week15-4up - 311 STA 2023 Final Exam Locations STA 2023 c...

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Unformatted text preview: 311 STA 2023 Final Exam Locations STA 2023 c D.Wackerly - Lecture 23 STA 2023 c D.Wackerly - Lecture 23 312 Assignments Wednesday, May 1, 10:00 am – 12:00 noon Conflicts – see Wackerly NOW Today : P. 389 – 396 For Tuesday : Exer. 9.29, 9.33, 9.35, 9.38, 9.39, 9.42, 9.94, 9.96, 9.101, 9.110, Final Exam Sections &&¢¡ §¤¢¡ §¤¤¢ ¥¥¥¨'££©££¡ ¡ §¡¤¡ §& ¦©¥"¡ &¡ § &¡ § ¤ "©'¦"(''¡ %£¥©¥£¥©£!¡ ¤¡¡ §&¡¡ §¡  ¥$!©£%¥£¡ ¡¡ §¤ ¡ §#¢ ¡¡¡¡ §¡¡¡ §¤¢ ¥¥¥©¨!¥©¥£¡ ¤¡¡¡ §##¢¡ §#¢ $¥¥£©¨¦¥£¡ ¤¢¡¡ § ¡ § £¥¥¨¦"©"¨! !¨©¥£¡ ¡¡ §¡ §¢ ¢¤¡ §¤¢¡ §¡¤¢ ©£©¨¦¥£¡ Terry Mashtare Pavlina Rumcheva Mathew Smeltzer Kelly Sodec CSE A101 Last Time : Small sample inferences about CSE A101 Assumptions: (p. 379) 1. : Same variances, McC C 100 TUR L007 TUR L007 CAR 100 STA 2023 c D.Wackerly - Lecture 23 313 Pooled estimator for common variance Bc  XT2 4 V@ U a !Y2 0 R@ ` Q b B Q 4 XT(WV@ 10 !TSR@ B 2 4 Q U GB 2 0 Q 4G 4 I PH G 4 # d.f. 3. : Independent samples. : Normal distributions. STA 2023 c D.Wackerly - Lecture 23 314 Ex. : #9.26, p. 389 Sea urchins were starved for 48 hours, then fed a 5 cm blade of turtle grass. 10 urchins were randomly selected and fed fresh grass, another independently selected 10 were fed decaying grass. The measurements were the time (in hours) necessary f y €v to ingest the grass blades.   w xv trrep h f e us'qig"d C.I. for 2. Saurabh Kumar problems on syllabus! McC C 100 B9 FE@ Donte Ford 9.60, 9.97, 9.100, 9.107 and rest of Chpt 9 CAR 100 B7 D@ David Finlay For Thursday : Exer. 9.46, 9.52–54, 9.56, 9.59, CAR 100 B6 CA@ Adam Glassman CAR 100 Wednesday : p. 402 – 406 97 ¨86 Shamshad Ali David Ashley Location 4) 20 531) Instructor Decayed Blades (1) (2) ‰ 4H !G 4 '‡…4  ‚0  ˆ †„ ƒ 2 Number of urchins 10 10 Mean Ingestion Time 3.35 2.36 Standard Deviation 0.79 0.47 ‘ 4 Q U 0Q (table value)(standard error) Test Statistic d¥™— 0 U ˜— –H 4 0 • !G 2 ” I„ estimator hyp. value (standard error) e “’ …2 4  2 0  I test stat Construct a f¢ ¦¥# ƒ estimator Green Blades confidence interval for the difference in mean ingestion times for urchins fed green and decaying grass. 315 STA 2023 c D.Wackerly - Lecture 23 STA 2023 c D.Wackerly - Lecture 23 316 90% CI : confidence level. All “plausible B¡ ¦…2 4 Q U 0 R@ Q 4 !Y2 4 R@ S0 !Y2 0 R@ B Q U GB Q 4G 4 e values” for the difference in means (green vs. 4H G I decayed) are . I In terms of this example, what are the assumptions I necessary for the above test and CI to be valid? 97 '86 e – : the POPULATION variances of 6  I ¢£ ¢„  ¤¡  ‘ 4 Q – ame for are approximately the grass. : the sea urchins in the study were 7 I , 6 c a ` b ¢ ¥& d.f. U  0Q ‰ 4H G I d.f. f¢ ¦¥# , at the assigned to green and 4 '‡…4  ‚0  ˆ †„ ƒ 2 ƒ decayed grass. 9 – or : the a 5 cm blade of sea grass is approximately ¢ £ ƒ (A bunch of “scholarly” words about sea urchins, green distributed for both green and decayed grass. grass and decayed grass) Is the average time it takes sea urchins to consume green grass larger than the average time to consume decayed grass? STA 2023 c D.Wackerly - Lecture 23 318 e Each individual subject had BP taken before AND Ex. : #9.101, P. 430 Study to assess the effect of Sample sizes are small. e 317 STA 2023 c D.Wackerly - Lecture 23 biofeedback exercises on blood pressure. Six subjects after learning biofeedback. were taught biofeedback. Blood pressure e What happens if someone is careful about diet and measurements (millimeters of mercury) were taken not overweight? before and after the training. Mean blood pressure after . e  ¢ I¥ learning biofeedback less than before? Use Is likely that both before and after BP measurement will be low. B X @ B¡ ¦E@ Before After e Samples are ¡ ¥" ¢ e Why take before and after BP’s on same subject? ¤ £¡ ¢ ¢ ¥" e What “factors” can impact BP?  ¥" && &  #¤ – ¢ £ ¢ ¡ £  – ¡ $!  & " ¡& –  ¥" #&  ¥" ¢& – e # ¥" & Subject independent!!! HOW do we analyse the data???  ¡  ¤  & 319 STA 2023 c D.Wackerly - Lecture 23 STA 2023 c D.Wackerly - Lecture 23 320 Method to consider : Objective : Compare durability of two brands of steel e Randomly select one tire of each brand to be belted radial tires. installed on the front of each car. Randomly select some tires of each brand – install on e How about left and right sides? cars – drive around – record mileages. e This strategy will control for (2) – (6). Problem : Samples are NOT independent Things that influence mileage: e ¡¡ 1. Quality of the tires Can’t analyse data using method of Section 9.1 PAIRED DIFFERENCE EXPERIMENT 2. Weight of cars 3. Speed driven e Treated wood last longer than untreated? 4. Road conditions and Surface e Two brands of solar collectors : one better? 5. Driving habits of drivers e Before/After Experiments, Biofeedback #9.101 6. Etc. e Why Pair? – Someone else collected the data in a paired Want: information about (1), compensating (controlling manner. for) (2) – (6). HOW?? – By design to control for other factors 321 STA 2023 c D.Wackerly - Lecture 23 STA 2023 c D.Wackerly - Lecture 23 322 Method of Analysis : do a one-sample “t” ON THE were taught biofeedback. Blood pressure measured —© (millimeters of mercury) were taken before and after the training. Mean blood pressure after learing biofeedback Before After  # ¥" & ¡ ! ¤ # £ £¡ &  #¤ ¡ £  ¤¢ $¡  & " ¡& & " ¢ & ¡ ¥ ¥ ¡ # ¤¤ £& ¥ ¡ &  && ¤ £¦#  ¥" ¢& ¤ £¡ ¢ ¡  ¥" &&  ¢ £ ¢ ¡ $!   ¥" #& “’ 2 I  ) ¨“   “’ ¤   & B is NOT required ¢ ¥"  I 4 5) 8) 20 I e  Q e I  ) e ) 4) 20  I 53‚8) ¨“ e  NORMALLY distributed ¢ £¡  I ASSUMPTION : the DIFFERENCES are approx. # ¡ ¤ £¦¤ . Diff. ¡ ¥" ¢ “2”. Diff. #¤ ¥ ¤ same as Subject @ 0) if differences are taken “1” B ! @ ¢¢ ££¢ mean of the population of DIFFERENCES . B¡ E@ ¥ ¦¤ 4) less than before? Use differences B §§ ¨¨§ depen. biofeedback exercises on blood pressure. Six subjects . . . Totals 4 ¢¢ ££¢ ¥ ¦¤ . . . I¥ §§ ¨¨§ Pairs © . . . Ex. : #9.101, P. 430 Study to assess the effect of 4 ¢¢ ££¢ ¥ ¦¤ Indep. §§ ¨¨§ depen. © depen. DIFFERENCES Difference  ¢ Popn. 2 0 Popn. 1 4 4" " I 0 #@ 4 !    323 STA 2023 c D.Wackerly - Lecture 23 STA 2023 c D.Wackerly - Lecture 23 324 P-value?  I B 4© ‘ ¡ ¢@  Q ¡ ¤ 4 £ ¢„ ¢ f v wv ƒ t C.I. for (table value)(standard error)  4) 20  I 53‚8) ¨“  G  I ¢ ¥4) 20 ¦538)    ) ¨“    ) 4 ¨† „ …  ˆ ƒ I ¤ 4 £¡ „ I £ ¤ I  ¢ I I¥  ¢ I¥ that the mean blood pressure reading is higher before ƒ ¡ " ¢ learning biofeedback. STA 2023 c D.Wackerly - Lecture 23 325 or  I ¢£ ¡ „ ¤ ƒ I„ I level there    £ ¤  Q ¨ ©§ G “’ …2   d.f. ©  Q Conclusion : At the (3.365) Q  G  T2 4G I I I (2.571) estimator d.f. 2.977  y  Mean blood pressure larger before than after?  v ‰ I ¤¤ ¢ ¥¤ $ 0 0 £ ¢„ ¡ 4© 2  d.f. STA 2023 c D.Wackerly - Lecture 23 326 Why were the data collected this way? Is there a “big” difference in the mean BP readings measurements? (and others) level.  ¢ I¥ e – Between f # mercury with and milliliters of confidence. 3. General physical condition 4. Weight of patients 5. Lifestyle – MORE information in CI, no more work!!! Why did we analyse the data using the f  paired difference . 2. The age of the patients test? The manner in which the data was collected 6. Stress level 7. Gender 8. Ethnic background 9. Etc. dictated the method of analysis Can assess the impact of B ! @ e CAN tell how big the difference is from the CI 1. Learing biofeedback  Can’t tell from hypothesis test. CAN say there is A  ¨ before and after learning biofeedback? difference at the What “factors” could have an impact on the BP , controlling for the others by collecting the data in this manner!!! e 327 STA 2023 c D.Wackerly - Lecture 23 STA 2023 c D.Wackerly - Lecture 24 328 Minitab? STA 2023 Final Exam Locations e Basic Statistics Paired Stat Wednesday, May 1, 10:00 am – 12:00 noon Conflicts – see Wackerly NOW „ e Punch in data values e Click in box labelled ”First”, double click on Variable 1 (Before in this case). Final Exam Instructor e Click Options, type in confidence level (for CI) e Choose alternative (Greater than in this case), null Sections Shamshad Ali Variable 2. (After in this case) &&¢¡ §¤¢¡ §¤¤¢ ¥¥¥¨'££©££¡ ¡ §¡¤¡ §& ¦©¥"¡ &¡ § &¡ § ¤ "©'¦"(''¡ %£¥©¥£¥©£!¡ ¤¡¡ §&¡¡ §¡  ¥$!©£%¥£¡ ¡¡ §¤ ¡ §#¢ ¡¡¡¡ §¡¡¡ §¤¢ ¥¥¥©¨!¥©¥£¡ ¤¡¡¡ §##¢¡ §#¢ $¥¥£©¨¦¥£¡ ¤¢¡¡ § ¡ § £¥¥¨¦"©"¨! !¨©¥£¡ ¡¡ §¡ §¢ ¢¤¡ §¤¢¡ §¡¤¢ ©£©¨¦¥£¡ e Click in box labelled ”Second”, double click on David Ashley Adam Glassman David Finlay value Donte Ford Click OK, OK. Saurabh Kumar e Paired T for Before - After N Mean StDev Before 6 166.27 22.00 After 6 156.07 16.64 Difference 6 10.20 8.39 Terry Mashtare SE Mean 8.98 6.79 3.43 Pavlina Rumcheva Mathew Smeltzer Kelly Sodec CAR 100 CAR 100 CAR 100 McC C 100 CSE A101 CSE A101 McC C 100 TUR L007 TUR L007 CAR 100 P-Value=0.015 STA 2023 c D.Wackerly - Lecture 24 329 95% CI for mean difference : (1.39, 19.01) T-Test of mean difference = 0 (vs > 0): T-Value = 2.98 Location STA 2023 c D.Wackerly - Lecture 24 330 Assignments Today : p. 402 – 406 For Thursday : Exer. 9.46, 9.52–54, 9.56, 9.59, Ex. # 9.54, p. 407 Managerial careers of men and # $ women. problems on syllabus! male managers, f¦&¥ ¡ £¥¡ 9.60, 9.97, 9.100, 9.107 and rest of Chpt 9 female managers from Fortune 500 corporations. of the male Last Time : Paired-Difference Experiments managers married, Assumptions: Differences appr. Normally dist. Find a 95% CI for difference in proportions of male and Method of Analysis : do a one-sample “t” ON THE female managers who are married. DIFFERENCES (Male = “1”, Female = “2”). I4Q 4Q I 4 0SQ 0 © ¡ ¢ I0 ¡ ¢ I4  Q © §  G “’ …2 I„ (table value)(standard error) I  G Q© 4 '† „ ƒ ˆ © ƒ estimator I0Q Confidence Interval: f ¦¤ Test Statistic: of females managers married. STA 2023 c D.Wackerly - Lecture 24 (P. 403) standard errors c ˆ a ` b @ ` 4 ¨† ƒ © b a ¨ w c ba ` ¢ IB 0 formula sheet table ¢ IB 4 ¡ ¡ ¢ ¢ 0Q U 0 $£ 0 ¡ ¡ ¢ ¢ 4 ¨† ƒ B 4 ˆ© 20@ I4 I0 # with attribute Ex. # 9.54, p. 407 Managerial careers of men and ¢ male managers, # $ ¢ 4 managers married, of the male of females managers married. Find a 95% CI for difference in proportions of male and ¢ 2 0 female managers who are married. (p. 402) (Male = “1”, Female = “2”). ¦¤ ¡ £¥¡ ¡ ¢ I0 ¡ ¢ I4 Q ¢ 2 0 ¢ ¡ I ™ H ¢˜ H ) e ¢ ¡ 20 ¢ 4Q 4£ ¥4 ¡ ¢ ¢ 4 I WQ ¢ 4 4 0Q U 0£ $¤0 ” ¡ ¡ I ™ H ¢˜ H " e ¡ ¡ ¡ e ¡ ¢ ¢ 4 20 333 &  ¡ I4 STA 2023 c D.Wackerly - Lecture 24 ’s I0Q ¢ is approx. normally dist’d when both 4 WQ I 4 0SQ I 0 (p. 402) # $ ¡ I0 ¡ ¢ 4 estimates female managers from Fortune 500 corporations. f ¦¤ estimates women. f¦&¥ ¡ £¥¡ 4Q 4 0SQ 0 estimates 0 4Q 0 SQ from pop 2, # with attribute are large. (p. 402) STA 2023 c D.Wackerly - Lecture 24 334 Confidence interval: Consider testing (p. 404) £’ ¢  ¢ I4 20 “ I ¤ 4 £¡ © a fixed particular value of difference versus @ £’ ©¥ † © ¢ score 20 2 %© smaller score @ © † 2  © or ¡ 4 ¨X2 ˆ†© 4 '$¥ © ˆ†©  © £’ ¢ ¡ ¡ ¢ 4 ¨† ƒ 4 ˆ© B ¢ 2 0 @ I ƒ ¢  4 2 0 ¢ I 4 ¡ 0 SQ U 0 $£ 0 U ¢ ¢ I $'¢ OR (tail area) B £’ I¥ ” OR  ¢ ¡ ¢ 4 ¥ ¡ ¥ ¢ 20 I ƒ f # female managers who are married by between larger 2 ©  £ ¡ ¢ 4 WQ 4 £ 4 I managers who are married exceeds the proportion of p-value B ¡ ¡ ¥¢ confidence level , the proportion of male RR  ˆ I 4 '† © f ¦# At the and ¡ from pop 1, ¡ @ Independent samples: 4Q 4 £ 4 @ attribute f formula sheet ¨ Pop 2 attribute estimator y Have: Two populations CI for c B Large Sample Independent Samples (p.402) Pop 1 ¦ph fe §qig"d Comparing Two Population Proportions 332 t s¨e rr 331 STA 2023 c D.Wackerly - Lecture 24 335 336 Use premature infants given inositol had an compensate for poorly developed lungs on standard diet had eye injuries. true prop. of premi’s given inositol with eye e I4 true prop. of premi’s not given inositol with ¢ 20 “ RR : ¡  I¥ e ? 4WQ U SQ 0 I U0 4   ¢ I # ¦" I I£’ I I ¡ e  ‘ 4 Q ¢ ¢ I4§ ¢ ¢ I4 I0 ¢ I U  0Q 4WQ U SQ 0 4 U0 ‰£ I ¢ I I0 ¢ I£’ ¢ I £ ’ ¢ ¢ ¢ ¢ I  ¢ 337  ‰ STA 2023 c D.Wackerly - Lecture 24 ¢ I© ” on formula sheet. 4 £ ¤ Is (2) 2 ‚0 ¢ ¢ ¢  standard error tail test, (1) ¢ 20 total sample size 4 e eye injuries total # “S” in experiment ¢ ¢ with ¢ 0Q U 0 $£ 0 4Q B ¤7 D@ £9 injuries , estimate this ¢  ¢ 4 ‘ I0 24 ¢ 4 "£ # £¡ e of ¢ "¥ £’ ¡ I© COMMON value of of . eye injury to to high oxygen levels used to , use the individual ’s , then ¤  e NULL HYPOTHESIS ¢ "¥ Formula Sheet reduce the risk of eye damage in premature infants? I¥ 2 standard error ¡ hypothesized value  ¢ e estimator If STA 2023 c D.Wackerly - Lecture 24 Ex. : # 9.107, p. 431 Does inositol (found in breast milk) TEST STATISTIC If STA 2023 c D.Wackerly - Lecture 24 STA 2023 c D.Wackerly - Lecture 24 338 e Test Statistic £’ -value larger than ¥ e ¢ 20 0SQ U ‰£  ¢ I© ¢  ¢ I© I¥e I¥e I¥e I ¤ ££¦¢ I¥ I p-value – ¢ ¥¢ I P-value : Lower tail test –  ¢ e oxygen levels for infants given inositol. ? ¥ of premature infants with eye injury due to high – Claim  ‘4WQ  24 evidence to claim a lower proportion ? ¢ ¥¢ ? – Claim level, there , claim a lower proportion with breathing irregs if given inositol. Not significantly lower for any Decision : at the ¢ ¥¢ For any e STA 2023 c D.Wackerly - Lecture 24 339 Minitab? e Stat e Click Radio button “Summarized data” e ”First Sample” in box labelled ”Trial”, type in # of Basic Statistics 2-Proportions trials (110 in Ex 9.107), in box labelled ”Successes” type in # successes (14 in Ex 9.107). e ”Second Sample” in box labelled ”Trial”, type in # of trials (110 in Ex 9.107), in box labelled ”Successes” type in # successes (29 in Ex 9.107). e Click ”Options” e Choose alternative (Less than in Ex 9.107) e Click in box ”Use pooled estimate for p for test”. e Click OK, OK Test and Confidence Interval for Two Proportions Sample 1 2 X 14 29 N 110 110 Sample p 0.127273 0.263636 Estimate for p(1) - p(2) : -0.136364 95% CI for p(1) - p(2): (-0.239604, -0.0331235) Test for p(1) - p(2) = 0 (vs < 0) : Z = -2.55 P=0.005 ...
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