Hypothesis Testing with the Binomial

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Testing Hypotheses with the Binomial Probability Distribution

A Binomial Experiment:

consists of n identical trials.
each trial results in one of two outcomes, a “success” or a “failure.”
the probabilities of success ( p) and of failure ( q= 1 p ) are constant across trials.
trials are independent, not affected by the outcome of other trials.
Y is the number of successes in n trials.
P(Y = y) may also be determined by reference to a binomial table.
The binomial distribution has:

Testing Hypotheses

State null and alternative hypotheses
the null hypothesis specifies the value of some population parameter. For example, p = .5 (twotailed, nondirectional; this coin is fair) or p .25 (onetailed; directional; student is merely guessing on 4choice multiple choice test).
the alternative hypothesis, which the researcher often wants to support, is the antithetical complement of the null. For example, p .5 (twotailed, the coin is biased) or p > .25 (onetailed, the student is not merely guessing, e knows tested material).
Specify the sampling (probability) distribution and the test statistic (Y). Example: the binomial distribution describes the probability that a single sample of n trials would result in (Y = y) successes (if assumptions of binomial are true).
Set alpha at a level determined by how great a risk of a Type I error (falsely rejecting a true null) you are willing to take. Traditional values of alpha are .05 and .01.

Specify a decision rule:
From the sampling distribution associated with a true null hypothesis, map out the rejection region, the values of the sample test statistic that would be obtained with a probability of  or less.
With directional hypotheses, the rejection region will be in one tail; with nondirectional hypotheses, in two tails (/ 2 in each tail)
if the test statistic falls in the rejection area, reject the null and assert the alternative hypothesis. If not, fail to reject the null.
The boundary between rejection and nonrejection areas is sometimes called the “critical value” of the test statistic.
Collect sample data, compute test statistic, and make decision.
An alternative and preferable decision rule is:
From the sample data, compute P, the significance level, the probability that one would obtain a sample result as or more contradictory to the null hypothesis than that actually observed.
Examples for a onetailed test:
H: p .5; H1: p > .5; n = 25, Y=18;
P(Y 18 | n = 25, p = .5) = .021
APA-style summary: “Mothers were allowed to smell two articles of infant’s clothing and asked to pick the one which was their infant’s. They were successful in doing so 72% of the time, significantly more often than would be expected by chance, exact binomial p (one-tailed) = .021.”
H: p .5; H1: p < .5; n = 25, Y = 18;
P(Y ≤ 18 | n = 25, p = .5) = .993 (note that the direction of the P(Y ≤) matches that of H1)
For a twotailed test
Compute a onetailed P and double it.
H: p = .5; H1: p .5; n = 25, Y = 18;
2P(Y ≥ 18) = 2(.022) = .044
H: p=.5; H1: p .5; n=25, Y=7;
2P(Y ≤7) = 2(.022) = .044 (the direction of the P(Y ≤) is that which gives the smaller pvalue; P(Y ≥ 7) = .993 and 2(.993) = 1.986, obviously not a possible p.
If p≤ alpha, reject H, otherwise fail to reject the null.
Reporting an exactpvalue gives the reader more information than does merely reporting rejection or no rejection of the null: p = .06 and p = .95 might both lead to no rejection, but the former nonetheless casts doubt on the null, the latter certainly does not. Likewise, p = .04 and p = .001 may lead to the same conclusion, but the latter would indicate greater confidence.

Normal Approximation of a Binomial Probability

If you do not have access to a computer or table to obtain a binomial probability and it just would be too hard to do it by hand, you may be able to approximate it using the normal curve.
If falls within 0 to n, then the binomial approximation should be good. For example, we want to find P(Y ≥ 18 | n = 25, p = .5). , which is contained within 0  25, so the normal approximation should be good.
We compute . Note that I reduced the value for the number of successes from 18 to 17.5. This is called the "correction for continuity." The reasoning behind it goes something like this: We are using a continuous distribution to approximate a discrete probability. Accordingly, we should think of '18' as being '17.5 to 18.5.' Since we want the probability of getting 18 or more successes, we find the z-score for getting 17.5 or more -- that way we include all of '18' in the obtained probability. If we wanted to estimate the probability of getting 18 or fewer successes, we would use '18.5,' again, to inlcude all of '18' in the obtained probability. On a practical note, the approximated probability will be closer to the exact binomial probability if the we use the correction for continuity than it would be if we did not use the correction.
We then use the normal curve table to get the probability from the z-score. For our z-score of 2.00, the probability is .0228.

The Binomial Sign Test

With matched pairs data we may simply compute, for each pair, whether the difference score is positive or negative and then test the null hypothesis that in the population 50% of the difference scores are positive.
For example, suppose we have pre and post blood pressure scores for each of ten subjects given an experimental drug. Our null hypothesis is that the drug has no effect on blood pressure.
We test 11 subjects. For 9 subjects pressure is lower after taking the drug, for 1 it is higher and for 1 it is unchanged.
Our test statistic is the larger of n+ (the number of positive difference scores) and n (the number of negative difference scores). Unless there are many difference scores of zero, we usually discard data from subjects with difference scores of zero (an alternative procedure would be to count zero difference scores as being included with the smaller of n+ and n).
Following the usual procedure, our test statistic would be n+ = 9, with n = 10. The exact two-tailed p value is 2P(Y  9 n = 10, p = .5) = .022, sufficiently low to reject the null hypothesis at the customary .05 level of significance.
Our APA-style summary statement might read like this: “An exact binomial sign test indicated that the drug significantly lowered blood pressure, 9 of 10 patients having post-treatment pressure lower than their pre-treatment pressure, p = .022.” Were we to use a normal approximation rather than an exact test, we would include the value of z in our summary statement.

Links

See this published example of a binomial sign test
Estimating Binomial Proportions
A little binomial humor