This is straightforward for the odds ratio. The three effect size measures described above can also be used for Fisher’s Exact test. The array formula =ODDS_RATIO(B5:C6,TRUE) returns the results shown in range B10:C13 of Figure 3. If lab = TRUE, then an extra column of labels is appended to the result (default FALSE) the default for alpha is. for ln OR and the lower and upper limits of the 1- alpha confidence interval. ODDS_RATIO(R1, lab, alpha): returns a column array with the odds ratio for the 2 × 2 contingency table in R1, s.e. Real Statistics Function: The following function is provided in the Real Statistics Resource Pack: The 95% confidence interval for the odds ratio of 2.52 in Example 1 is (1.15, 5.53), as shown in Figure 3.įigure 3 – 95% confidence interval for odds ratio Worksheet Functions Thus, an estimate of the 1 – α confidence interval is Observation: For a 2 × 2 contingency table with entries the standard error of the natural log of the odds ratio is (2000), the odds ratio can be reinterpreted as a Cohen’s d effect size by using the formula The ratio 2.52 is the odds ratio.Īccording to Chinn, S. 8947/.3548 = 2.52 times greater for therapy 2 than for therapy 1. This means that the odds of remaining uncured is. The odds of a person who took therapy 2 is 51 to 57 or. The odds of a person who took therapy 1 remaining uncured is 11 to 31 or. This is a meaningful measure of effect size, called the risk ratio or relative risk.Ī related measure of effect size is the odds ratio. 4722/.2619 = 1.80 times as likely as those taking therapy 1 to remain uncured. This shows that those taking therapy 2 are. In fact, 11/42 = 26.19% of the people who took therapy 1 were not cured, while 51/108 = 47.22% of those who took therapy 2 were not cured. Odds Ratioįor a 2 × 2 contingency table, we can also use the odds ratio measure of effect size as described in the following example.Įxample 1: Calculate the odds ratio for the data in Example 2 of Independence Testing.Īs we saw in Example 2 of Independence Testing, there is a significant difference between those taking therapy 1 and those taking therapy 2. 21 (with df* = 2), which should be viewed as a medium effect. These guidelines are equivalent to those shown in Figure 1.Īs we saw in Figure 4 of Independence Testing, Cramer’s V for Example 1 of Independence Testing is. The guidelines (Cohen) are the same as for the equivalent phi value i.e. Where df* = min( r – 1, c – 1) and r = the number of rows and c = the number of columns in the contingency table. Cramer’s VĬramer’s V is an extension of the phi effect size for non 2 × 2 contingency tables, and is calculated as Commonly phi is denoted w when used in this way. Phi is the measure of effect size that is used in power calculations even for contingency tables that are not 2 × 2 (see Power of Chi-square Tests). Phi is equivalent to the correlation coefficient r, as described in Correlation. Phi φįor a 2 × 2 contingency table, phi is the commonly used measure of effect size, and is defined by We also describe the effect size for Fisher’s exact test. > pchisq(2.8302,df=1,lower.We review three different measures of effect size for the chi-square goodness-of-fit and independence tests, namely Phi φ, Cramer’s V, and the Odds Ratio. The test statistic is correct, but the df isn't (because the function doesn't know there's an estimated parameter) So the test is for $(M,MN,N)$ being the counts of the 0,1,2 outcomes from a Binom $(2,p)$ distribution (with estimated $p$) vs the alternative of a general trinomial distribution.Ĭhisq.test() is not ideally suited to this: it's designed for either a contingency table or a fixed set of probabilities.Īnd now, finally, the test > chisq.test(c(M, MN, N), p=c(M.exp, MN.exp, N.exp), rescale.p=TRUE) In this case, the $M$ group has MM genotype and the $N$ group has NN genotype (would have been helpful to say this up front). Statistically, its the condition that the number of alleles of each type for an individual has a Binom $(2,p)$ distribution. Genetically, that's the condition that the two alleles at a locus are independent. The question is about testing for Hardy-Weinberg equilibrium.
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