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Does too much sleep impair intellectual performance? Researchers examined this commonly held belief by comparing the performance of subjects on the mornings following (a) two normal night’s sleep and (b) two nights of “extended sleep.” In the morning they were given a number of tests of ability to think quickly and clearly. One test was for vigilance where the lower the score, the more vigilant the subject, (vigilance = alertness). The following data were collected: Subject 1 2 3 4 5 6 7 8 9 10 Normal 8 9 14 4 12 11 3 26 3 11 – L1 Extended 8 9 15 2 21 16 9 38 10 11 – L2 (L3 -> Difference of L2-L1) 1. If this experiment was properly designed, explain how the assignment of subjects to treatments could have been carried out. 2. Carry out an appropriate test to help answer the researchers’ question. Sleep and Performance 1. The subjects should all be tested under both conditions, but randomly assigned to one of 2 groups: Group 1: 2 nights of “normal sleep” followed by “extended sleep” Group 2: 2 nights of “extended sleep” followed by “normal sleep” and each should get a break in between to avoid any effects from the previous sleep impairment. Sleep and Performance 2. P: μ = mean difference in vigilance score (extended – normal) H: Ho: μ = 0, Ha: μ>0 A: SRS, Normality, Indep T: x-bar = 3.8, s = 4.60, n = 10, df = 9 Table: .01 <p<.02 Calc: p = .015 C: Reject the Ho; there is significance evidence that the scores for vigilance test were higher, on average, after extended sleep. ie: Too much sleep impairs intellectual performance. 12.2 Inference for a Population Proportion • We are interested in the unknown proportion p of a population that has some outcome – call the outcome we are looking for a “success.” • The statistic that estimate the parameter p is the sample proportion p-hat = count of successes in sample count of observations sample Recall: The sampling distribution of p-hat… • The mean of the sampling distribution is p. • The sample proportion p-hat is an unbiased estimator of the population proportion p. • The standard deviation of p-hat is p(1 p) n Only problem: we don’t know st. dev. of p-hat since don’t know p Solution to not knowing p To test the null hypothesis H 0 : p p0 that the unknown p has a specific value p0 , replace p by p0 in the test statistic: z pˆ p p (1 p ) n z pˆ p0 p0 (1 p0 ) n p: p̂ will to get In a confidence interval for be close to p if n is large, so replace the std. dev. by the standard error of p̂ . pˆ z * pˆ (1 pˆ ) n Assumptions for inference about a proportion • The data are an SRS from the population of interest • The population is at least 10 times as large as the sample (Independence). • For a test of H 0 : p p0, the sample size n is so large that: 10 np0 and 10 n(1 p0 ) (Normality) • For a confidence interval, n is so large that and 10 n(1 pˆ ) 10 npˆ (successes) (failures) Example • A coin that is balanced should come up heads half the time in the long run. The population for coin tossing contains the results of tossing the coin forever. The parameter p is the probability of a head, which is the proportion of all tosses that give a head. The tosses we actually make are an SRS from this population. The French naturalist Count Buffon (1707-1788) tossed a coin 4040 times. He got 2048 heads. The sample proportion of heads is 0.5069. That is more than one-half. • Is this evidence that Buffon’s coin was not balanced? • Find the 95% confidence interval. Buffon’s Coin • P: p = probability of a head p-hat = 2048/4040 = .5069 • H: Ho: p = .5, Ha: p ≠ .5 • A: SRS, Indep, Normality • T: Stat, Tests, 1-PropZTest (po:.5, x:2048, n:4040, prop≠po) Calculate => z = .88, p-val = .3783 Stat, Tests, 1-PropZInt(x:2048, n:4040, C-level: .95) Calculate => (.49151, .52235) • C: Fail to reject Ho; we are 95% confident that the probability of a heads is between above CI. Recall: n for a desired margin of error • To determine the sample size n that will yield a level C confidence interval for a population proportion p, solve the following for n: * p* (1 p* ) z n m • z * is the standard normal critical value for level of confidence we want • p* is a guess (or is from a study). Many colleges that once enrolled only male or only female students have become coeducational. Some administrators and alumni were concerned that the academic standards of the institutions would decrease with the change. One formerly all-male college undertook a study of the first class to contain women. The class consisted of 851 students, 214 of whom were women. An examination of first-semester grades revealed that 15 of the top 30 students were female. 1) What is the proportion of women in the class? Call this value p-nought. 2) Assume that the number of females in the top 30 is approximately a binomial random variable with n=30 and unknown probability p of success. In this case success corresponds to the student being female. What is the value of p-hat? 3) Are women more likely to be top students than their proportion in the class would suggest? State hypotheses that ask this question, carry out a significance test, and report your conclusion in non- technical language. College Study: 1. po = 214/851 = .2515 2. p-hat = 15/30 = .5 P: p = proportion of women who are top students 3. H: Ho: p = .2515, Ha: p>.2515 A: SRS, Normality, Indep z = (p-hat – po) / √[(po(1-po))/n] = 3.14 p-val=.00085 C: There is sufficient evidence to reject Ho and conclude that a higher % of the women are among the top 30 students than would be expected. Methods of Poll Explained • The Times-Dispatch/12 News poll was conducted by the research department of Medial General, Inc., parent company of the Times-Dispatch. Based on telephone interviews October 23 through Wednesday with 502 respondents who identified themselves as registered voters, the survey had a sampling error of plus or minus 4.5 percentage points. In other words, one could say with 95 percent certainty that the results of the poll would vary 4.5 percent in either direction if the entire adult population of Virginia had been polled. • Verify that the newspaper’s margin of error is correct. What is the exact margin of error? Methods Example: Margin of error = z* √[(p*(1-p*))/n] No p* given so use .5 m = 1.96 √[(.5)(.5)/502] = .044 => The newspapers margin of error is essentially correct.