Definition 1: If x is a random variable with binomial distribution B(n, p) then the random variable y = x/n is said to have the proportion distribution.
Property 1: Where y has a proportion distribution as defined above
By Property 3b of Expectation
Theorem 1: Provided n is large enough – generally when np ≥ 5 and n (1 – p) ≥ 5 – then N(μy, σy) is a good approximation for the proportion distribution for y with
Proof: This is a consequence of Theorem 1 of the Binomial Distribution.
One Sample Hypothesis Testing
From the theorem, we know that when sufficiently large samples of size n are taken, the distribution of sample proportions is approximately normal, distributed around the true population proportion mean π, with standard deviation (i.e. the standard error)
We can use this fact to do hypothesis testing as was done for the normal distribution. In addition when a two-tailed test is performed a confidence interval can be calculated where
where we use the sample mean p as an estimate for the population mean when calculating the standard error. This introduces additional error, which is acceptable for large values of n.
Example 1: A company believes that 50% of their customers are women. A sample of 600 customers is chosen and 325 of them are women. Is this significantly different from their belief?
H0: π = 0.5; i.e. any difference in the number of men and women is due to chance
H1: π ≠ 0.5
Method 1: Using the binomial distribution, we reject the null hypothesis since:
BINOMDIST(325, 600, .5, TRUE) = 0.981376 > 0.975 = 1 – α/2 (2-tailed)
Method 2: By Theorem 1 we can also use the normal distribution
The observed mean is 325/600 = 0.541667. Based on the null hypothesis, we can assume that the mean p = .5 and the standard error
NORMDIST(.541667, .5, .020412, TRUE) = 0.979387 > 0.975 = 1 – α/2 (2-tailed)
And so we reach the same conclusion, namely to reject the null hypothesis.
Example 2: A survey of 1,100 voters showed that 53% are in favor of the new tax reform. Can we conclude that the majority of voters (from the population) are in favor?
H0: π < 0.5
Since people are not surveyed twice, we essentially have a hypergeometric distribution instead of a binomial distribution; i.e. we are selecting without replacement. But for large n the hypergeometric distribution is approximately binomial (i.e. it is not so likely that you will select the same person twice).
Since p = .53 and n = 1100, np > 5 and n (1 – p) > 5, and so we can approximate the distribution as a normal distribution. We will also use the sample p as an estimate of π in calculating the standard error.
NORMDIST(.53, .5, 0.01505, TRUE) = .976889 > .95, and so we can reject the null hypothesis and conclude with 95% confidence that the population will vote in favor of the tax reform.
We determine the 95% confidence interval as follows:
zcrit = NORMSINV(1 – α/2) = NORMSINV(0.975) = 1.96
and so the 95% confidence interval is
Thus we conclude with 95% confidence that between 50.1% and 55.9% of the population will be in favor. If however we are looking for a 99% confidence interval then
zcrit = NORMSINV(1 – α/2) = NORMSINV(0.995) = 2.58
And so the 99% confidence interval is
This means that with 99% confidence, between 49.1% and 56.9% of the population will be in favor. Since 50.0% is in this interval this time we cannot conclude (with 99% confidence) that the population will vote in favor of the tax reform.
Example 3: In conducting a survey of potential voters, how big does the sample need to be so that with 95% confidence the actual result (i.e. the population mean) will be within 2.5% of the sample mean? (i.e. how big a sample is necessary to have a 2.5% margin of error?)
This time we are looking for the value of n such that
zcrit · s.e. = 2.5%
As we saw in the previous example for 95% confidence zcrit = 1.96. The question is for any value of n when is s.e. the maximum? For any n, s.e. = is maximum when p(1–p) is maximum. It is easy to see that this occurs when p = .5. Thus the maximum s.e. = = . It now follows that
Solving for n yields n = 1536.584. Thus a sample of size 1,537 is sufficient. Using a similar calculation, to achieve 99% confidence requires a sample of size of 2,654.
Two Sample Hypothesis Testing
Theorem 2: 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 ni (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 4: A company which 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 2, 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 . The observed value of x1 – x2 is .60 – .50 =.10, and so we have (two-tail test):
p-value = NORMDIST(.1, 0, .0497, TRUE) = .978 > .975 = 1 – α/2
Thus, we reject the null hypothesis, and conclude there is a significant difference between the two types of bulbs.
We reach the same conclusion since critical value of x1 – x2 = NORMINV(.975,0,.0497) = .0975 < .1 = observed value of x1 – x2.