Theorem 1: Let x₁ and x₂ be random variables with proportional distributions with mean π₁ and π₂ respectively. Let p₁ be the proportion of successes in n₁ trials of the first distribution and let p₂ be the proportion of successes in n₂ trials of the second distribution. When the number of trials n₁ and n₂ are sufficiently large, usually when n_i π_i ≥ 5 and n_i (1 –π_i) ≥ 5, the difference between the sample proportions p₁ – p₂ will be approximately normal with mean π₁ – π₂ and standard deviation

Proof: Based on Theorem 2 of the Binomial Distribution, x_i has approximately the distribution

Since x₁ and x₂ are independently distributed, by the linear transformation property of the normal distribution, x₁ – x₂ has distribution

Example 1: A company that manufactures long-lasting light bulbs sells halogen and compact florescent bulbs. They ran an experiment in which they ran 100 halogen and 100 florescent bulbs continuously for 250 days. After 250 days they found that half of the halogen bulbs were still working while 60% of the florescent bulbs were still operating. Is there a significant difference between the two types of bulbs?

Let x₁ = the percentage of halogen bulbs that are functional after 250 days and x₂ = the percentage of florescent bulbs that are functional after 250 days. The presumption is that the distributions for each of these are proportional. We now test the following null hypothesis:

H₀: π₁ = π₂

Assuming the null hypothesis is true, by Theorem 1, x₁ – x₂ will be approximately normal with mean π₁ – π₂ = 0 and standard deviation

where the common value of the mean is denoted π and both samples are of size n. Since the value for π is unknown, we estimate its value from the sample, namely, 50 + 60 = 110 successes out of 200, i.e. π ≈ 0.55, Thus, the mean of x₁ – x₂ is 0 (based on the null hypothesis) and the standard deviation is approximately  = .704. The observed value of x₁ – x₂ is .60 – .50 =.10, and so we have (two-tail test):

p-value = NORMDIST(.1, 0, .704, TRUE) = .922 < .975 = 1 – α/2

Thus, we can’t reject the null hypothesis and so we cannot conclude there is a significant difference between the two types of bulbs. More precisely

p-value = 2*(1–NORM.DIST(.1, 0, .0703, TRUE)) = .155 > .05 = α

Alternatively, we can reach the same conclusion via the following test:

critical value of x₁ – x₂ = NORMINV(.975,0,.0703) = .138 > .1 = observed value of x₁ – x₂