Chapter 19 For Explaining Psychological Statistics, 4th ed. by B. Cohen 1 A dichotomous outcome is one that has only two possibilities (e.g., pass or fail; heads or tails). The possible outcomes of a series of dichotomous events (i.e., coin flips) can be summarized in a binomial distribution that displays their relative probabilities of occurrence. To understand the construction of the binomial distribution, you need to know the basic rules of probability as applied to discrete events. Chapter 19: The Binomial Distribution

Chapter 19 For Explaining Psychological Statistics, 4th ed. by B. Cohen 2 Defining Probability –Definition: P(event) = number of ways the specified event can occur divided by the total number of possible events. P cannot be less than 0 or greater than 1. –Odds: The ratio of the number of unfavorable outcomes to the number of favorable outcomes (all outcomes being equally likely).

Chapter 19 For Explaining Psychological Statistics, 4th ed. by B. Cohen 3 The Rules of Probability The Probability of A or B occurring: P(A or B) = P(A) + P(B) – P(A and B) where P(A and B) represents the probability that both events will occur If A and B are mutually exclusive (i.e., if one occurs, the other cannot): P(A or B) = P(A) + P(B) The Probability of A and Then B: P(A and then B) = P(A) x P(B) –This rule assumes that the two events are independent. Conditional Probability –If two events are not independent: the probability of the second event is changed by the first event P(A given B) ≠ P(A) x P(B)

Chapter 19 For Explaining Psychological Statistics, 4th ed. by B. Cohen 4 The Binomial Distribution Definition: –A probability distribution in which events or observations can be classified into one (and only one) of two categories, each with some probability of occurrence. The two probabilities associated with each event can be symbolized as P and Q. P and Q must add up to 1, therefore, the two probabilities can be referred to as P and 1 – P. Commonly, P = Q =.5 (e.g., tosses of a fair coin), but this is not necessarily the case (e.g., the probabilities that a random person will be right- handed or left-handed). –Total number of trials = N. –The binomial distribution is a function of both P and N. –Number of trials that fall into the first (i.e., target) category = X (with probability P).

Chapter 19 For Explaining Psychological Statistics, 4th ed. by B. Cohen 5 Constructing the Binomial Distribution with N = 4 –Example: Consider all families in which there are exactly four children. Assume that the probabilities of having a boy or a girl in the family are equal: P = Q =.5 –Because there are two possible outcomes for each child, there are a total of 2 × 2 × 2 × 2 = 16 different sequences that can occur in a family: GGGG GGGB GGBG GGBB GBGG GBGB GBBG GBBB BGGG BGGB BGBG BGBB BBGG BBGB BBBG BBBB Let’s focus on the number of girls in each family: How many of the above sequences have just one girl? Ans: 4 (i.e., GBBB BGBB BBGB BBBG)

Chapter 19 For Explaining Psychological Statistics, 4th ed. by B. Cohen 6 The Binomial Distribution With N = 4 (cont.) Classifying all sequences by the number of girls (X) they contain: –0 girls: only 1 sequence –1 girl: 4 –2 girls: 6 –3 girls: 4 –4 girls: 1 Because P = Q, the probability of any one sequence is the same as any other: –e.g., P(GBBG) = P × Q × Q × P = (.5)(.5)(.5)(.5) =.0625 Another way to arrive at this result is to note that because the 16 possible sequences are equally likely the probability of any one of them occurring is 1/16 =.0625.

Chapter 16 For Explaining Psychological Statistics, 4th ed. by B. Cohen 7 The Binomial Distribution With N = 4 (cont.) So, what is the probability of getting just one girl out of the four children? –You can use the addition rule for mutually exclusive events, and simply add the proba- bilities of the four sequences that each have one girl: =.25, or –Because all the sequences are equally likely, you can divide the number of sequences with one girl by the total number of sequences: 4/16 =.25. –As listed in the previous slide, there are five possibilities for the number of girls in each family (i.e., 0 to 4). To construct the binomial distribution, we need to find the probability associated with each of those values of X. For each X, you can divide the number of sequences by 16 (because P = Q). The result is shown in the next slide.

Chapter 19 For Explaining Psychological Statistics, 4th ed. by B. Cohen 8

Chapter 19 For Explaining Psychological Statistics, 4th ed. by B. Cohen 9 The Normal Approximation to the Binomial Distribution –When P is higher or lower than.5, the binomial distribution will be skewed. –Fortunately, as N increases, the binomial distribution resembles the normal distribution (ND) more closely, though as P differs more from.5, a larger N is needed to resemble the ND. –We can use the resemblance to the ND to test null hypotheses about P by creat- ing a z score, given that the parameters of the corresponding ND are: Mean = NP; Standard deviation = √NPQ Therefore, the z formula is:

Chapter 16 For Explaining Psychological Statistics, 4th ed. by B. Cohen 10 The z Score Formula for Proportions –p is the proportion you observe in your sample. –π is the population proportion under the null hypothesis. –You must convert percentages to proportions to use this formula.

Chapter 19 For Explaining Psychological Statistics, 4th ed. by B. Cohen 11 The Sign Test for Matched Samples –Uses the binomial distribution to test hypotheses about the differences between matched pairs of subjects. –If N is large enough (e.g., at least 20), the normal distribution approximation can be used. –Example: Twenty-eight dance students are matched into pairs based on their prior abilities. Members of each pair are then randomly assigned to either imagery training to improve performance, or to a control condition. The pairs of students are later compared by an independent judge as to who performed better. We can define P as the probability that the imagery-trained member of the pair will be rated more highly. –The usual hypotheses are: H 0 : P =.5 H A : P ≠.5

Chapter 19 For Explaining Psychological Statistics, 4th ed. by B. Cohen 12 Analyzing Data for the Sign Test In our example, each pair is given a plus (+) if the imagery dancer is better and a minus (–) if the control dancer is judged superior. –X is the total number of + signs (N = 28). –Suppose that the Imagery participants were judged better in 18 cases and the controls were judged better in 10 cases. –We then have 18 pluses and 10 minuses, so X = 18. –Using the normal approximation formula, we have: –Statistical Decision: For a.05, two-tailed test, z = +/– Because 1.51 < 1.96, we cannot reject the null hypothesis.

Chapter 19 For Explaining Psychological Statistics, 4th ed. by B. Cohen 13 When to Use the Sign Test The sign test is particularly appropriate when you cannot make quantitative measurements, but you can determine the direction in which a variable changes, or which participant exhibits a higher level of the variable. The sign test is also used as an alternative to the matched-pairs t test when the shape of the underlying distribution is questionable. However, because it ignores all of the quantitative information, the sign test has much less power than the corresponding matched t test. If you can rank order the differences, you should use the Wilcoxon signed- ranks test, which has much more power than a corresponding sign test.