Solved – Difference estimate & confidence intervals for $\chi^2$ test between 2 proportions

chi-squared-testconfidence intervalhypothesis testingproportion;

I am using a chi-sq test between two proportions (http://statistic-on-air.blogspot.com/2009/07/comparison-of-two-proportions.html). I am using this over the z-test for proportions because I do not think my data is normal.

My question is how do I get the difference (shift) estimate & confidence intervals for this test.

For example, 

Data1: 193/252=.77  
Data2: 154/227=.68

P-value=.032 (I used the formula in the link above. I will also add my exact code below)

Shift Estimate: I assume its the difference between proportions, ie .77-.68=.09. Is this correct?

Confidence Interval: ??????????????????

Code for generating P-value (matlab): 

% Pooled estimate of proportion
 p0 = (n1+n2) / (N1+N2);
% Expected counts under H0 (null hypothesis)
 n10 = N1 * p0;
 n20 = N2 * p0;
% Chi-square test, by hand
 observed = [n1 N1-n1 n2 N2-n2];
 expected = [n10 N1-n10 n20 N2-n20];
 chi2stat = sum((observed-expected).^2 ./ expected);
 p = 1 - chi2cdf(chi2stat,1);
 H=0; if(p<.05), H=1; end

Best Answer

Just for statistically intimidated people (wait, it should be 'intimidated by statistics') that can identify themselves with this joke:

A patient asks his surgeon what the odds are of him surviving an impending operation. The doctor replies that the odds are usually 50-50. "But there is no need to worry," the doctor explains.

"The first fifty have already died.

It is meant to be a quick reference for those confused with the terminology and overlapping tests - I am not addressing confidence intervals.

COMPARING PROPORTIONS BETWEEN SAMPLES:

I will use the following toy tabulated data:

Antacid <- matrix(c(64, 178 - 64, 92, 190 - 92), nrow = 2)    
Antacid_marginals <- matrix(c(64, 178 - 64, 92, 190 - 92), nrow = 2)
    Antacid_marginals <- rbind(Antacid_marginals, margin.table(Antacid_marginals,2))
    Antacid_marginals <- cbind(Antacid_marginals, margin.table(Antacid_marginals,1))
    dimnames(Antacid_marginals) = list(Symptoms = c("Heartburn", "Normal","Totals"),
                                       Medication = c("Drug A", "Drug B", "Totals"))

This is what it looks like (with marginals):

           Medication
Symptoms    Drug A Drug B Totals
  Heartburn     64     92    156
  Normal       114     98    212
  Totals       178    190    368

So we have 368patients: 178 on Drug A, and 190 on Drug B and we try to see if there are differences in the proportion of heartburn symptoms between drug A and B, i.e. $p1 = 64/178$ vs $p2 = 92/190$.

1. FISHER EXACT TEST: There is a discussion on Wikipedia about "Controversies". Based on the hypergeometric distribution, it is probably most adequate when the expected values in any of the cells of a contingency table are below 5 - 10. The story of the RA Fisher and the tea lady is great, and can be reproduced in [R] by simply grabbing the code here. [R] seems to tolerate without a pause the large numbers in our data (no problem with factorials):

 Antacid <- matrix(c(64, 178 - 64, 92, 190 - 92), nrow = 2)
    fisher.test(Antacid, alternative = "two.sided")
    Fisher's Exact Test for Count Data
data:  Antacid
p-value = 0.02011
alternative hypothesis: true odds ratio is not equal to 1
95 percent confidence interval:
 0.3850709 0.9277156
sample estimates:
odds ratio 
 0.5988478

2. CHI-SQUARE TEST OF HOMOGENEITY: Otherwise known as the goodness of fit Pearson's chi squared test. For larger samples (> 5 expected frequency count in each cell) the $\chi^2$ provides an approximation of the significance value. The test is based on calculating the expected frequency counts obtained by cross-multiplying the marginals (assuming normal distribution of the marginals, it makes sense that we end up with a $\chi^2$ distributed test statistic, since if $X\sim N(\mu,\sigma^)$, then $X^2\sim \chi^2(1))$:

                    Medication
Symptoms       Drug A                   Drug B               
Heartburn     156 * 178 / 368 = 75      156 * 190 / 368 = 81 
Normal        212 * 178 / 368 = 103     212 * 190 / 368 = 109

The degrees of freedom will be calculated as the {number of populations (Heartburn sufferers and Normals, i.e. 2) minus 1 } * {number of levels in the categorical variable (Drug A and Drug B, i.e. 2) minus 1}. Therefore, in a 2x2 table we are dealing with 1 d.f. And crucially, a $\chi^2$ of $1\,df$ is exactly a squared $N \sim (0,1)$ (proof here), which explains the sentence "a chi-square test for equality of two proportions is exactly the same thing as a z-test. The chi-squared distribution with one degree of freedom is just that of a normal deviate, squared. You're basically just repeating the chi-squared test on a subset of the contingency table" in this post.

The Test Statistic is calculated as:

$\chi^2=\frac{(64-75)^2}{75} + \frac{(92-81)^2}{81} +\frac{(114-103)^2}{103} + \frac{(98-109)^2}{109} = 5.39$

This is calculated in R with the function prop.test() or chisq.test(), which should yield the same result, as indicated here:

prop.test(Antacid, correct = F)

2-sample test for equality of proportions without continuity correction

data:  Antacid
X-squared = 5.8481, df = 1, p-value = 0.01559
alternative hypothesis: two.sided
95 percent confidence interval:
 -0.22976374 -0.02519514
sample estimates:
   prop 1    prop 2 
0.4102564 0.5377358

or..

chisq.test(Antacid, correct = F)

Pearson's Chi-squared test

data:  Antacid
X-squared = 5.8481, df = 1, p-value = 0.01559

3. G-TEST: The Pearson's chi-test statistic is the second order Taylor expansion around 1 of the G test; hence they tend to converge. In R:

library(DescTools) 
GTest(Antacid, correct = 'none')
Log likelihood ratio (G-test) test of
independence without correction

data:  Antacid
G = 5.8703, X-squared df = 1, p-value = 0.0154

4. Z-TEST OF PROPORTIONS: The normal distribution is a good approximation for a binomial when $np>5$ and $n(1-p)>5$. When the occurrences of successes are small in comparison with the total amount of observations, it is the actual number of expected observations that will determine if a normal approximation of a poisson process can be considered ($\lambda \geq 5$).

Although the post hyperlinked is old, I haven't found in CV an R function for it. This may be due to the fact explained above re: $\chi^2_{(df=1)}\sim \, N_{(0,1)}^2$.

The Test Statistic is:

$ \displaystyle Z = \frac{\frac{x_1}{n_1}-\frac{x_2}{n_2}}{\sqrt{p\,(1-p)(1/n_1+1/n_2)}}$ with $\displaystyle p = \frac{x_1\,+\,x_2}{n_1\,+\,n_2}$, where $x_1$ and $x_2$ are the number of "successes" (in our case, sadly, heartburn), over the number of subjects in that each one of the levels of the categorical variable (Drug A and Drug B), i.e. $n_1$ and $n_2$.

In the linked page there is an ad hoc formula. I have been toying with a spin-off with a lot of loose ends. It defaults to a two-tailed alpha value of 0.05, but can be changed, as much as it can be turned into a one tailed t = 1:

zprop = function(x1, x2, n1, n2, alpha=0.05, t = 2){
  nume = (x1/n1) - (x2/n2)
  p = (x1 + x2) / (n1 + n2)
  deno = sqrt(p * (1 - p) * (1/n1 + 1/n2))
  z = nume / deno
  print(c("Z value:",abs(round(z,4))))
  print(c("Cut-off quantile:", 
    abs(round(qnorm(alpha/t),2))))
  print(c("pvalue:", pnorm(-abs(z))))
}

In our case:

    zprop(64, 92 , 178, 190)
    [1] Z value:          2.4183  
    [1] Cut-off quantile: 1.96             
    [1] pvalue:           0.0077

Giving the same z value as the function in the R-Bloggers: z.prop(64, 178 , 92, 190) [1] -5.44273

OK... Hope it helps somebody out there, and I'm sure mistakes will be pointed out...

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