# Two-sample Proportion Testing

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 ni πi ≥ 5 and n(1 –πi) ≥ 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, xi 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 $sqrt{frac{2(.55)(.45)}{100}}$ = .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

Statistics with R for Business Analysts – Normal Distribution

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