Chapter 17: Probability Models

Transcription

1 Chapter 17: Probability Models So far, probability models were developed by considering the physical aspects of simple random phenomena such as coin tossing or randomly sampling of population. There are a handful of probability models that are important in statistical applications. The normal and binomial are the most commonly used. Most of this discussion focuses on the binomial probability model and the binomial random variable. The binomial probability model is used when the random phenomenon is a sequence of independent Bernoulli trials. A Bernoulli trial is a random phenomenon with the following properties: 1. There are only two possible outcomes (generically referred to as successes and failures). 2. The probability of a success p is the same for every trial. Examples: 1. coin toss H or T. 2. true/false question correct or incorrect. 3. ask opinion of a randomly selected individual favor or oppose. 4. generic Bernoulli trial S or F. Independence Random sampling of a finite population and observing a binary response variable on each sample unit does not produce a sequence of Bernoulli trials. Even though the draws are done independently, the probability of a success p changes because random sampling is done without replacement. Example: Draw 10 cards without replacement from a standard deck of 52 cards and observe whether or not each card is an ace. p changes with the trial: Trial p = P (Ace) 1 4/52 2 4/51 (if the first drawn not an Ace) or 3/51 (if the first drawn is an Ace) 3 4/50 or 3/50 or 2/

2 Example: Draw 10 cards with replacement from a standard deck of 52 cards 1 and observe whether or not each card is an ace. p does not change with the trial (p = 1/13 for each trial), and the 10 draws comprise a set of 10 independent Bernoulli trials. Suppose that n independent observations on a binary variable are obtained by randomly sampling a population. These observations can be treated as if they were realizations of n Bernoulli trials without a significant loss of accuracy if the 10% condition holds. The 10% condition: The sample size n is less than 10% of the population size. Example: A random sample of n = 100 Missoula drivers is collected. Each is asked whether they support a ban on cell phone use while driving. Are the responses independent Bernoulli trials? 2 Is the 10% condition satisfied? 3 Example: A random sample of 20 U.S. Senators is identified and each senator in the sample is asked whether or not they support a certain bill. Can the sample responses be analyzed by treating them as if they were realizations of 20 independent Bernoulli trials? 4 The Binomial Model A binomial model describes the distribution of number of successes when the random phenomenon is a sequence of independent Bernoulli trials. Suppose that n independent Bernoulli trials each with probability of success p are observed. Let X count the number of successes in the n trials. The random variable X has a binomial distribution, expressed as X Binom(n, p). The binomial distribution has two parameters: n and p. n is the number of trials and sometimes called the binomial denominator, and p is the probability of success on any trial. Example: Bioassay is an experiment designed to assess the potency of an agent when administered to a living organism. Smith 5 describes a bioassay of an anti-pneumococcus serum 6. 1 Replace each card after drawing it and shuffle the deck 2 No since the 100 drivers are a random sample. 3 Yes since 100 is 10% of 1000, and there are far more than 1000 drivers in Missoula. 4 No since there are 100 US senators and 20% is too large of a sampling fraction. 5 Smith, W The titration of an antipneumococcus serum. J. Pathology, 35, Pneumococcus is the bacterium responsible of pneumonia 125

3 200 mice were randomly assigned to one of five groups, and each group received a dose of serum and a dose of culture pneumococci. The numbers of deaths due to pneumonia within 7 days of inoculation were observed. The number that die at a particular dosage has a binomial distribution with parameters n = 40 and unknown p. The observed value of X at dosage.0028 was 35, so we write X = 35. Presumably, p depends on dosage, so all five counts (one count per dosage) do not follow the same binomial model. The binomial distribution Let X count the number of successes in n independent Bernoulli trials, each with a common probability of success p. X has a discrete distribution, and so there are two lists that comprise the distribution: all possible realizations of X, and the probabilities of each realization. The possible realizations of X are 0, 1, 2,..., n, where n is the number of trials. The probability of a particular realization (e.g., P (X = 0)) can be computed using the following formula. First, let q = 1 p denote the probability of a failure. Then, the probability of x successes in n independent Bernoulli trials is and P (X = x) = ( ) n p x q n x, x = 0, 1,..., n, x ( ) n n! where = x x!(n x)!, x! = { x(x 1) 2 1 if x > 0 1 if x = 0. The formula will be explained after a few more facts are presented. The expected value and standard deviation of the binomial random variable X are: µ = E(X) = np and σ = SD(X) = npq. Example: Suppose that a particular trait (e.g. eye color) is determined by one gene pair and that there are two alleles, d and r that define the gene. The dominant allele is d and the recessive allele is r. The dominant allele will be expressed if the gene pair possessed by an organism is dd, dr or rd. 126

4 Suppose that a child is equally likely to inherit either of the two alleles from each parent and that both parents possess both alleles (both parents are hybrid). Then a child will inherit the gene pair dd, dr, rd, and rr with probabilities 1/4 each. The probability that the dominant allele is expressed in the child is thus p = 3/4. Suppose that four children belong to a family where both parents are hybrid. What is the probability that 3 children express the dominant allele? Each child represents an independent Bernoulli trial resulting in either a expression of the dominant allele d or r. If X counts the number of children expressing d, then X Bin(4, 3/4), and we can use the binomial model to compute P (X = 3): P (X = 3) = ( ) ( ) 3 ( ) = = = The expected number of dominant expressions among the n = 4 children is E(X) = np = = 3. The standard deviation of the number of dominant expressions among the four children is SD(X) = npq = =.866. Using R and the function dbinom 7, the remaining probabilities can be computed quickly. The probability distribution of X is x P (X = x) Binomial probabilities can be computed using StatCrunch: Stat Calculators Binomial Choose from {,, <, >, =} Compute The computation of P (X = 3) was P (X = 3) = ( ) 4 p 3 q 1 3 = The function call is dbinom(0:4,size=4,p=.75). 127

5 The term = is the probability that the alleles of the four children (ordered by age) was (d, d, d, r). But there are 4 = ( 4 3) ways that X = 3 may occur: (d, d, d, r), (d, d, r, d), (d, r, d, d), or (r, d, d, d). Each outcome has the same probability, namely, , so P (X = 3) is The term ( n x) counts the number of ways that x successes can be arranged among n positions and the term p x q n x is the probability of obtaining any one of the arrangements. Example: The probability there are no more than 3 children that express the dominant allele is P (X 3) = P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) = 1 P (X = 4) = =.684. If the sample size n in a binomial model is large, say n > 500, then n! may be too large for a computer or calculator to compute. In such situations, the normal model usually can be used to approximate the binomial probabilities. More importantly, the normal approximation of the binomial distribution leads to several important inferential methods (confidence intervals and hypothesis tests). The normal approximation to the binomial distribution Suppose that X Binom(n, p) and that np 10 and nq 10. Then, the distribution of X can be accurately approximated by a normal distribution with mean µ = np and standard deviation npq. The approximation is expressed as X N(np, npq). The condition np 10 and nq 10 is called the success/failure condition. Example: Most human traits are not determined by a single gene but by multiple genes. If multiple genes control the trait, then the theorized dominant allele will not be expressed in 75% of the offspring when both parents are hybrids. An analysis of a large number of humans with hybrid parents can confirm or refute the supposition that the trait is controlled by a single gene. 128

6 Suppose that n = 490 individuals are sampled and tested to determine whether the dominant or recessive allele is expressed. 1. How many recessive expressions are expected if the single gene supposition is correct? The random selection of an individual with hybrid parents is a random phenomenon and observing whether or not the recessive allele is expressed is a Bernoulli trial. The observed number recessive allele expressions (out of 490) is a realization of a binomial random variable parameters p =.25 and n = 490. The question can be answered with this knowledge. Suppose that X Binom(490,.25). The expected number of recessive expressions is E(X) = np = = Also SD(X) = npq = = Suppose that X = 150 is the observed number recessive allele expressions. X = 150 is larger than expected, but is it sufficiently large to refute the single-gene supposition? It would be sufficiently large if the probability of obtaining as many or more than 150 recessive allele expressions was very small. A very small probability indicates that either an extremely improbable sample was collected, or the supposition is incorrect and p is larger than.25. We suppose that the sample is not the result of an extremely improbable draw. The probability of obtaining as many or more than 150 recessive allele expressions is P (X 150) = P (X = 150) + P (X = 151) + +P (X = 490). Calculating all of these binomial probabilities can be avoided by using the normal approximation. However, the normal distribution is accurate only if np 10 and nq 10. The normal approximation is accurate since np = and nq = Wile the distribution of X is exactly binomial, it is is also approximately normal with mean µ = and standard deviation σ = In shorthand, X N(122.5, 9.585). 129

7 Returning to the question, P (X 150) ( ) = P Z > = P (Z 2.869) = 1 P (Z < 2.869) = = Using R, a near-exact calculation of the probability is Since the single gene supposition implies that the observed event X 150 is highly unlikely, the conclusion is that supposition is incorrect. Exercise on page 423 : A wildlife biologist examines frogs for a genetic trait he suspects may be linked to sensitivity to industrial toxins in the environment. Previous research has established that this trait is usually found in one of every 8 frogs. Suppose he decides to collect and examine 150 frogs. 1. Determine the mean and standard deviation of the number of frogs with the trait he should expect to find in his sample. (a) Let X Binom(n = 150, p =.125) represent the number in his sample with the trait. (We assume that his sample is a random sample from a very large population of frogs. Then, the number of frogs with the trait satisfies the 10% condition and the binomial distribution is an appropriate model for the count of frogs with the trait). (b) Compute the mean and standard deviation: E(X) = np = = SD(X) = npq = = Can a normal model be used to approximate the distribution of the number of frogs with the trait? (a) Test whether np > 10 and nq > 10. Compute np = = 18.75, nq = =

8 (b) Both numbers are larger than 10, so the answer is yes, a normal distribution can be used as an approximation of the binomial distribution. The N(18.75,4.0505) is an accurate approximation of the binomial distribution. 3. He found the trait in 22 of his frogs. Is X = 22 significant evidence that the trait has become more common? (a) How do we answer the question? If 22 or more frogs is very unlikely to be observed when the expected number of frogs is µ = 18.75, then the conclusion is that the mean is actually larger than because the trait has become more common. (b) Compute the probability of observing 22 or more frogs. That is compute, P (X 22) where X N(18.75, ) P (X 22) = 1 P (X < 22) ( ) = 1 P Z < = 1 P (Z <.8024) = = (c) It is not highly unlikely to have observed 22 or more. While the observed value of 22 is suggestive, it cannot be considered significant evidence that the mean is larger than and that the trait has become more common. 4. Why is the statistical evidence the probability of 22 or more with the trait instead of 22 with the trait? (a) Because all values of X are unlikely. After all, there are n + 1 = 151 possible realizations of X (0, 1,..., 150) and the most likely value is 19 (because µ = 18.75; specifically, P (X = 19) =.0973). (b) A conservative measure of evidence is adopted because scientists are reluctant to contradict existing theories. The conservative measure of evidence is P (X 22), as this probability includes all other outcomes that are as least as unlikely as 22 and suggest that µ > (namely, 23, 24,..., 150). By adding in these additional outcomes to the probability calculation, the scientist is protected from rushing to an incorrect conclusion of significance. 131