# Tag Info

9

...the relationship is nonlinear yet there is a clear relation between x and y, how can I test the association and label its nature? One way of doing this would be to fit $y$ as a semi-parametrically estimated function of $x$ using, for example, a generalized additive model and testing whether or not that functional estimate is constant, which would ...

7

It is best to get the data into a normalized form where smoking and heart attack are separate, parallel columns and other columns provide the identifying (key) fields: country year smoking heart 1 Congo 1988 1200 900 2 Congo 1984 1146 400 3 Congo 2010 675 550 4 Nigeria 1988 1100 950 5 Nigeria 1984 ...

6

I am going to interpret your question as one regarding a hypothesis on the population quantity $\tau$. If this is not what you intended, please comment to that effect and I will revise the answer accordingly. Definition and equivalent expressions Let $(X,Y)$ by a bivariate random vector with a continuous joint distribution function and let ...

6

Arguably, the question is not very precise. Rather than enumerating all measures of association for $2\times 2$ tables, I shall concentrate on the way such measures may be constructed and how to select the one that is most appropriate with respect to hypothesis or constraints relevant to a cross-classification. The very first questions to ask are: what ...

6

Conover (1999:202) suggested that the expected values can be "as small as 0.5, as long as most are greater than 1.0, without endangering the validity of the test." He also provides a "rule of thumb" from Cochran (1952) which suggested that if expected values are less than 1 or if more than 20% are less than 5, the test may perform poorly. However, Conover ...

5

The $\chi^2$-test was originally devised by Pearson as an approximation to the log-likelihood ratio, due to the fact that log-likelihoods were too computationally intensive for the time. Pearson's G is defined as $G = 2\sum_{ij}O_{ij}\ln(O_{ij}/E_{ij})$. It follows the same distribution as the corresponding $\chi^2$-test. (Forgot to mention originally: G ...

5

How many variables are present in your cross-classification will determine the degrees of freedom of your $\chi^2$-test. In your case, your are actually cross-classifying two variables (period and country) in a 2-by-3 table. So the dof are $(2-1)\times (3-1)=2$ (see e.g., Pearson's chi-square test for justification of its computation). I don't see where you ...

4

A few thoughts: There are many different binary-binary and ordinal-ordinal measures of association. SPSS provides names and algorithms for many of them under proximities and crosstabs. I'm also intrigued by tetrachoric (binary-binary) and polychoric (ordinal-ordinal) correlations that aim to estimate the correlation between theorised latent continuous ...

4

Linear or monotonic trend tests--$M^2$ association measure, WMW test cited by @GaBorgulya, or the Cochran-Armitage trend test--can also be used, and they are well explained in Agresti (CDA, 2002, §3.4.6, p. 90). The latter is actually equivalent to a score test for testing $H_0:\; \beta = 0$ in a logistic regression model, but it can be computed from the ...

4

For a moment, let's ignore the continuous/discrete issue. Basically correlation measures the strength of the linear relationship between variables, and you seem to be asking for an alternative way to measure the strength of the relationship. You might be interested in looking at some ideas from information theory. Specifically I think you might want to ...

4

I am a little confused; your title says "correlation" but your post refers to t-tests. A t-test is a test of central location - more specifically, is the mean of one set of data different from the mean of another set? Correlation, on the other hand, shows the relationship between two variables. There are a variety of correlation measures, it seems that ...

4

We know that Jaccard (computed between any two columns of binary data $\bf{X}$) is $\frac{a}{a+b+c}$, while Rogers-Tanimoto is $\frac{a+d}{a+d+2(b+c)}$, where a - number of rows where both columns are 1 b - number of rows where this and not the other column is 1 c - number of rows where the other and not this column is 1 d - number of rows where both ...

4

This is just a different definition of the statistic $T$. Call your statistic $T_1$ and the other $T_2$. Note the $T_2 = T_1/N$ and that is the reason that the variance of $T_2$ differs from $T_1$ by a factor of $1/N^2$. However you should note that the chi square stitistic is the same in either case. For $T_2$ there is a factor of $1/N^2$ in the ...

4

If the nonlinear relationship had been monotonic rank correlation (Spearman's rho) would be appropriate. In your example there is a clear small region where the curve changes from monotoncally increasing to montonically decreasing like a parabola would do at the point where the first derivative equals $0$. I think if you have some modeling knowledge (beyond ...

3

You can't, unless you know the sample size. A $\chi^2$ test is a test of whether observed data are plausibly generated from an underlying population with no association between the two variables. Unless you know how many data were actually observed you can't make any conclusions about whether they are plausibly from that null-hypothesis distribution. If ...

3

Let's simplify things. With N = 4,000 for cholesterol level, you should have no problem with your results being biased by outliers. Therefore you can use correlation itself, as implied by your initial sentence. It will make little difference whether you assess correlation via the Pearson, Spearman, or Point-Biserial method. If instead you really need to ...

3

On a 2x3 contingency table where the three-level factor is ordered you may use rank correlation (Spearman or Kendall) to assess association between the two variables. You may also think about the data as an ordered variable observed in two groups. A corresponding significance test could be the Mann-Whitney test (with many ties). This has an associated ...

3

As a follow-up to my comment, if independence.test refers to coin::independence_test, then you can reproduce a Cochrane and Armitage trend test, as it is used in GWAS analysis, as follows: > library(SNPassoc) > library(coin) > data(SNPs) > datSNP <- setupSNP(SNPs,6:40,sep="") > ( tab <- xtabs(~ casco + snp10001, data=datSNP) ) ...

3

Technically to compute a dis(similarity) measure between individuals on nominal attributes most programs first recode each nominal variable into a set of dummy binary variables and then compute some measure for binary variables. There are many measures for binary variables, however, not all of them logically suit dummy binary variables, i.e. former nominal ...

3

There exist many such coefficients (most are expressed here). Just try to meditate on what are the consequences of the differences in formulas, especially when you compute a matrix of coefficients. Imagine, for example, that objects 1 and 2 similar, as objects 3 and 4 are. But 1 and 2 have many of the attributes on the list while 3 and 4 have only few ...

2

One way to incorporate the ordering of the column factor into your analysis is to use the cumulative frequencies instead of the cell frequencies. So in your table you have: $$f_{ij}=\frac{n_{ij}}{n_{\bullet\bullet}}\;\;\;\; i=1,2\;\;j=1,2,3$$ where a "$\bullet$" indicates sum over that index. So I suggesting modeling instead: ...

2

Try measuring dependences of $(s, b)$ and $(s,c)$ for fixed $a$. For your case ($g$ and $h$ are monotonic real-valued functions) perhaps the best way is to use Spearman's rank correlation coefficient. So you get the coefficients $\rho_{s,b}(a)$ and $\rho_{s,c}(a)$ as functions of $a$. If there is such switching you get $$\max\{|\rho_{s,b}(a)|, ... 2 This is correct. In the epidemiology or social science, we would like to find the causal association between, say exposure and outcome. Then the most important thing is to identify the confounding factors, which need to be adjusted in your multivarate model settings. This does not necessarily mean to fit a model well, but only for adjustment purpose which ... 2 If the categorical variable is ordinal and you bin the continuous variable into a few frequency intervals you can use Gamma. Also available for paired data put into ordinal form are Kendal's tau, Stuart's tau and Somers D. These are all available in SAS using Proc Freq. I don't know how they are computed using R routines. Here is a link to a presentation ... 2 Eta is about the proportion of variance explained. If you have an ordinal outcome, you don't have a variance, so I'd say no. Here's some more explanation. Variance is about how different the scores are. So: 1.1, 1.2, 1.3 has a small difference , hence a large variance. 1, 101, 201 has larger differences, hence larger variance. 1, 2, 10001 has even ... 2 Here's how you'd normally set these things out: Original table of counts: Column variable 1 2 3 4 Row 1 1 3 10 6 variable 2 2 3 10 7 3 1 6 14 12 4 0 1 9 ... 2 Since your outcome appears to be dichotomous in nature (absence or presence) and you have numerous predictors of interest, why not calculate unadjusted odds ratios by performing simple logistic regression between each of your potential predictor variables of interest with the outcome (1 = absence; 0 = presence, or vice versa). If your sample size permits ... 2 You can simulate the first variable independently i.e. based on the marginal distribution provided by table(mtcarsgear) as you have done. However, the second variable should then be simulated conditioned on the value of the first i.e. based on table(mtcarsvs[which(mtcarsgear==gearSim[i])]) , which uses only the subset of vs values for the correct ... 1 In a word yes. In a Bayesian framework, a good feature (or feature set) X is one where P(Y | X ) depends significantly on the specific value of X observed; which can be reasonably interpreted as the variables being correlated. One way to quantify this is in terms of the relative entropy between the distributions. 1 The correlation coefficient between two variables X and Y is just Cov(X,Y)/[√Var(X)√Var(Y)] and Cov(X,Y) = E[(X-m_1)(Y-m_2)] where m_1 and m_2 are the respective means for X and Y. Given paired observations (X_i,Y_i) for i=1,2,..,n Cov(X,Y) is estimated by ∑ (X_i-m_1$$_b$) (Yi-m$_2$$_b)/n where m_1$$_b$= ∑X$_i\$/n and ...

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