# Equivalence test for binominal data

I want to apply an equivalence test on my sample to infer whether they are equivalent or not.

Since my data are bionominal [0,1] I don’t know whether the TOST procedure (tost() in R) can handle my problem or not.

My data consists of two groups (G1 and G2) which are not equal in numbers of samples. E.g., G1= 164 people and G2=280 people. The samples are binominal, such that G1 includes the number of people who finished and failed in completing a game “in our case study”, and specified with 1 and 0, respectively. E.g., G1 includes 55 players who could finish the game and the rest fail the game. The same for G2 with different numbers. My question is, what is the best equivalence test for this type of data to infer whether they are equivalent? I implemented tost() test in r and got the result, but I am not sure whether the test is correct, since the function automatically calculates mean and SD. For example tost() calculates the mean for the above example G1, m= 0.35, but, when I calculate the mean following n*p(0.5) formula, I obtained m=82.

The # of success in G1 is 58 out of 164, and for G2 is 113 out of 280. Here is my code:

    equivalence::tost(G1,G2,paired = FALSE, epsilon=0.15, var.equal=FALSE,conf.level = 0.95, alpha = 0.05)


And here is the result that I got:

    Welch Two Sample TOST

data:  G1 and G2
df = 348.24
sample estimates:
mean of G1 mean of G2
0.3536585 0.4035714

Epsilon: 0.15
95 percent two one-sided confidence interval (TOST interval):
-0.12840509  0.02857931
Null hypothesis of statistical difference is: rejected
TOST p-value: 0.01809221

• Can you show your output from tost? And give the number of successes for G2? Are you testing $H_0: p_1 = p_2$ vs $H_a: p_1 \ne p_2?$ Unclear whether this is mainly about how to use tost or mainly about testing a hypothesis. – BruceET Aug 5 '18 at 18:18

While one can use the t test to test for proportion difference, the z test is a tad more precise, since it uses an estimate of the standard deviation formulated specifically for binomial (i.e. dichotomous, nominal, etc.) data. The same applies to the z test for proportion equivalence.

First, the z test for difference in proportions of two independent samples is pretty straightforward:

About z tests for unpaired proportion difference
The null hypothesis is $$H_{0}\text{: }p_{1} - p_{2} = 0$$ (i.e. $$H_{0}\text{: }p_{1} = p_{2}$$), with $$H_{\text{A}}\text{: }p_{1} - p_{2} \ne 0$$.

$$z = \frac{\hat{p}_{1}-\hat{p}_{2}}{\sqrt{\hat{p}\left(1-\hat{p}\right)\left[\frac{1}{n_{1}} + \frac{1}{n_{2}}\right]}}$$,
where:
$$\hat{p}_{1}$$ and $$\hat{p}_{1}$$ are the sample proportions in group 1 and group 2;
$$n_{1}$$ and $$n_{2}$$ are the sample sizes in group 1 and group 2; and
$$\hat{p}$$ is the estimate of the sample means if $$H_{0}$$ is true, the best guess of which is simply the overall sample proportion (i.e. of all the data, ignoring which group an observation is from).

You might want to consider a continuity correction. For example, Hauck and Anderson's (1986) correction gives:

$$c_{\text{HA}} = \frac{1}{2\min{(n_{1},n_{2})}}$$, and a redefined $$s_{\hat{p}}$$:

$$s_{\hat{p}}= \sqrt{ \frac{\hat{p}_{1}(1-\hat{p}_{1})}{n_{1}-1} + \frac{\hat{p}_{2}(1-\hat{p}_{2})}{n_{2}-1}}$$, so that

$$z = \frac{\left|\hat{p}_{1} - \hat{p}_{2}\right| - c_{\text{HA}}}{\sqrt{ \frac{\hat{p}_{1}(1-\hat{p}_{1})}{n_{1}-1} + \frac{\hat{p}_{2}(1-\hat{p}_{2})}{n_{2}-1}}}$$

The appropriate $$p$$-value for this $$z$$-statistic is then calculated or looked up in a table, and compared to $$\alpha/2$$ (two-tailed test).

About z tests for unpaired proportion equivalence
Because all differences are "statistically significant" given a large enough sample size, it is a good idea to decide beforehand what the smallest relevant difference in proportions is to you, and then look for evidence of such relevance. You find such evidence by combining the inferences from the test for difference just described, with a test for equivalence.

Suppose you decide beforehand that a meaningful difference in proportion for your purposes is on that is at least 0.05 (i.e. $$|p_{1} - p_{2}| \ge 0.05$$), then the corresponding test for equivalence of proportions for two independent groups is:

$$H^{-}_{0}\text{: }|p_{1} - p_{2}| \ge 0.05$$, which translates into two one-sided null hypotheses:

1. $$H^{-}_{01}\text{: }p_{1} - p_{2} \ge 0.05$$
2. $$H^{-}_{02}\text{: }p_{1} - p_{2} \le -0.05$$

These two one-sided null hypotheses can be tested with (these test statistics have been constructed both for upper tail one-sided tests):

1. $$z_{1} = \frac{0.05 - \left(\hat{p}_{1}-\hat{p}_{2}\right)}{\sqrt{\hat{p}\left(1-\hat{p}\right)\left[\frac{1}{n_{1}} + \frac{1}{n_{2}}\right]}}$$, and
2. $$z_{2} = \frac{\left(\hat{p}_{1}-\hat{p}_{2}\right)+0.05}{\sqrt{\hat{p}\left(1-\hat{p}\right)\left[\frac{1}{n_{1}} + \frac{1}{n_{2}}\right]}}$$.

With a continuity correction $$z_{1}$$ and $$z_{2}$$ instead become (see Tu, 1997):

1. $$z_{1} = \frac{0.05 - \left(\hat{p}_{1}-\hat{p}_{2}\right) + c_{\text{HA}}}{\sqrt{ \frac{\hat{p}_{1}(1-\hat{p}_{1})}{n_{1}-1} + \frac{\hat{p}_{2}(1-\hat{p}_{2})}{n_{2}-1}}}$$, and
2. $$z_{2} = \frac{\left(\hat{p}_{1}-\hat{p}_{2}\right)+0.05-c_{\text{HA}}}{\sqrt{ \frac{\hat{p}_{1}(1-\hat{p}_{1})}{n_{1}-1} + \frac{\hat{p}_{2}(1-\hat{p}_{2})}{n_{2}-1}}}$$.

If you reject both $$H^{-}_{01}$$ and $$H^{-}_{02}$$ (both tested at $$\alpha$$, not $$\alpha/2$$, and both tested with right tail rejection regions), then you can conclude that you have evidence of equivalence.

Finally... if you combine inference from tests of $$H_{0}$$ and $$H^{-}_{0}$$ (i.e. test for difference and test for equivalence), then you get one of the following possibilities:

1. reject $$H_{0}$$ and reject $$H^{-}_{0}$$: conclude trivial difference between proportions (i.e. yes there is a difference, but it's too small for you to care about because it is smaller than 0.05);
2. reject $$H_{0}$$ and not reject $$H^{-}_{0}$$: conclude relevant difference between proportions (i.e. larger than 0.05);
3. not reject $$H_{0}$$ and reject $$H^{-}_{0}$$: conclude equivalence of proportions; or
4. not reject $$H_{0}$$ and not reject $$H^{-}_{0}$$: conclude indeterminate (i.e. underpowered tests).

R code

First the test for difference:

Assume g1 and g2 are vectors containing the binomial data for group 1 and group 2 respectively.

n1 <- length(g1) #sample size group 1
n2 <- length(g2) #sample size group 2
p1 <- sum(g1)/n1 #p1 hat
p2 <- sum(g2)/n2 #p2 hat
n <- n1 + n2 #overall sample size
p <- sum(g1,g2)/n #p hat
cHA <- 1/(2*min(n1,n2))

# without continuity correction
z <- (p1 - p2)/sqrt(p*(1-p)*(1/n1 + 1/n2)) #test statistic
pval <- 1 - pnorm(abs(z)) #p-value reject H0 if it is <= alpha/2 (two-tailed)

# with continuity correction
zHA <- (abs(p1 - p2) - cHA)/sqrt((p1*(1-p1)/(n1-1)) + (p2*(1-p2)/(n2-1))) #with continuity correction
pvalHA <- 1 - pnorm(abs(zHA)) #p-value reject H0 if it is <= alpha/2 (two-tailed)


Next the test for equivalence:

Delta <- 0.05 #Equivalence threshold of +/- 5%.
# You will want to carefully think about and select your own
# value for Delta before you conduct your test.


Again, assume g1 and g2 are vectors containing the binomial data for group 1 and group 2 respectively.

n1 <- length(g1) #sample size group 1
n2 <- length(g2) #sample size group 2
p1 <- sum(g1)/n1 #p1 hat
p2 <- sum(g2)/n2 #p2 hat
n <- n1 + n2 #overall sample size
p <- sum(g1,g2)/n #p hat
cHAeq <- sign(p1-p2)* (1/(2*min(n1,n2)))

# without continuity correction
z1 <- (Delta - (p1 - p2))/sqrt(p*(1-p)*(1/n1 + 1/n2)) #test statistic for H01
z2 <- ((p1 - p2) + Delta)/sqrt(p*(1-p)*(1/n1 + 1/n2)) #test statistic for H02
pval1 <- 1 - pnorm(z1) #p-value (upper tail) reject H0 if it is <= alpha (one tail)
pval2 <- 1 - pnorm(z2) #p-value (upper tail) reject H0 if it is <= alpha (one tail)

# with continuity correction
zHA1 <- (Delta - abs(p1 - p2) + cHAeq)/sqrt((p1*(1-p1)/(n1-1)) + (p2*(1-p2)/(n2-1))) #with continuity correction
zHA2 <- (abs(p1 - p2) + Delta - cHAeq)/sqrt((p1*(1-p1)/(n1-1)) + (p2*(1-p2)/(n2-1))) #with continuity correction
pvalHA1 <- 1 - pnorm(zHA1) #p-value (upper tail) reject H0 if it is <= alpha (one tail)
pvalHA2 <- 1 - pnorm(zHA2) #p-value (upper tail) reject H0 if it is <= alpha (one tail)


References

Hauck, W. W. and Anderson, S. (1986). A comparison of large-sample confidence interval methods for the difference of two binomial probabilities. The American Statistician, 40(4):318–322.

Tu, D. (1997). Two one-sided tests procedures in establishing therapeutic equivalence with binary clinical endpoints: fixed sample performances and sample size determination. Journal of Statistical Computation and Simulation, 59(3):271–290.

• Thanks for the code and the nice explanation. Do I have to consider my case as "with continuity correction" or "without continuity correction"? Although "with continuity correction " the formula dos not work in my case, since this line (p1*(1-p1)/n1-1) + (p2*(1-p2)/n2-1)" provides negative result and $sqrt$ dos not work! – nahid khosh Aug 6 '18 at 9:06
• @nahidkhosh Whoops! Order of operations got me! Should be fixed now with the added pair of parentheses. Because you're using an $\alpha=0.05$, have $p_{1}$ & $p_{2}$ that are not too extreme, and have a not tiny $\Delta$, you should have a large enough sample size for the Hauck-Anderson continuity correction to not make much difference. (Although $\Delta=0.15$ is huge in my opinion... that's like saying 0.2 is equivalent to 0.05... but that's your business, not mine. :) – Alexis Aug 6 '18 at 14:25
• Great answer! I wonder, if there is a package in R for it? – user1700890 Jan 11 '19 at 16:00
• @user1700890 Not yet!* There is a Stata package, though: tost.$$\phantom{0}$$ * There are several packages that perform equivalence tests on CRAN, but I haven't seen one that specifically performs the test I described above. – Alexis Jan 11 '19 at 17:22
• @Alexis, I came across TOSTER package, not sure if I should use it. – user1700890 Jan 11 '19 at 19:10