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PSTAT 120B Probability and Statistics - Week 10 Fang-I Chu University of California, Santa Barbara June 7, 2013 Fang-I Chu PSTAT 120B Probability and Statistics Topic for review Exercise #10.55 Exercise #10.82 Exercise #10.88 Exercise #10.92 Fang-I Chu PSTAT 120B Probability and Statistics Exercise 10.55 10.55 A check-cashing service found that approximately 5% of all checks submitted to the service were bad. After instituting a check-verification system to reduce its losses, the service found that only 45 checks were bad in a random sample of 1124 that were cashed. Does sufficient evidence exist to affirm that the check-verification system reduced the proportion of bad checks? What attained significance level is associated with the test? What would you conclude at the α= .01 level? Fang-I Chu PSTAT 120B Probability and Statistics Exercise 10.55 Solution: Known: approximately 5% of all checks submitted to the service were bad. 45 checks were bad in a random sample of 1124 after instituting a check-verification system. Goal: decide whether check-verification system reduced the proportion of bad checks. i.e. Hypothesis: H0 : p = .05 versus Ha : p < .05 Way to approach: 45 Compute p̂ = 1124 = 0.040 Formula for test statistic: Z = q p̂−p0 p0 (1−p0 ) n ∼ N (0, 1) 0.04−0.05 = −1.538 Under H0 , test statistic is Z = √ 0.05×0.95 1124 The p-value= P(Z < −1.538) = 0.0616. Fail to reject H0 with α = .01 There is not enough evidence to conclude that the proportion of bad checks has decreased from 5%. Fang-I Chu PSTAT 120B Probability and Statistics Exercise 10.82 10.82 Exercises 8.83 and 10.73 presented some data collected in a 1993 study by Susan Beckham and her colleagues. In this study, measurements of anterior compartment pressure (in millimeters of mercury) were taken for ten healthy runners and ten healthy cyclists. The researchers also obtained pressure measurements for the runners and cyclists at maximal O2 consumption. The data summary is given in the accompanying table. condition Rest 80%maximal O2 consumption Maximal O2 consumption Fang-I Chu Runners Cyclists Mean s Mean s 14.5 3.92 11.1 3.98 12.2 3.49 11.5 4.95 19.1 16.9 12.2 PSTAT 120B Probability and Statistics 4.67 Exercise 10.82 10.82 (a) (a)Is there sufficient evidence to support a claim that the variability of compartment pressure differs for runners and cyclists who are resting? Use α = .05. Solution: Known: Let σ12 , σ22 denote the population variances for compartment pressure for resting runners and cyclists, respectively. Goal: test whether the variability of compartment pressure differs for runners and cyclists. i.e. Hypothesis: H0 : σ12 = σ22 , Ha : σ12 6= σ22 Fang-I Chu PSTAT 120B Probability and Statistics Exercise 10.82 10.82 (a) (a)Is there sufficient evidence to support a claim that the variability of compartment pressure differs for runners and cyclists who are resting? Use α = .05. Solution: Way to approach: Note s12 = 3.982 , s22 = 3.922 , as sample variance for runners and cyclists, respectively. S2 Formula for F- statistic: F = S12 ∼ Fn1 −1,n2 −1, α2 2 2 Compute F-statistics: F = 3.98 3.922 = 1.03 With α = .05, F9,9,.025 = 4.03 (look up from table) 1.03 < 4.03, we fail to reject H0 with α = .05 There is no evidence to conclude that the variability of compartment pressure is different for runners and cyclists. Fang-I Chu PSTAT 120B Probability and Statistics Exercise 10.82 10.82 (b) (b) i. What can be said about the attained significance level using a table in the appendix? (b) ii. Applet Exercise What can be said about the attained significance level using the appropriate applet? Solution: (i) From Table 7, p-value > .1. (ii) Using the Applet, 2P(F > 1.03) = 2(.4828) = .9656 Fang-I Chu PSTAT 120B Probability and Statistics Exercise 10.82 10.82 (c) (c)Is there sufficient evidence to support a claim that the variability in compartment pressure between runners and cyclists differs at maximal O2 consumption? Use α = .05. Known: Let σ12 , σ22 denote the population variances for compartment pressure for 80% maximal O2 consumption for runners and cyclists, respectively. Goal: test whether the variability of compartment pressure differs for runners and cyclists. i.e. Hypothesis: H0 : σ12 = σ22 , Ha : σ12 6= σ22 Fang-I Chu PSTAT 120B Probability and Statistics Exercise 10.82 10.82 (c) (c)Is there sufficient evidence to support a claim that the variability in compartment pressure between runners and cyclists differs at maximal O2 consumption? Use α = .05. Solution: Way to approach: Note s12 = 16.92 , s22 = 4.672 , as sample variance for runners and cyclists, respectively. S2 Formula for F- statistic: F = S12 ∼ Fn1 −1,n2 −1, α2 2 2 Compute F-statistics: F = 16.9 4.672 = 13.096 With α = .05, F9,9,.025 = 4.03 (look up from table) 13.096 > 4.03, we reject H0 with α = .05 There is evidence to conclude that the variability of compartment pressure between runners and cyclists exits. Fang-I Chu PSTAT 120B Probability and Statistics #10.82 10.82 (d) (d)i. What can be said about the attained significance level using a table in the appendix? (d)ii. Applet Exercise What can be said about the attained significance level using the appropriate applet? Solution: (i) From Table 7, p-value< 0.005. (ii) Using the Applet, 2P(F > 13.096) = 2(.00036) = .00072 Fang-I Chu PSTAT 120B Probability and Statistics Exercise 10.88 10.88 Refer to Exercise 10.2. Find the power of the test for each alternative in (a)-(d). (a) p = .4 (b) p = .5 (c) p = .6 (d) p = .7 (e) Sketch a graph of the power function. Fang-I Chu PSTAT 120B Probability and Statistics Exercise 10.2 10.2 An experimenter has prepared a drug dosage level that she claims will induce sleep for 80% of people suffering from insomnia. After examining the dosage, we feel that her claims regarding the effectiveness of the dosage are inflated. In an attempt to disprove her claim, we administer her prescribed dosage to 20 insomniacs and we observe Y , the number for whom the drug dose induces sleep. We wish to test the hypothesis H0 : p = .8 versus the alternative, Ha : p < .8. Assume that the rejection region y ≤ 12 is used. Fang-I Chu PSTAT 120B Probability and Statistics Exercise 10.88 10.88 Refer to Exercise 10.2. Find the power of the test for each alternative in (a)-(d). Known: Y is binomial with parameter n = 20 and p Hypothesis: H0 : p = .8 versus Ha : p < .8 Rejection Region: reject H0 if y ≤ 12 definition of power for a test: P( reject H0 |Ha true) Goal: Find the power of the test for alternative p = .4 Note: Power of a test =1- type II error Fang-I Chu PSTAT 120B Probability and Statistics Exercise 10.88 Solution (a) p = .4 Way to approach: P( reject H0 |Ha true) = P(Y ≤ 12|p = .4) 12 X 20 = 0.4y 0.620−y y y =0 = 0.979 (b) p = .5 Way to approach: P( reject H0 |Ha true) = P(Y ≤ 12|p = .5) 12 X 20 = 0.5y 0.520−y y y =0 = 0.86 Fang-I Chu PSTAT 120B Probability and Statistics Exercise 10.88 Solution (c) p = .6 Way to approach: P( reject H0 |Ha true) = P(Y ≤ 12|p = .6) 12 X 20 = 0.6y 0.420−y y y =0 = 0.584 (d) p = .7 Way to approach: P( reject H0 |Ha true) = P(Y ≤ 12|p = .7) 12 X 20 = 0.7y 0.320−y y y =0 = 0.228 Fang-I Chu PSTAT 120B Probability and Statistics Exercise 10.88 (e) Sketch a graph of the power function. Fang-I Chu PSTAT 120B Probability and Statistics Exercise 10.92 10.92 Consider the situation described in Exercise 10.91. What is the smallest sample size such that an α = .05-level test has power at least .80 when µ = 8? 10.91 Let Y1 , Y2 , . . . , Y20 be a random sample of size n = 20 from a normal distribution with unknown mean µ and known variance σ 2 = 5. We wish to test H0 : µ = 7 versus Ha : µ > 7. Fang-I Chu PSTAT 120B Probability and Statistics Exercise 10.92 Solution Known: Yi follows normal distribution with unknown mean µ and known σ 2 = 5, sample size is 20. Hypothesis:H0 : µ = 7 versus Ha : µ > 7 Goal: Find the smallest sample size such that an α = .05-level test has power at least .80 when µ = 8 i.e. Ha : µ = 8 Fang-I Chu PSTAT 120B Probability and Statistics Exercise 10.92 Solution Way to approach: from 10.91(a), we obtain rejection region as reject H0 if Ȳ > 7..82. (Check this!) Using the definition of power, we write P(reject H0 |Ha true) = P(Ȳ > 7.82|µ = 8) 7.82 − 8 = P(Z > q ) = .80 5 n √ 5 = −.84 From table, we have z.80 = −.84. So we have 7.82−8 n Solve this equation, we obtain n = 108.89. After rounding up, at least 109 observations must be taken such that an α = .05-level test has power at least .80 when µ = 8. Fang-I Chu PSTAT 120B Probability and Statistics