Week10-4up - 224 Thought: A committee is a group of people...

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Unformatted text preview: 224 Thought: A committee is a group of people who, STA 2023 c D.Wackerly - Lecture 17 STA 2023 c D.Wackerly - Lecture 17 225 Last Time: to within “B” units with ¡ Estimate and decide that nothing can be done. £© § ¥ £ ¨¦¤¢ individually, can do nothing, but collectively can meet, confidence. 20 31) ' (& and solve for 5 ¡ 06 870 Today : P. 328 – 332 ¡ otherwise use if you have one, . A9 B@£ $"  %#! Use “ballpark” value for . (p. 309) 4 Assignments Parts of a statistical test. (p. 322) For Tomorrow : Exercises 8.18, 8.21–23, 8.25, 8.27 5 The hypothesis of MAIN INTEREST is the ALTERNATIVE or RESEARCH hypothesis, – D EC light bulb ex. Thursday : Exercises 8.29, 8.33, 8.34, 8.38–41, 8.59, , ©UAT£ R PH D C QQ1SQIGF¢ Wednesday : P. 334 – 338, 347–351, . (p. 322) What we are “trying to prove” in an objective, fair 8.61, 8.67–69 manner Monday 11/4: OPTIONAL review 5 The “other” hypothesis is called the NULL P QH V C W¢ 226 0 STA 2023 c D.Wackerly - Lecture 17 , light bulb ex. (p. 322) ©UAT QQ£ DISCUSSION SECTION V EC HYPOTHESIS, – Tuesday 11/5: EXAM 2 – During your STA 2023 c D.Wackerly - Lecture 17 Errors: p. 325 227 Parts of a Statistical Test (p. 326) Reality Ha true Correct Type II error Reject Ho Type I error Correct 2. Alternative Hypothesis : . 3. Test Statistic : (TS) 5 computed from the sample data using a formula forms the basis for our decision. (p. 323), SIGNIFICANCE h true Then 5 § X 30 V EC c e D § C ¢ Y `5 X 30 e fY 5 ca db§ 5 § gX I30 5 Do experiment 5 Get data 5 Compute value of TS 5 Make decision h g IX In our lightbulb example, P UAT£ R QpiP gX I30 5 § The test that we will discuss have the SMALLEST § for the ’s that we pick. Y 0 when hUAT QQ£ 5 § 0 saying what we “want” to say when we should not saying depends on the choice of ’s r s§ Y a when gives values of TS for which 5 accepting © and/or 4. Rejection Region : (RR) (p. 325) 5 Type II error C ¢ LEVEL of the test. V © Type I error 5 Accept Ho . D qC Ho true V EC 1. Null Hypothesis : Decision is REJECTED 228 STA 2023 c D.Wackerly - Lecture 17 STA 2023 c D.Wackerly - Lecture 17 229 Decision : 5 If the value of the TS is in the REJECTION and C D REGION, we . V GC 5 If the value of the TS is NOT in the REJECTION C . because we usually do V C – We do not ACCEPT V REGION, we DO NOT REJECT Courtroom Analogy not know the probability of making an error if we 5 5 Innocent : 5 Guilty : 5 Put burden of proof on Prosecutor : Experimenter 5 Proof “beyond a reasonable doubt” : , what kind of error could we small. C usually D depends on the value of the parameter in § – What is the probability of a TYPE II error? C V EC make? D – If we accept V EC do so. Prosecutor : Experimenter that is really true V C Don’t want to accept , so we reserve judgement 230 STA 2023 c D.Wackerly - Lecture 17 STA 2023 c D.Wackerly - Lecture 17 231 Large Sample Hypothesis Testing ¤ ¦§ ¨¤ §¦ 9 0 ¨¤ ¥ ©¥ ¦ ¥ VP 4 4 9 §¦ 9 0 0 ¨¤ VP ©¥ ¦ ¥ P QH V GC 5 UAT Q£ VP ©¥ ¦ ¥ ¨¤ 0 0 V C C D EC P QH GC V UAT£ R P Q1SQH  0 V D C 0VP 5  5 5 has a ¥ 0 RP iQH P QH Test Statistic, TS §¦ VP ¨¤ is true, VP ¨¤ 5 0 VP VP P 5 P is a TEST STATISTIC If 4 P is true . Need Rejection Region, RR ¥ ¥ 0 if P P is a distribution 4 ¥¦ 0 ¡ 5 In this case, has an APPROXIMATE That is Lightbulb Example : 4¢ a fixed particular value of ¢ £ Recall : Large Sample © T about a Population Mean, distribution § 4 9 §¦ 9 RP iQH  H 0  D  R H C ¦ §¦ ¢ £¡ VP VP P P “Upper Tail test”, “One Tail Test” (p. 329) in favor of 5 V GC D EC § 0 ¥ 0    T £ ¤ 0 ¨ ©0 4 0 A r T T  0 STA 2023 c D.Wackerly - Lecture 17 235 AT THE C D in favor of printed circuit boards claims that “product can inspect, “refute the claim” based on data for P . level of ? V and H C V 5 C U Q r § 0  U  r  U £ r HD IGC 0 § – RR : If we are interested in : 4 § ¥ 9 ¥ 9 PH D QIGC 0  £ r H ¦ §¦ H £ r UAT QQ£ confidence ). level” ( or P §¦ ¢ ¨ ¤  £¡  VP V C    with at the ¥¤4 P bulbs is larger than ¨¤ claim that the mean lifelength of all C  – In terms of this problem “ V 0 level P QH ? 0  U B£ r at the R – T T r A – Conclusion : Is VP 5 5 UAT £ confidence ). one-second runs? Use D GC 5 significance” ( or with at the , is § 5 “ Claim that the mean lifelength of all bulbs, 0 X In terms of this problem: U Q r g V GC on average, at least 10 boards per second”. Evidence to LEVEL!! 4 5 Ex. : #8.24, p. 333 Manufacturer inspection equip. for Conclusion : ¨ 0 D GC 5 234 TT BA 5 § 0 RP SQH STA 2023 c D.Wackerly - Lecture 17 level test, RR : U Q r  R h 0 C D zα UAT£ R P QpiQH Data : 0 RP SQH P α P QH P 5 Rejection region : UAT Q£ 5 P “something” type I error 0 true mean lifelength of ALL BULBS 0 ¤ 5 P QH EC V 0  Rejection region : If we wanted ¨¤ . Ex. : Lightbulb Example If we are interested in Want than VP Should is probably ¥ The true value of by a “lot” of standard errors. 4 ¥ ¤ P than ¨¤ is ¥ ¤ is POSITIVE and LARGE P P If 233 V GC true value is. P QH , whatever that 0 is close to the true value of STA 2023 c D.Wackerly - Lecture 17 VP FACT: 232 STA 2023 c D.Wackerly - Lecture 17 α – NOTE: This DOES NOT mean that is true!! − Zα 0 “Lower Tail test”, (p. 329) UAT Q(£ 0 P QH C 5 236 STA 2023 c D.Wackerly - Lecture 17 STA 2023 c D.Wackerly - Lecture 17 Back to #8.24: Minitab? Hypothesis test for a mean. 8 9 13 9 10 9 9 9 7 12 6 9 10 11 12 10 0 10 11 12 9 8 9 6 10 11 10 12 8 10 8 7 9 7 9 9 10 Must have actual data (not just 1-second runs. Data : 48 actual numbers Minitab File Open Worksheet; find M8-24.mtw in MiniData (double click). Stat Basic Statistics Display Descriptive Statistics, select (double click) variable, Click OK Descriptive Statistics T Q£ r A Variable JntsInsp ¥ 0 A  A r ©0  ¤ c Minimum 0.000 Median 9.000 Maximum 13.000 TrMean 9.432 Q1 9.000 StDev 2.103 SE Mean 0.304 Q3 10.000 Stat Basic Statistics 1 Sample Z, select (double click) variable; Click radio button Test Mean; Type in null value for mean; Select Alternative; Type StDev in box labelled Sigma, Click OK ¤ r T r A ¥ T Z-Test Test of mu = 10.000 vs mu < 10.000 The assumed sigma = 2.10 0  U Q r ¨  T r 0 5  0 U Q r 5 § 0  r ¥ level of significance. In terms of this 5 application: “There Mean 9.292 ¤ 4 0 RR : N 48 Variable JntsInsp level test and ). Ex. # 8.24 Number of solder joints inspected in 48 ¤ 9 ¤ 11  £ 10 ¤ 7 ¤ 12 9  11 ¡¡ £¢¡ 10  £ 10 5 9 ¤ 10 ¥ 5 Data: at the 237 enough evidence at the level to indicate that the mean number of circuit Variable JntsInsp N 48 Mean 9.292 StDev 2.103 SE Mean 0.304 Z -2.33 P 0.0099 U Q r boards inspected per second is less than 10 .” STA 2023 c D.Wackerly - Lecture 18 239 UNKNOWN population mean VP  © ¦¦ VP ¥¦¦ R SP   ¥   © © VP  § ¦¦¨ D GC 0 P hypothesized value standard error  ¥ estimator 0 VP 4 ¥ §¦ 9 ¤  0 Estimator and Standard Error from Formula Hypothesized Value from NULL hypothesis 5 Tuesday 11/5: EXAM 2 – During your DISCUSSION SECTION RR Test Statistic 8.61, 8.67–69 Monday 11/4: OPTIONAL review R OR Today: P. 334–338, 347–351 Thursday : Exercises 8.29, 8.33, 8.34, 8.38–41, 8.59,  5 V GC Assignments ¦¦  P QH 0 0 makes a pretty small package – (John Ruskin) P 5 Thought: When a man is wrapped up in himself, he Sheet   Last Time: Large Sample Hypothesis Testing about P 238 STA 2023 c D.Wackerly - Lecture 18 240 STA 2023 c D.Wackerly - Lecture 18 STA 2023 c D.Wackerly - Lecture 18 241 Ex. : pH of 7 is neutral, over 7 is alkaline, under 7 water specimens Back to pH example: ¨¤ 0  T level test: £ r 5 4 0 £ r § 0 ? Tr 5 § 0 V H and level? ¥ pH is NOT that of neutral water at the 0 from a recreational lake. Can we claim that the mean r Ar   T indicates acidity. Randomly select D GC C H V D GC VP  at the level of significance. In terms £ r V C of this application: “There § ¨¤ ¥¤4 P ¥ 9  R VP ¥ ¨¤ §¦ " # 9 P QH 0  0 ¥ α/2 the £ r P " #!  4 5 Reject 0 V P QH 0 D GC ¦ §¦ ¢ £¡ H   C H α/2 or 0 5 C RR : How? enough evidence at level to indicate that the mean pH reading is not .” 0 α/2 z α/2 “Two Tailed test” (p. 331) STA 2023 c D.Wackerly - Lecture 18 242 -z STA 2023 c D.Wackerly - Lecture 18 243 Hypothesis Testing If the mean pH is NOT , what is it? to reject – Provides . CONFIDENCE in our 0 decision to reject in favor of V EC £¡ ¤ ¢ c 0 ¥ 4 C § D C The p-value or observed significance level 5 Recall the lightbulb example UAT Q£ P QH 0 UAT£ R P Q1SQH  § c A 0 §c   r 0 § B£ ¥ ¤ ¥ ¦ § H U B£ r 0  V C D GC ¨ ¥ r 0 P Tr Agrees with two-tailed test!! the 99% could ? (P. 335) ? 5 confidence interval for which V    be rejected in favor of — the value “ ” is ( IF we DO SO). What is the SMALLEST value of Do you think that C P – D 5 Smaller V EC . § 5 confidence interval for § Construct a is chosen BEFORE the test is performed 5 244 STA 2023 c D.Wackerly - Lecture 18 STA 2023 c D.Wackerly - Lecture 18 rejection region . V EC . ¡ ¡ V GC . V GC . ¢ 5 § § . § 0 § 0 § 0 0  5 T  r  U BQ r A R T T r A A Q r  R r £ r UAT Q(£ § P QH UAT£ R PH D Q(pSQIGC 0 V GC 0 C indicative that D 5 Larger z-values are Instead of “imposing” YOUR CHOICE of is on a person § C ¨ R – REJECT V  R Probability of a z-value . V GC § 0 U Q r – ¡¢£  r ¡£Q r A ¡¤ r T ¡£Q r U ¡ ¢£ r  5  § – . V EC r£ pR ¨ U In our case, p-value = .0256 – the one observed CANNOT reject C p-value REJECT V p-value – p-value 245 who might be interested in your conclusions, the true. p-value allows him/her to assess the “rareness” of the p-value = observed event. © R  ¢ X 5 0 246 STA 2023 c D.Wackerly - Lecture 18 STA 2023 c D.Wackerly - Lecture 18 247 Ex. : #8.24, P. 333 TWO - Tailed Test  £ V P H 0 C  PH D IGC £ p  5 T T r A ¥ Probability of a z-value and DOUBLE IT. EX. : Have done a two-tailed test: P QH  . £ 0 P claim that P 0.0099 with § Z -2.33 ? 0 V 5 C 5 V GC 5 – SE Mean 0.304 . U  r § 0 § § StDev 2.103 U Q r 5  Q r ¢ r  p Mean 9.292 value = ¥¡      Test of mu = 10.000 vs mu < 10.000 The assumed sigma = 2.10 N 48 0 5 0 C . Z-Test Variable JntsInsp U B£ r D that is 0 5 for any See page 234 of notes: D qC that is £ V GC for any P QH p-value is true. 0 Smaller z-values are more indicative that £ 0 the one observed  0 p-value Find the area in whichever “tail” the -value is in . STA 2023 c D.Wackerly - Lecture 18 249 Large Sample Tests About 248 STA 2023 c D.Wackerly - Lecture 18 (Section 8.5) Interested in a POPULATION that contains an Ex. : #8.68, p. 352 In a “Pepsi Challenge”, 100 Diet UNKNOWN but FIXED PROPORTION of items with a Coke drinkers were given unmarked cups of both Diet particular attribute ¡ ¢ ¢¢ £ ¤ indicated that they preferred © 5 Recall the BINOMIAL EXPERIMENT. the proportion of Diet Coke drinkers who ¡ indicate that a majority of the Diet Coke drinkers will select Diet Pepsi in a blind taste test. select Diet Pepsi in a blind taste test? ¡ true proportion of Diet Coke drinkers who the proportion of batteries that fail before 0 U the taste of Diet Pepsi. Is there sufficient evidence to 0 Coke and Diet Pepsi. . guarantee expires. 5 ¡ 0 would select Diet Pepsi in a blind taste test. 5 number of trials in the trials in the sample r ¤ ¢¡ ¤ ¦ £ ¤ ¡ 0 4 0 ¥ 0 4 sample size C 5 ¡ H V GC 5 if 0 is “large” 251 is the null hypothesis, TEST STATISTIC standard error Estimator and Standard Error from Formula Hypothesized Value from NULL hypothesis Sheet ¥ d¡ V6V ¡  ¥ ¡ ¥ d¡ §&   0 ¦ 0© §& 4 4 , has a STANDARD V C 5 NORMAL distribution Rejection Regions (RR): ¦¦   ¦¦ ¦¦  R ¦ ¦¦ © V¡ ¥¦¦ ¦¦ ¡ ¦ ¦¦ OR  RR ¦¦ ¥ ¦¦ ¦¦ ¦¦ ¦ or ¦¦    " #!  " #!  ©© V¡  ¦¦ R ¥    ©©© ¦¦ ¦¦ 0 ¦¦ V¡ 0 ¦¦ ¡ ©© § ¡ OR ¦¦ 5 © ¦¦ R ¡ V¡ ¦¨ ¦¦ D GC © © © V¡ ¡ ¦¦ ¦¦ § ¦¦  0 ¡ 0 ¡ H 0 ¦¦ R V¡ V¡ ¥¦¦ OR ¦¦ ¦¦ OR ¡¢ ¦¦¨ ¡ V GC versus is true 0 a fixed particular value of If ©V¡ Consider testing  0 ¨P §&  0 ¦ ¡ distribution. V¡  6¡ has an approximate hypothesized value   5 ¡ That is estimator 0 ¥ 4 distribution ¥ has an   5 ¡ STA 2023 c D.Wackerly - Lecture 18 V¡ V H If 250 0 STA 2023 c D.Wackerly - Lecture 18 ¢¡ Estimate for How??? # of £ ¤ # of trials ; 4 HD IGC (2) based on a “large” ¡ GOAL : Test hypotheses about (1) D C 252 STA 2023 c D.Wackerly - Lecture 18 STA 2023 c D.Wackerly - Lecture 18 253 Ex. : #8.68, p. 352 In a “Pepsi Challenge”, 100 Diet Coke drinkers were given unmarked cups of both Diet indicated that they preferred the taste of Diet Pepsi. Is there sufficient evidence to 5 Conclusion : § 5  BU r  BU r (4) confidence ) to indicate that the majority of Diet  QU  R ¡ H 5 ¥¡ 5 is “large” 4 0 C 0 ¥ 0 U Q r 5 § 0 U Q r ¡  £ 0 4 0 5 5  0 254 Minitab? 5 Stat 5 Click radio button “Summarized Data”, type in 1 Proportion ¤ ¤ Number of trials, Number of Successes 5 Click Options, Select Alternative, Type in Null Value 5 Click Box “Use test and interval based on normal distribution”, OK, OK Test and Confidence Interval for One Proportion Test of p = 0.5 vs p > 0.5 90% CI (0.462710, 0.657290) Z-Value 1.20 P-Value 0.115 value? ¥¡ 5 ¡ 0 D EC ¡ H V 0 test. STA 2023 c D.Wackerly - Lecture 18 Sample p 0.560000 level of significance” ( or with Coke drinkers will select Diet Pepsi in a blind taste SAMPLE of all Diet Coke drinkers. Note: N 100 0 the the the Pepsi Challenge are a X 56 claim that there is sufficient evidence at (3) Assumptions : the 100 individuals participating in Sample 1 AT THE LEVEL!! “ level test, RR : Basic Statistics in favor of In terms of this problem: care reform is the leading priority Data : U Q r true proportion of all voters who think health V GC reject select Diet Pepsi in a blind taste test? C indicate that a majority of the Diet Coke drinkers will D U Coke and Diet Pepsi. value = ...
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