# Means and Proportions with two populations

August 30, 2009
71 Views

Statistical inference about means and proportions with two populations seems to be one of the most commonly used applications in the field of analytics – comparing campaign response rates between 2 groups of customers, pre and post campaign sales, membership renewal rates, etc.

Call it chance or whatever, but whenever these kind of tasks came up I hear people talking about the t-tests only. No issues as long as you want to compare means or when your target variable is a continuous value. But how or why do people talk about the t-test when they want to compare ratios or proportions? Whatever happened to the Chi-Square tests or the Z-test for difference in proportions?

I did a bit of research on the net, a bit of calculation using pen and paper [very good exercise for the brain in this age of calculators and spreadsheets 🙂 ], read a very good article by Gerard E. Dallal, and I found the answers.

Going back to our introductory class in statistics, let’s check out the formulae for the t-tests.

1. Assuming that the population variances are equal,
T = (X1 – X2)/sqrt (Sp2(1/n1 + 1/n2) ……….Equation 1

where
X1, X2 = means of sample 1 and 2
n1, n2 = size of sample 1 and 2
Sp = pooled

Statistical inference about means and proportions with two populations seems to be one of the most commonly used applications in the field of analytics – comparing campaign response rates between 2 groups of customers, pre and post campaign sales, membership renewal rates, etc.

Call it chance or whatever, but whenever these kind of tasks came up I hear people talking about the t-tests only. No issues as long as you want to compare means or when your target variable is a continuous value. But how or why do people talk about the t-test when they want to compare ratios or proportions? Whatever happened to the Chi-Square tests or the Z-test for difference in proportions?

I did a bit of research on the net, a bit of calculation using pen and paper [very good exercise for the brain in this age of calculators and spreadsheets 🙂 ], read a very good article by Gerard E. Dallal, and I found the answers.

Going back to our introductory class in statistics, let’s check out the formulae for the t-tests.

1. Assuming that the population variances are equal,
T = (X1 – X2)/sqrt (Sp2(1/n1 + 1/n2) ……….Equation 1

where
X1, X2 = means of sample 1 and 2
n1, n2 = size of sample 1 and 2
Sp2 = pooled variance = [((n1-1)S12+(n2-1)S22)/(n1+n2-2)]

2. Assuming that the population variances are not equal,
T = (X1 – X2)/sqrt(S12/n1 + S22/n2) ……….Equation 2

We have also been taught that the test statistic Z is used to determine the difference between two population proportions based on the difference between the two sample proportions (P1 – P2).

And the formula for the Z statistic is given by
Z = (P1 – P2)/ sqrt(P(1-P)(1/n1 + 1/n2)) ……….Equation 3

where
P1, P2 = proportions of success (or target category) in samples 1 and 2
S1, S2 = variances for samples 1 and 2
n1, n2 = size of samples 1 and 2
P = pooled estimate of the sample proportion of successes =(X1 + X2) / (n1 +n2)
X1, X2 = number of successes (or target category) in samples 1 and 2

The test statistic Z (equation 3) is equivalent to the chi- square goodness-of-fit test, also called a test of homogeneity of proportions.

But how different is the proportions from means? The proportion having the desired outcome is the number of individuals/observations with the outcome divided by total number of individuals/observations. Suppose we create a variable that equals 1 if the subject has the outcome and 0 if not. The proportion of individuals/observations with the outcome is the mean of this variable because the sum of these 0s and 1s is the number of individuals/observations with the outcome.

Let’s suppose there are m 1s and (n-m) 0s among the n observations. Then, XMean (=P) =m/n and is equal to (1-m/n) for m observations and 0-m/n for (n-m) observations. When these results are combined, the final result is

∑(Xi – XMean)2 = m(1-m/n)2 + (n – m) (0 – m/n)2
= m(1 – 2m/n + m2/n2) + (n – m) m2/n2
= m – 2(m2/n2) + (m3/n2) + (m2/n) – (m3/n2)
= m – (m2/n)
= m(1-m/n)
= nP(1-P)

So, variance = ∑(Xi – XMean)2/n = P(1-P)

Substituting this in the equation 3 (for Z statistic), we get
(P1 – P2)/ sqrt(Variance/n1 + Variance/n2)), which is not so different from equation 2 (the formula for the “equal variances not assumed” version of t test).

As long as the sample size is relatively large, the distributional assumptions are met, and the response is binomial – the t test and the z test will give p-values that are very close to one another.

And in the case where we have only two categories, the z test and the chi-square test turn out to be exactly equivalent, though the chi-square is by nature a two-tailed test. The chi-square distribution for 1 df is just the square of the z distribution.

The various tests and their assumptions as listed in Wikipedia are given below:
1. Two-sample pooled t-test, equal variances
(Normal populations or n1 + n2 > 40) and independent observations and σ1 = σ2 and (σ1 and σ2 unknown)

2. Two-sample unpooled t-test, unequal variances
(Normal populations or n1 + n2 > 40) and independent observations and σ1 ≠ σ2 and (σ1 and σ2 unknown)

3. Two-proportion z-test, equal variances
n1 p1 > 5 and n1(1 − p1) > 5 and n2 p2 > 5 and n2(1 − p2) > 5 and independent observations

4. Two-proportion z-test, unequal variances
n1 p1 > 5 and n1(1 − p1) > 5 and n2 p2 > 5 and n2(1 − p2) > 5 and independent observations

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