhythmia (+)Arrhythmia (-) Arrhythmia (+)10313 Arrhythmia (-)13922 Total231235

The significance test for the difference between two dependent population / McNemar test H 0 : P 1 =P 2 H a : P 1  P 2 Critical z value is ±1.96 Reject H 0

McNemar test approach:  2 (1,0.05) =3.841<5.1 p<0.05; reject H 0.

Evaluation of arrhythmia patients using these two devices will provide significantly different results. Further research is required to understand which one is better for diagnosis.

A researcher wants to know whether the mothers age is affecting the probability of having congenital abnormality of neonatals or not. The collected data is given in the table: Congenital abnormality Total PresentAbsent Age groups≤ > Total N x M Chi Square

H 0 : There is no relation between the age of mother and presence of congenital abnormality. Under the assumption that null hypothesis is true: (Expected count) Congenital abnormality PresentAbsent Age groups≤253 (7.2)22 (17.8) (12.1)34 (29.9) >3518 (9.8)16 (24.2)

Reject H 0

Congenital abnormalityχ2χ2 PresentAbsent Age groups ≤253 (7.2)22 (17.8) (12.1)34 (29.9)1,95 >3518 (9.8)16 (24.2)9,64 Omit the >35 age group

Congenital abnormality PresentAbsent Age groups ≤ H 0 is accepted

At the end of the analysis, we should conclude that the risk of having a baby with congenital abnormality is significantly higher for >35 age group. However, risk is not differing significantly between <= 25 age group and age group

Attention In N x M contingency tables, if the proportion of cells those have expected frequencies less than 5 is above 20%, then it is not possible to perform any statistical analysis

EXAMPLE: Researcher wants to know if there is any significant difference among education groups in terms of their alcohol consumption rates

At the end of the analysis, since the proportion of cells which have expected count <5 is 50%, we must conclude that this hypothesis cannot be tested under this circumstances. The samples size in the study is not high enough. Calculated p value is not valid.