Exam 3. Instructor: Cynthia Rudin TA: Dimitrios Bisias. November 22, 2011

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1 Exam 3 Istructor: Cythia Rudi TA: Dimitrios Bisias November 22, 2011 Gradig is based o demostratio of coceptual uderstadig, so you eed to show all of your work. Problem 1 A compay makes high-defiitio televisios ad does ot like to have defective pixels. Historically, the mea umber of defective pixels i a TV is 20. A MIT egieer is hired to make better TV s that have fewer defective pixels. After her first week of work she claims that she ca sigificatly improve the curret method. To check her claim you try her ew method o 100 ew televisios. The average umber of defective pixels i those 100 TV s is Assume that the ew method does t chage the stadard deviatio of defective pixels, which has always bee Test if the ew method is sigificatly better tha the old oe at the α = 0.05 level. 2. Usig the ew method, assume that the mea umber of defective pixels is actually 19. What is the chace that your test from part 1 will coclude that the ew method is statistically more effective? 3. How may televisios will you have to check so that the test you did i part 1 will coclude that the ew method is effective, with 95% probability? Please assume agai that the mea umber of defective pixels is actually 19. Solutio 1. The hypotheses are: H 0 : µ = 20 H 1 : µ < 20 We perform a z-test. Uder the ull hypothesis, the z-statistic is approximately stadard ormal. The observed value of the z-statistic is: x µ z = = = 2.25 σ 4/10 1

2 Sice this value is less tha z α = z 0.05 = 1.645, the ull hypothesis is rejected at the α = 0.05 level. 2. The questio asks for the power of the test at 19. The rejectio regio is z < 1.645, or x < = The power of the test is 100 π(19) = P (Test rejects H 0 µ = 19) = P ( x < µ = 19) x µ = P < = σ/ 4/ 100 Usig Eq.(7.8), we get the same result: (µ 0 µ) (20 19) 100 π(µ) = π(19) = Φ z α + = Φ = σ The rejectio regio for a sample of tires is x < We wat such that Sice P (Z < 1.645) = 0.95, we have The least ( ) P x < µ = / 4 Usig Eq.(7.10), we get the same result: 2 = , so 174 televisios should be tested for the power to be at (z α + z β )σ (z z )4 ( )4 = = = µ 0 µ so agai 174 televisios should be tested for the power to be at least

3 Problem 2 Two methods of memorizig words are to be compared. You choose two groups of 5 people, where the first perso i the first group has the same characteristics as the first perso i the secod group (they have the same educatioal level, age, etc.). Same thig about the secod perso i each group - they are also similar to each other i terms of educatio, age, etc. Same thig for the third, fourth ad fifth people from each group. The first group is assiged to the first method of memorizatio ad the secod group to the other method. The umber of words recalled i a memory test after a week s traiig with these two methods is show below. Method 1 Method 2 Pair 1 Pair 2 Pair 3 Pair 4 Pair Test the hypothesis that the first method is better tha the secod method at the 0.05 level. You may assume ormality of the data. Solutio We will perform a paired t-test. The hypotheses are: H 0 : µ 1 = µ 2 H 1 : µ 1 > µ 2 We have: d 1 = = 4, d 2 = = 10, d 3 = = 1, d 4 = = 9, d 5 = = 12. It is: d = 6.8 ad i=1(d(i) d ) s 2 d = = Uder the ull hypothesis, the t-statistic d t = s d / has a t 4 distributio. Sice 6.8 t = = 2.89 > t 4,0.05 = / 5 we coclude that the first method is sigificatly better tha the secod method. 3

4 Problem 3 After your MIT graduatio you decide to go to Mote Carlo for vacatio. You visit a casio ad decide to gamble; specifically, you wat to play roulette. Before you start bettig, you watch 500 roulette games at the casio, ad you fid that red is hit 260 times. Ca you determie whether the roulette is fair? I case you do t kow how roulette is played i Mote Carlo, it ivolves spiig a wheel that has may slots aroud it. There is 1 gold slot, 18 red slots, ad 18 black slots. 1. Set up the hypotheses. 2. Calculate the P-value. Ca you reject ull hypothesis at the 0.05 level? 3. Fid a 95% CI for the proportio of red results. Solutio 1. We let p be the probability of a red slot o a sigle spi, which is = if the roulette is fair. The alterative hypothesis for this test is two-sided because we just wat to check if p is differet from 18 The hypotheses are: 37. H 0 : p = p 0 = H 1 : p = p 0 = Sice = 500, we do a large-sample test. We are give the total umber of red hits is Y = 260, so the sample proportio is Y 260 pˆ = = = Uder the ull hypothesis, ˆp is approximately ormally distributed with mea p 0 = = p 0 q ad variace = 500 = Thus the followig test statistic is approximately stadard ormal: pˆ p 0 Z =. p 0 q 0 / The observed value of the test statistic is z = = = The p-value is 2 P (Z > 1.5) = So o, we caot reject H 0 at the 0.05 level because > To compute a cofidece iterval, we estimate the variace of ˆp by pˆqˆ. A 95% CI for the proportio of red hits is pˆqˆ pˆ ± z α/2 = 0.52 ± 1.96 = [0.4762, ]. 500 Notice that is i this iterval, so we do ot reject H 0. 4

5 Problem 4 The weight of a object is measured usig a electroic scale that reports the true weight plus a radom fluctuatio that is ormally distributed with zero mea. The maufacturig compay of the electroic scale claims that the stadard deviatio of the fluctuatio is 2 milligrams. Assume that the fluctuatios are idepedet. The compay measures the weight of oe object 8 times, ad observes these values: 100.9, 98.2, 101.5, 102.2, 105.1, 99.4, 93.6, You believe that the stadard deviatio of the fluctuatio is greater tha what the compay claims. Is there statistically sigificat evidece for your belief at α = 5% level? 2. Compute a 95% cofidece iterval for the variace of the fluctuatio. Solutio 1. We perform a chi-square test with hypotheses: Whe H 0 is true, the statistic H 0 : σ = 2 H 1 : σ > 2 ( 1)s 2 χ 2 = has a χ 2 7 distributio. The observed value of the chi-square statistic is: σ 2 ( 1)s 2 χ 2 = = 21.1 σ 2 The P-value is P (χ 2 7 > 21.1) = ad therefore there is statistically sigificat evidece for our belief at α = 5% level. 2. A 95% two-sided cofidece iterval for the variace of the fluctuatio is give by: 2 2 ( 1)s ( 1)s, = [5.27, 49.94] χ 2 χ 2 7, ,

6 Problem 5 You are a ivestmet maager of a asset maagemet compay ad you wat to evaluate the performace of two hedge fuds. For this you record their past performace for the last 10 years. The returs for both of the hedge fuds i differet years are idepedet ad ormally distributed with ukow mea ad variace. The mea retur for Hedge Fud 1, averaged over the 10 years, is X = 22.15% ad the stadard deviatio calculated from those 10 measuremets is 2.79%. For Hedge Fud 2 the mea retur, averaged over the 10 years, is Y = 20.6% ad the stadard deviatio calculated from those 10 measuremets is 2.46%. 1. Test equality of the variaces for the two hedge fuds usig α = Test for the equality of the meas of the two hedge fuds performaces, usig α = 0.05 with a suitable test. Would you recommed usig a t-test calculated from a pooled variace or the method for uequal variaces? If you rejected the test i (1) the you may assume the variaces are uequal, otherwise you may assume that the variaces are equal. Solutio 1. We perform a F test with hypotheses: The F -statistic is H 0 : σ 2 = σ H 1 : σ 2 = σ s2 1 2 F = = 1.29 s 2 Uder H 0 this statistic has F 9,9 distributio. From the table i the book we see that this is less tha the critical value 3.18, therefore we do ot reject the ull hypothesis. 2. The hypotheses are: H 0 : µ 1 = µ 2 H 1 : µ 1 = µ 2 Sice we do ot reject the hypothesis that σ 2 = σ 2 2, we compute the pooled variace. 1 ( )s 1 + ( 2 1)s s = = = ( 1 1) + ( 2 1) Uder the ull hypothesis that the mea returs of the two hedge fuds are equal,the t-statistic: t = X Ȳ S

7 has a t = t 18 distributio. Its observed value is: x ȳ t = = s Sice t is less tha t 18,0.025 = 2.1, we do ot reject the ull hypothesis at α = 5% level. 7

8 MIT OpeCourseWare J / ESD.07J Statistical Thikig ad Data Aalysis Fall 2011 For iformatio about citig these materials or our Terms of Use, visit:

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