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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
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GC 5 UAT
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0 0 V
C C
D
EC P
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V UAT£ R P
Q1SQH
0 V D C
0VP 5 5
5 has a ¥ 0 RP
iQH P
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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 ). onesecond 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 1second runs. Data : 48 actual numbers
Minitab File Open Worksheet; ﬁnd M824.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 ZTest
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
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§¦
¢
£¡ 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 pvalue 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 twotailed 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 zvalues are Instead of “imposing” YOUR CHOICE of
is on a person
§ C ¨ R – REJECT V
R Probability of a zvalue . 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, pvalue = .0256
– the one observed CANNOT reject C pvalue REJECT V pvalue – pvalue 245 who might be interested in your conclusions, the true. pvalue allows him/her to assess the “rareness” of the pvalue = 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 zvalue and DOUBLE IT.
EX. : Have done a twotailed 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 . ZTest 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 pvalue is true.
0 Smaller zvalues are more indicative that £ 0 the one observed 0 pvalue 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¡ ¥¦¦
¦¦
¡ ¦
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RR ¦¦ ¥ ¦¦
¦¦
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¦ or ¦¦
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#!
"
#! ©© V¡ ¦¦ R ¥
©©© ¦¦
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§ ¡ OR ¦¦ 5 ©
¦¦ R ¡ V¡ ¦¨ ¦¦ D
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© V¡ ¡ ¦¦
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0 ¡
H
0 ¦¦ R V¡ V¡ ¥¦¦ OR ¦¦
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¡ 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) ZValue
1.20 PValue
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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