One-Sample Test for a Binomial ProportionTest based on pˆ, sample proportionHo: p = poH1: p  po (2-tailed)Assume normal approximation to the binomial distribution—n*po*qo > 5 where po isthe hypothesized proportion in the populationnqp,p~poooNˆWe will standardize pˆ by subtracting the expected or hypothesized value, po and dividing by the standard error of pˆ under Ho, nqpoo, i.e.),~N(nqpppzooo10ˆFor 2-tailed testreject Ho if z > z1-/2 or z < z/2accept Ho if z/2 < z < z1-/2For example, for  = 0.05, reject Ho if z > z0.975 (1.96) or z < z0.025 (-1.96)p-valueIf pˆ po, p-value = 2* area to left of z under N(0, 1) curveIf pˆ po, p-value = 2* area to right of z under N(0, 1) curveExample: Suppose the annual incidence of asthma in the general population of children 0-4 years of age is 1.4% for boys. For 500 boys 0-4 years old with mothers who smoke, 10 new cases were observed in one year. Is the incidencein this group compatible with the general population?Ho: p = 0.014H1: p  0.014Let  = 0.05Note: npoqo = 500 * 0.014 * 0.986 = 6.902Reject Ho if z > 1.96 or z < -1.961.14.00525430.0065000.986*0.0140.0140.02nqpppz0.0250010poooˆˆDecision: Do not reject Ho. We have no evidence that the incidence of asthma in boys whose mothers smoke differs from the incidence in boys of the same agein the general population. p-value: pˆ po, therefore p-value = 2 * the area in N(0, 1) to the right of 1.14.Using Table 3, column B, p = 2* 0.1271 = 0.2542.One-tailed tests:Ho: p = poor Ho: p = poH1: p < poH1: p > poReject Ho ifz < zz > z1-For example, if  = 0.05 reject Ho ifz < z0.05 = -1.645 z > z0.95 = 1.645p-value:area to left of z area to right of zSuppose n*po*qo < 5Use exact binomial probabilities (Table 1)X ~ Bi(n, po)pˆ= X/n where X = number of events and n = number of trialsCalculate exact p-valueIf pˆ pop = 2 * (Pr  X successes in n trials | Ho) knokoXkqpkn*02If pˆ> pop = 2 * (Pr  X successes in n trials | Ho)knokonXkqpkn*2Each represents the sum of the probabilities of all events that are as extreme or more extreme than the one that occurred. Example: Suppose that the failure rate for a certain electrical device is 25%. For a sample of 15 devices developed by a new method, 6 failed. Is this result statistically significant?2-tailed,  = 0.05, po = 0.25, qo = 0.75n*po*qo = 15*0.25*0.75 = 2.8125 < 5pˆ = 6/15 = 0.40 > 0.25Therefore from Table 1, n = 15 29660148302000000131003930091702750250152156..*.......*.. k*pknkkWe have no evidence to state that the failure rate by the new method is significantly different than the failure rate by the usual method.Note: One-tailed tests—specify direction before conducting the study. For calculating the p-value, use the same procedures except do not multiply by 2.Supplemental Problem:Suppose that three infants in a sample of 20 infants did not complete the hepatitis B vaccine series. Can we conclude on the basis of these data that, in the population, fewer than 40% have completed the series? Let  = 0.05. Give the appropriate null and alternative hypotheses, find the critical value(s), test the null hypothesis, give a decision, and write a conclusion. Please work problem and we will go over solution in the next class.PowerUse normal approximationHo: p = poH1: p = p1  pooooα/oooooα/ooqpn|p|pzqpqpzqpn|p|pzqpqpΦ12111211PrPowerExample: Suppose we want to test the hypothesis that boys ages 0-4 years whose mothers are smokers have a different incidence rate for asthma than is true in the general population. The incidence rate in the general population is 1.4% whereas we believe the rate is 2% for boys whose mothers are smokers.We will test 500 boys whose mothers are smokers. What is the power of such a study assuming that we conduct a 2-tailed test with  = 0.05? = 0.05, po = 0.014, qo = 0.986, p1 = 0.02, q1 = 0.98, n = 500     245106906865081810839201419196183920986001405000060961980020986001401211..Φ.Φ.*.Φ...Φ.*.*...*..*.Φqpn|p|pzqpqpΦoooα/ooPowerTherefore, there is only a 24.5% chance of finding a significant difference based on asample of 500 boys. Sample SizeHo: p = poH1: p = p1  po  212011121ooβα/ooppqpqpzzqpnHow many boys should we test to have 80% power if a two-sided significance test with  = 0.05? = 0.05, po = 0.014, qo = 0.986, p1 = 0.02, q1 = 0.98, 1-β = 80%              36237361300003607671801380400060191586184096101380400060986001409800209860014022228009750,or.,.............zz..n..Therefore, 3,362 boys should be tested to have an 80% change of detecting a difference of 0.006 using a 2-tailed test with  = 0.05. Note: For one-sided tests, substitute z1- for z1-/2 in sample size formula and z for z/2 in power