Aoucemets MidtermII Review Sta 101 - Sprig 2017 Duke Uiversity, Departmet of Statistical Sciece Whe: Tuesday, April 4 - I class What to brig: Scietific calculator (graphig calculator ok, No Phoes!) Writig utesils Cheat sheet (ca be typed!) Tables will be provided PS 5 due March 31 at 11:55 PM. PA 5 is due April 2 at 11:55 PM. Dr. Mukherjee Slides posted at http://www2.stat.duke.edu/courses/sprig17/sta101.002/ 1 Exam Format Uit 4.1 - Iferece for Numerical Variables Covers HT from Uit 3, Uits 4 ad Uit 5 3 writte questios 70 pts 10 multiple choice questios - 3pts each Two mea testig problems Idepedet meas Paired (depedet) meas Coditios Idepedece Skew or Approximate Normality 2 3
All other details of the iferetial framework is the same... Uit 4.2 - Bootstrappig Oe mea: df = 1 HT: H 0 : µ = µ 0 T df = x µ s CI: x ± t df s HT : test statistic = poit estimate ull SE CI : poit estimate ± critical value SE Paired meas: df = diff 1 HT: H 0 : µ diff = 0 T df = x diff 0 s diff CI: x diff ± t df diff s Idepedet meas: df = mi( 1 1, 2 1) HT: H 0 : µ 1 µ 2 = 0 T df = x 1 x 2 s 2 1 + s2 2 1 2 CI: x 1 x 2 ± t df s 2 1 1 + s2 2 2 Bootstrappig works as follows: (1) take a bootstrap sample - a radom sample take with replacemet from the origial sample, of the same size as the origial sample (2) calculate the bootstrap statistic - a statistic such as mea, media, proportio, etc. computed o the bootstrap samples (3) repeat steps (1) ad (2) may times to create a bootstrap distributio - a distributio of bootstrap statistics The XX% bootstrap cofidece iterval ca be estimated by the cutoff values for the middle XX% of the bootstrap distributio, OR poit estimate ± t SE boot 4 5 Bootstrap iterval, stadard error Uit 4.4: Power For a radom sample of 20 Horror movies, the dot plot below shows the distributio of 100 bootstrap medias of the Rotte Tomatoes audiece scores. The media of the origial sample is 43.5 ad the bootstrap stadard error is 4.88. Estimate the 90% bootstrap cofidece iterval for the media RT score of horror movies usig the stadard error method. 30 35 40 45 50 55 bootstrap medias Decisio fail to reject H 0 reject H 0 H 0 true 1 α Type 1 Error, α Truth HA true Type 2 Error, β Power, 1 β Type 1 error is rejectig H 0 whe you should t have, ad the probability of doig so is α (sigificace level) Type 2 error is failig to reject H 0 whe you should have, ad the probability of doig so is β (a little more complicated to calculate) Power of a test is the probability of correctly rejectig H 0, ad the probability of doig so is 1 β I hypothesis testig, we wat to keep α ad β low, but there are iheret trade-offs. 6 7
SEROTONIN (NG/GM) HEART DISEASE CONTROLS 8 12 yq 3,840 5,310 SE 850 640 (a) For these data, the SE of (Y 1 - Y 2 ) is 1,064 ad df = 14.3 (which ca be rouded to 14). Use a t test Appilcatio Exercise 1.344.4 t I a study of the periodical to compare cicada the meas (Magicicada at the 5% sigificace septedecim), level. Use a = 0.05. (b) Verify the value of SE researchers measured the hid tibia (Y1 - Ylegths 2) give i part (a). of the shed skis of 110 7.2.8 I a study of the periodical cicada (Magicicada idividuals. Results septedecim), for males researchers ad females measured the are hid show tibia i the legths of the shed skis of 110 idividuals. Results for accompayig table. males ad females are show i the accompayig table. 9 TIBIA LENGTH ( m) GROUP MEAN SD Males 60 78.42 2.87 Females 50 80.44 3.52 (a) Use a t test to ivestigate the associatio of tibia legth o geder i this species. Use the 5% sigificace level. [Note:Formula(6.7.1) yields 94.3 df.] (b) Give the precedig data, if you were told the tibia legth of a idividual of this species, could you make a fairly cofidet predictio of its sex? Why or why ot? (c) Repeat the t test of part (a), assumig that the meas ad stadard deviatios were as give i the table, but that they were based o oly oe-teth as may idividuals (6 males ad 5 females). [Note: Formula (6.7.1) yields 7.8 df.] 2.28 4.68 2.24 3.50 2.17 10 8 yq 2.51 5.14 s 0.60 0.84 Use a test to ivestigate the effect of hypoxia o MBF. 7.2.10 I a study of the developmet of the thymus glad, researchers weighed the glads of 10 chick embryos. Five of the embryos had bee icubated 14 days ad 5 had bee icubated 15 days. The thymus weights were as show i the table. 11 [Note: Formula (6.7.1) yields 7.7 df.] We wat to test to see if there is a icrease i the average tibia legth of male ad female cicadas. Calculate the power of the test to detect a icrease of 2 µm. THYMUS WEIGHT (MG) 14 DAYS 15 DAYS 29.6 32.7 21.5 40.3 28.0 23.7 34.6 25.2 44.9 24.2 5 5 yq 31.72 29.22 s 8.73 7.19 (a) Use a t test to compare the meas at a = 0.10. (b) Note that the chicks that were icubated loger had a smaller mea thymus weight. Is this backward result surprisig, or could it easily be attributed to chace? Explai. 8 Achievig desired power There are several ways to icrease power (ad hece decrease Type 2 error rate): 1. Icrease the sample size. 2. Decrease the stadard deviatio of the sample, which is equivalet to icreasig the sample size (it will decrease the stadard error). With a smaller s we have a better chace of distiguishig the ull value from the observed poit estimate. This is difficult to esure but cautious measuremet process ad limitig the populatio so that it is more homogeous may help. 3. Icrease α, which will make it more likely to reject H 0 (but ote that this has the side effect of icreasig the Type 1 error rate). 4. Cosider a larger effect size. If the true mea of the populatio is i the alterative hypothesis but close to the ull value, it will be harder to detect a differece. 9 Uit 4.3: Aalysis of VAriace (ANOVA) ANOVA tests for some differece i meas of may differet groups ANOVA tests for some differece i meas of may differet groups Coditios 1. Idepedece: (a) withi group: sampled observatios must be idepedet, i.e., SRS + 10% rule (b) betwee group: groups must be idepedet of each other 2. Approximate ormality: distributio should be early ormal withi each group 3. Equal variace: groups should have roughly equal variability 10 Null hypothesis: H 0 : µ placebo = µ purple = µ brow =... = µ peach = µ orage. Clicker questio Which of the followig is a correct statemet of the alterative hypothesis? (a) For ay two groups, icludig the placebo group, o two group meas are the same. (b) For ay two groups, ot icludig the placebo group, o two group meas are the same. (c) Amogst the jelly bea groups, there are at least two groups that have differet group meas from each other. (d) Amogst all groups, there are at least two groups that have differet group meas from each other. 11
To idetify which meas are differet, use t-tests ad the Boferroi correctio F-statistic: F = k: # of groups; : # of obs. SSG / (k 1) SSE / ( k) = MSG MSE Df Sum Sq Mea Sq F value Pr(>F) Betwee groups k 1 SSG MSG F obs p obs Withi groups k SSE MSE Total 1 SSG+SSE Note: F distributio is defied by two dfs: df G = k 1 ad df E = k Thep-valuewillbegiveoexam, comparewiththestadard α level. If the ANOVA yields a sigificat results, ext atural questio is: Which meas are differet? Use t-tests comparig each pair of meas to each other, with a commo variace (MSE from the ANOVA table) istead of each group s variaces i the calculatio of the stadard error, ad with a commo degrees of freedom (df E from the ANOVA table) Compare resultig p-values to a modified sigificace level α = α K where K = k(k 1) 2 is the total umber of pairwise tests 12 13 To idetify which meas are differet, use t-tests ad the Boferroi correctio Uit 4.3: ANOVA ApplicatioExercise4.3 Youwillotbeaskedtoperformtheactualtests, butyou shouldkow: How to compute the adjusted Boferoi sigificace level α. How to compute the stadard error for this test. The associated degrees of freedom for the test statistic. Df Sum Sq Mea Sq F p- value Rak 2 1.59 0.795 2.74 0.066 Residuals 460 135.07 0.29 Total 462 136.66 Whatpercetofthetotalvariabilityievaluatioscoresis explaiedbyistructorrak? 14 15
Uit 4.3: ANOVA Uit 5.1: Iferece for a Sigle Proportio ApplicatioExercise4.3 Df Sum Sq Mea Sq F p- value Rak 2 1.59 0.795 2.74 0.066 Residuals 460 135.07 0.29 Total 462 136.66 Whatsigificacelevelshouldbeusedforapair-wisepost hoctestcomparigtheevaluatioscoresofteachig professorsadteuredprofessors? Distributioof ˆp Cetral limit theorem for proportios: Sample proportios will be early ormally distributed with mea equal to the populatio mea, p (1 p) p, ad stadard error equal to. ( ) p (1 p) ˆp N mea = p, SE = Coditios: Idepedece: Radom sample/assigmet + 10% rule At least 10 successes ad failures 16 17 Uit 5.1: Iferece for a Sigle Proportio HT vs. CI foraproportio Success-failure coditio: CI: At least 10 observed successes ad failures HT: At least 10 expected successes ad failures, calculated usig the ull value Stadard error: CI: calculate usig observed sample proportio: SE = HT: calculate usig the ull value: SE = p 0 (1 p 0 ) ˆp(1 ˆp) Clicker questio Suppose p = 0.05. What shape does the distributio of ˆp have i radom samples of = 100. (a) uimodal ad symmetric (early ormal) (b) bimodal ad symmetric (c) right skewed (d) left skewed 18 19
Recap o simulatio methods Radomizatio Test If the S-F coditio is ot met HT: Radomizatio test simulate uder the assumptio that H 0 is true, the fid the p-value as proportio of simulatios where the simulated ˆp is at least as extreme as the oe observed. CI: Bootstrap iterval resample with replacemet from the origial sample, ad costruct iterval usig percetile or stadard error method. Clicker questio A report o your local TV statio says that 60% of the city s residets support usig limited city fuds to hire ad trai more police officers. A secod local ews statio has picked up this story, ad they claim that certaily less tha 60% of residets support the additioal hirig ad traiig. I order to test this claim the secod ews statio takes a radom sample of 100 residets ad fids that 57 of them (57%) support the use of limited fuds to hire additioal police officers. 20 21 Clicker questio Which of the followig is the correct set-up for calculatig the p-value for this test? (a) Roll a 10-sided die (outcomes 1-10) 100 times ad record the proportio of times you get a 6 or lower. Repeat this may times, ad calculate the proportio of simulatios where the sample proportio is 57% or less. (b) Roll a 10-sided die (outcomes 1-10) 100 times ad record the proportio of times you get a 6 or lower. Repeat this may times, ad calculate the proportio of simulatios where the sample proportio is 60% or less. (c) I a bag place 100 chips, 57 red ad 43 blue. Radomly sample 100 chips, with replacemet, ad record the proportio of red chips i the sample. Repeat this may times, ad calculate the proportio of samples where 57% or more of the chips are red. (d) Radomly sample 100 residets of a earby city, record how may of the them who support the hirig ad traiig of additioal police officers. Repeat this may times ad calculate the proportio of samples where at least 57% of the residets support additioal hirig ad traiig. Clicker questio Is owig a pet associated with lower stress levels? Researchers measured blood cortisol levels (a marker for stress) of 20 pet owers ad 20 people who do ot have pets. I the pet ower group 8 people had above ideal blood cortisol levels (over-stressed) ad i the o pet ower group 10 did. To test whether the observed differece could be attributed to chace, a statistics studet coducted a radomizatio test. She wrote over-stressed o 18 (10 + 8) out of 40 (20 + 20) idex cards, shuffled them, ad dealt them ito two groups of 20 cards. Which of the followig best describes the outcome? (a) If pet owership has a effect o stress, the differece betwee the umbers of over-stressed cards i the two stacks is expected to be more tha 2. (b) The differece betwee the umbers of over-stressed cards i the two stacks is expected to be 0. (c) The differece betwee the umbers of over-stressed cards i the two stacks is expected to be more tha 2. 22 23
Uit 5.2: Iferece for Two Proportios Uit 5.2: Iferece for Two Proportios CLT alsodescribesthedistributioof ˆp 1 ˆp 2 (ˆp 1 ˆp 2 ) N mea = (p 1 p 2 ), SE = Coditios: + p 2(1 p 2 ) 1 p 1 (1 p 1 ) Idepedece: Radom sample/assigmet + 10% rule Success-failure coditio: At least 10 successes ad failures 2 ForHT where H 0 : p 1 = p 2, pool! As with workig with a sigle proportio, Whe doig a HT where H 0 : p 1 = p 2 (almost always for HT), use expected couts / proportios for S-F coditio ad calculatio of the stadard error. Otherwise use observed couts / proportios for S-F coditio ad calculatio of the stadard error. Expected proportio of success for both groups whe H 0 : p 1 = p 2 is defied as the pooled proportio: ˆp pool = total successes total sample size = suc 1 + suc 2 1 + 2 24 25 Summary Uit 5.3: χ 2 Tests Type Parameter Estimator SE SampligDist. Oe mea µ x s/ t 1 Two meas µ diff x diff s d / t 1 Paired data Two meas t df s µ 1 µ 2 x 1 x 2 1 2 + s2 2 1 2 for df use Idepedet mi{ 1 1, 2 1} ˆp(1 ˆp) C.I. Oe prop p ˆp Z p0 (1 p H.T. 0 ) ˆp1 (1 ˆp C.I. 1 ) + ˆp 2 (1 ˆp 2 ) 1 2 Two prop p 1 p 2 ˆp 1 ˆp 2 Z ˆppool (1 ˆp H.T. pool ) + ˆp pool (1 ˆp pool ) 1 2 Categoricaldatawithmoretha2levels χ 2 oe variable: χ 2 test of goodess of fit, o CI two variables: χ 2 test of idepedece, o CI Coditiosfor χ 2 testig 1. Idepedece: I additio to what we previously discussed for idepedece, each case that cotributes a cout to the table must be idepedet of all the other cases i the table. 2. Sample size / distributio: Each cell must have at least 5 expected cases. 26 27
The χ 2 statistic χ 2 statistic: Whe dealig with couts ad ivestigatig how far the observed couts are from the expected couts, we use a ew test statistic called the chi-square (χ 2 ) statistic: χ 2 = k (O E) 2 i=1 Importatpoits: E where k = total umber of cells Use couts (ot proportios) i the calculatio of the text statistic, eve though we re truly iterested i the proportios for iferece Expected couts are calculated assumig the ull hypothesis is true The χ 2 distributio The χ 2 distributio has just oe parameter, degrees of freedom (df), which iflueces the shape, ceter, ad spread of the distributio. For χ 2 GOF test: df = k 1 For χ 2 idepedece test: df = (R 1) (C 1) Degrees of Freedom 0 5 10 15 20 25 2 4 9 28 29