week13-4up - 1 STA 2023 c D.Wackerly - P-value Analogy STA...

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Unformatted text preview: 1 STA 2023 c D.Wackerly - P-value Analogy STA 2023 c D.Wackerly - Lecture 19 252 Garage sale, I want to go for coffee and tell you that “the Thought: Eagles may soar, but weasels aren’t sucked smallest amount that I will take for this jacket is $6”. If into jet engines. leave. What do you do if someone: ¡ Offers $15? ¡ Offers any amount ¡ Offers any amount Assignments Today : P. 347–351 (Sec. 8.5), $6? £ ¡ Offers $5? ¢ ¡ Offers $7? $6? Tuesday: Exer. 8.59, 8.61, 8.67–69, 8.116, 8.119 Wednesday : P. 288 – 393 (Sec. 7.2), P. 341 – 345 (Sec. 8.4) Same scenario, but I say that I want at least “something between $5 and $10” Choices: Sell, Not Sell, Can’t tell. For Thursday: Exer. 7.27, 7.30, 7.33, 7.80, 7.81, 8.49, 8.50, 8.53, 8.54, 8.56, 8.57, 8.105 – 108, 8.111, 8.117 LAST QUIZ ¡ ¢ ¡ ¡ Offers $4? Tuesday, April 9, regular 20 point quiz OPTIONAL Up to 20 BONUS POINTS - will ¡ Offers any amount between $5 and $10? ¡ Offers any amount ¡ Offers $7? $10? contain 5 EXTRA questions from Exam 2, 4 pts each, no computation. Answers same, but re-ordered. Pts. earned added to exam score. £ $5? 253 STA 2023 c D.Wackerly - Lecture 19 STA 2023 c D.Wackerly - Lecture 19 254 Ex. : #8.68, p. 352 In a “Pepsi Challenge”, 100 Diet Coke and Diet Pepsi. indicated that they preferred indicate that a majority of the Diet Coke drinkers will ¤ ! "  ¤ © ¨¥ ¤  ©§ ¥ ¦¤ ¨¦¡ $ ¨# ¤ £ ¤ ! (% £ $ )# ¤ 0  1¤ 4 529!  42 53! (% £ p-value select Diet Pepsi in a blind taste test? would select Diet Pepsi in a blind taste test. ' 7 86 FD A G 3EC @  A BA ¤ % A Estimator and Standard Error from Formula Sheet Hypothesized Value from NULL HYPOTHESIS Large Sample Tests About (Section 8.5) Interested in a POPULATION that contains an UNKNOWN but FIXED PROPORTION of items with a particular attribute . Recall the BINOMIAL EXPERIMENT. SH SH % & (tail area) ¡ score true proportion of Diet Coke drinkers who XRWPUV T¨¥ S©  % & smaller score ' or larger  )¡ S OR Test Statistic Coke drinkers were given unmarked cups of both Diet the taste of Diet Pepsi. Is there sufficient evidence to RR OR ¤ Last Time: Large Sample Tests about ' fBc$ e db ¡ Offers any amount a XRWPU` Y© § ¥ S ¡ Offers $10? Q RP ¡ Offers $15? the proportion of Diet Coke drinkers who select Diet Pepsi in a blind taste test. the proportion of batteries that fail before guarantee expires. H H IH ¡ 255 STA 2023 c D.Wackerly - Lecture 19 STA 2023 c D.Wackerly - Lecture 19 256 Consider testing versus estimator hypothesized value standard error A  G ed fBb ¡  @  S ¦ §S ¥ STATISTIC %A ¡ H ¢¤ S  £¤ H S H G¡ G ¢¤ C  S % S H has an approximate STANDARD NORMAL Estimator and Standard Error from Formula H IH  ¢¤ C That is is the null hypothesis, TEST A BA S § S  T© § ¥ ¡ if has an APPROXIMATE NORMAL distribution § S 0 S S GH is “large” OR   @ If OR U ed fBb sample size trials ©©© G in the sample ©©  § S £ S ©©©   ¥  ©© § S  S ¨©©©©©© in the a fixed particular value of  ¡ Estimate for # of S # of trials ; S § S  T© § ¥ number of trials based on a “large” S GOAL : Test hypotheses about Hypothesized Value from NULL hypothesis Sheet G § ¦ § S ¥   §S % S distribution. 257 STA 2023 c D.Wackerly - Lecture 19 STA 2023 c D.Wackerly - Lecture 19 258 Ex. : #8.68, p. 352 In a “Pepsi Challenge”, 100 Diet Coke drinkers were given unmarked cups of both Diet is true ' § S  S$ § ¨¥ ¡ If , has a STANDARD Q RP Coke and Diet Pepsi. indicated that they preferred the taste of Diet Pepsi. Is there sufficient evidence to NORMAL distribution indicate that a majority of the Diet Coke drinkers will Rejection Regions (RR): select Diet Pepsi in a blind taste test? © ©©© © ! (% £ ©©© would select Diet Pepsi in a blind taste test. © ©©© ©©© ©©© P UX  !¡ ¡ ©  or (1) (2) level test, RR : Assumptions : the 100 individuals participating in the the Pepsi Challenge are a SAMPLE of all Diet Coke drinkers. Note: G ©©© RR X RWPU  ST¨¥ ©§ XS RWPUV T©  ¥   § S 0  S ©©©©© ©©© ©©©  ¨¥   § S £ S  ©© ©©© § S  S ¨©©©©©©  ! " OR  )¡ S OR true proportion of Diet Coke drinkers who is “large” 4 523!  42 53! (% £ ¡ STA 2023 c D.Wackerly - Lecture 19 259  S  P UX Minitab? ¡  Click radio button “Summarized Data”, type in ¡ Click Options, Select Alternative, Type in Null Value  ¨¥ AT THE Number of trials, Number of Successes ¡  ¡ §¥ P UX  ¡ In terms of this problem: ¡ claim that there is sufficient evidence at level of significance” ( or with P the ¢ £¡ “ 1 Proportion ¡ in favor of ¤ X IX  G reject Basic Statistics ¤ Stat Conclusion : LEVEL!! 260  Data : STA 2023 c D.Wackerly - Lecture 19 Click Box “Use test and interval based on normal distribution”, OK, OK confidence ) to indicate that the majority of Diet Test and Confidence Interval for One Proportion Coke drinkers will select Diet Pepsi in a blind taste Test of p = 0.5 vs p > 0.5 test. Sample 1 X 56 N 100 Sample p 0.560000 90% CI (0.462710, 0.657290) Z-Value 1.20 P-Value 0.115 % )¡ S % )¡ S value? value = STA 2023 c D.Wackerly - Lecture 19 STA 2023 c D.Wackerly - Lecture 19 261 Summary: Large Sample Hypothesis Tests (tail area) or Sample size for each test is reject .5 A BA 76 4 529!  42 53! (% £ hypothesized value A %A standard error score S or score 4 .6 §¥ ! (% £ $ )# £ 0 estimator % & Test Statistic smaller ; RR : ' ! $ ¨#  ©¥ ©§¥ ¨¦¡ value larger %  value OR param. p-value ' RR OR param. X U  !¡ WPU  T© § ¦¡ S¥ value value param. Computer Study: ¡ §¥ X I§¨ G G WP8WPU U7 WPU % ¥  S P ¥¦WQU % £ P ¦ UQ  ¥ WPU 0  T© ¨¥ S param. 262 not reject tests Prop. rejects 46 50 .08 12 38 50 .24 .7 32 18 50 .64 .8 48 2 50 .96 .2 47 3 50 .94 .1 50 0 50 1.00  Estimator and Standard Error from Formula Sheet Hypothesized Value from NULL HYPOTHESIS H H IH ¡ S .7 .8 § ¨¥ 21 §¥ .6 44 ¡ 6 not reject X IQ  G .5 STA 2023 c D.Wackerly - Lecture 20 263 Sample size for each test is reject STA 2023 c D.Wackerly - Lecture 19 264 Thought: People will accept your ideas much more readily if you tell them that Benjamin Franklin said it first. tests Prop. rejects 50 .12 29 50 .42 48 2 50 .96 50 0 50 1.00 Assignments Today : P. 288 – 393 (Sec. 7.2), P. 341 – 345 (Sec. 8.4) For Thursday: Exer. 7.27, 7.30, 7.33, 7.80, 7.81, 8.49, What do we see? 8.50, 8.53, 8.54, 8.56, 8.57, 8.105 – 108, 8.111, S ¡ For each fixed sample size, as the value of 8.117 moves away from .5, ( and the null becomes “less true”) — GOOD! §¥ we REJECT Monday : P. 374 – 383 (Sec. 9.1, “Large Sample”) a greater percentage of the time. For Tuesday: Exer. 9.1, 9.7, 9.13, 9.15–17, 9.19, 9.22, 9.24, 9.25 . Big STA 2023 c D.Wackerly - Lecture 20 OPTIONAL Up to 20 BONUS POINTS- see explanation on intro page for Lecture 19. is “better”. G X ¢ greater percentage of the time for larger Tuesday, April 9, regular 20 point quiz a G , we REJECT LAST QUIZ ¡ ¡ WPU 0  S ' X U  $ . approx. §¥ we reject ¡ WPU  S of the time. G ¡ For each fixed value of , for each §¥ When STA 2023 c D.Wackerly - Lecture 20 265 266 Summary: Large Sample Hypothesis Tests value A BA A %A 0  Sheet Hypothesized Value from NULL HYPOTHESIS ¡ £ Estimator and Standard Error from Formula mean level of phosphorus is less than . Sample size ' 6 $ 4 529!  42 53! (% £ standard error (measurements in parts per © ¨¥ § 76 hypothesized value and measurements, obtaining billion [ppb]). Can the EPA support the claim that the ©  ¦¡ ¥ P UX  (tail area) of park, EPA makes P ! (% £ $ )# estimator % & Test Statistic or ' param. score value OR smaller of concern to the EPA in the Everglades. In one section U6  @ value Ex. Phosphorus content is a water quality index that is § param. score ¡ OR larger ¥ 9P  U 6  G ! $ ¨#  ©¥ ©§¥ ¨¦¡ value param. p-value % RR '  param. is small! How??? ppb? Use H H IH ¡ ¡ Bell-shaped . (like the -distribution) @ % scores – Variability depends on degrees of freedom. – Variability ¡ does not have a standard normal dist. as d.f. . ¤ ¡ can’t use More variable (heavy-tailed) than the -distribution ¥ ¡X ¢I§ £ G can’t use CLT to get NORMALITY of the sampling distribution of 268 Properties of the -distribution: Symmetric about 0. (like the -distribution) ¡ Small Sample Inferences about STA 2023 c D.Wackerly - Lecture 20 £ 267 STA 2023 c D.Wackerly - Lecture 20 G FD ¤%@ ¡ – Becomes more and more like the -distribution However: as d.f. . ¥ ¡ If the POPULATION is approximately NORMALLY distributed t with 2 df t with 8 df G FD ¤ %@ £ (looks a lot like !!!) Std Normal ¡ has a sampling distribution called the “degrees of freedom”, -4 0 2 4 STA 2023 c D.Wackerly - Lecture 20 270 1 3.078  5 £ 6.314   £  12.706 31.821 63.657 2 1.886 2.920 4.303 6.965 9.925 3 1.638 2.353 3.182 4.541 5.841 4 1.533 2.132 2.776 3.747 4.604 5 1.476 2.015 2.571 3.365 4.032 6 1.440 1.943 2.447 3.143 3.707 7 1.415 1.895 2.365 2.998 3.499 8 1.397 1.860 2.306 2.896 3.355 9 1.383 1.833 2.262 2.821 3.250 10 1.372 1.812 2.228 2.764 3.169 11 1.363 1.796 2.201 2.718 3.106 12 1.356 1.782 2.179 2.681 3.055 13 1.350 1.771 2.160 2.650 3.012 14 1.345 1.761 2.145 2.624 2.977 15 1.341 1.753 2.131 2.602 2.947 4 ¡ ¡ Note: d.f. = denominator in calculating : 4  B £ d.f.    £  269 %£ STA 2023 c D.Wackerly - Lecture 20 -2    £ %G d. f. distribution with G FD ¤ %@ £ @ % @ G  4' % $ ¦ 4 ¡ ¡ Thus ¡ used in its calculation. X § X U U§P U§ U§ IWXUY¨P IWXY§ X T©I6 X T©P X T¨X U  ! £  ' !"  ¨# $ !  ' ! £  £ )# $ !£ ¡ Define ¡ Table VI (p. 811) gives C has the same number of d.f. at the estimator for so that so that ) -values for and U P II UX XX (Remember: 271 2.898 2.878 2.539 2.861 20 1.325 1.725 2.086 2.528 2.845 21 1.323 1.721 2.080 2.518 2.831 22 1.321 1.717 2.074 2.508 2.819 23 1.319 1.714 2.069 2.500 2.807 24 1.318 1.711 2.064 2.492 2.797 25 1.316 1.708 2.060 2.485 2.787 26 1.315 1.706 2.056 2.479 2.779 27 1.314 1.703 2.052 2.473 2.771 28 1.313 1.701 2.048 2.467 2.763 29 1.311 1.699 2.045 2.462 2.756 1.282 1.645 1.960 2.326 2.576   5 £ df=5 df=10 df=20 df=30 df= Note : When d.f. 273 STA 2023 c D.Wackerly - Lecture 20 ¡ Hypothesis Tests (p. 342) versus ¨©©© ¤  ¤ ©  ¤ £ ¤ ©©  ¨¥ !£ % £ £ © ©©© ©©© Assumption : POPULATION approx. NORMALLY dist. C Test statistic : RR or 42 53! £ G FD ¤ % @ £ (looks just like !!) ¡ !£ ¡ ¢ G F 59! 42 @ G F 59! £ @ 42 ¡ ¢ Small Sample (p. 292): ¡ %G Large Sample: ©© Confidence Interval : OR ©©© dist. ©©© d.f. instead of £ dist. with © Small sample situation similar to large, except use OR ¤ 0  1¤ Small Sample Inferences About 274 !£  £ STA 2023 c D.Wackerly - Lecture 20 t  2.093 4 529! £  £ 42 53! £ % £ £ 1.729 ©© 1.328 ©©© 19 .025 ©©© 2.552 ©©© 2.101 © 1.734 ©©© 1.330 ©©© 18 ©©© © 2.567 ¤ ©§  ¤ ¨¥ 2.110 1.740  1.333 17 !  !£§    £  2.921  4  £ 2.583 ¡   £  2.120 d.f. ( B £ 4 6 ¥ X 96 ( B £ U 4 Q X 96   B  £ U 4   B  £ 4  4 ( B £  5 £ 1.746 272 X 1.337 STA 2023 c D.Wackerly - Lecture 20  B  £ 4 16 STA 2023 c D.Wackerly - Lecture 20 (and (new) ) depends on (like before) AND #d.f. ¡ STA 2023 c D.Wackerly - Lecture 20 Ex. Phosphorus content is a water quality index that is of concern to the EPA in the Everglades. In one section  P £ ¡ 42  G F 59! P ¦¤ ¨¥  ©§ ¢ @  P £ ¤ ©  ¦¡ ¥ P UX  G 6§ U %  £   G ¤ F %D @  £ §U 6 @ §¥   ( £  ! £ £ if 6§ U %  U 9P  ¡ ¡ Test statistic: ¥ ¡ reject . 42 (4  £  53! £ U 96  @ , , § Rejection Region: Lower tail test. d.f. 95% CI is U 9P  U6  @ § . ppb? Use ¥ ¡ mean level of phosphorus is less than 6 G billion [ppb]). Can the EPA support the claim that the (measurements in parts per ¥ 9P  U 6  G and Ex. Give a 95% CI for the mean phosphorus index in the section of the Everglades measurements, obtaining of park, EPA makes 276 ¡ 275 STA 2023 c D.Wackerly - Lecture 20 Note: In last example (both test and CI), we are assuming that population from which the sample is ¡ taken is (approx) normally distributed ¡ Conclusion: Since is in the rejection region, CANNOT reject Ho . There is P X ` U evidence to conclude, at the level of significance, that the mean level of are (approx) normally distributed ppb. STA 2023 c D.Wackerly - Lecture 20 277 P phosphorus is less than That is, that STA 2023 c D.Wackerly - Lecture 20 278 How about -values? a ©©© © ' ©©© © From the table, H H  £ H  76  ¤ 0  ¤  H H ©©© ¡ 6§ U ¡ 6§ U %  £ H ) to Ho . Ho . ¡ $ ¨# %£  Closest values in table ( with d.f. and Can’t be any more precise using these tables! , Look at table, are Thus, in this case, best we can say is that § UX  P UX  Q I UX  UX  X U  P U  ©© ©©© ' p-value p-value ¡ ©©© ¨©  ¤ £ ¤ ©© ©©© ©  ¥  ©  ¤  ¤ ©©©©©© $ )# %£ Table does not allow us to get exact p-values. In Everglades example, lower tail test, d.f. (tail area) score ¡  3§ I§ U % £ £ )# 'Q $ ¡  ' Q ¡ U  £ )#  3I¡ U % £ £ )# $ 'Q $ OR ¡ OR score From the table, Ho . Ho . ¡ ¡ 279 STA 2023 c D.Wackerly - Lecture 20 STA 2023 c D.Wackerly - Lecture 20 Exercise 10.52 Pouring temperature of molten iron. 280 Target setting is 2550 degrees. Actual mean pouring 2620 2559 2562 2553 2552 2553 Is 2560 temperature differs from 2550 at the .01 level of significance. Minitab CI? ¡ £ ¤ ¤ N 10 Type in Confidence Level OK Variable C1 ¡ ¡ ¡ T-Test of the Mean Variable C1 interval” T Confidence Intervals Test of mu = 2550.00 vs Select Alternative, Type in Null Value, OK vs Mean 2558.70 mu not = 2550.00 StDev 22.75 SE Mean 7.19 T 1.21 P 0.26 1-Sample Select variable and click Radio button “Confidence ¡ ¡ Select variable and click Radio button “Test mean” Test of mu = 2550.00 ¡ 1-Sample Basic Statistics ¤ ¡ Punch in data values Basic Statistics Stat ¤ XPP6 IIIV0  ¤ © ¨¦¡ ¥ Minitab? Stat ? claim that the mean pour XP RP I6  ¤ © § ¥ Want : two tailed test ¡ 2544 £ X  G 2541 Conclusion at .01 level? . ¡ WXU  2543 measurements ¡ temperature different that target? Use N 10 Mean 2558.70 mu not = 2550.00 StDev 22.75 SE Mean 7.19 99% CI (2535.32, 2582.08) ...
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This note was uploaded on 12/15/2011 for the course STA 2023 taught by Professor Ripol during the Spring '08 term at University of Florida.

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