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Môn: Statistics for Business (BAO8OIU)
Trường: Trường Đại học Quốc tế, Đại học Quốc gia Thành phố Hồ Chí Minh
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Môn:
Statistics for Business (BAO8OIU)
Trường:
Trường Đại học Quốc tế, Đại học Quốc gia Thành phố Hồ Chí Minh
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International University IU
STATISTICS FOR BUSINESS [IUBA] CHAPTER 08 s
THE COM PARISON OF TW O POPULATIONS n tio la u p o P
STRUCTURE OF PAPER: o w f T o
PART I – THE COM PARISON OF TW O POPULATION M EANS n o ris a p M ETHOD 01 m o
THE COM PARISON OF TWO POPULATION M EANS USING PAIRED-OBSERVATION C e h : T M ETHOD 02 8 r 0
THE COM PARISON OF TWO POPULATION M EANS USING INDEPENDENT RANDOM SAM PLES te p a h C
PART II – THE COM PARISON OF TW O POPULATION PROPORTIONS ss | e sin
PART III – THE COM PARISON OF TW O POPULATION VARIANCES u r B s fo tistic ta S 1
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International University IU PART I
COM PARISON OF TW O POPULATION M EANS s n tio M ETHOD 01 la u p o
COM PARISON OF TW O POPULATION M EANS USING PAIRED-OBSERVATION P o w HYPOTHESIS TESTING f T
Two – tailed Test ing
Right – tailed Testing
Left – tailed Test ing o PROCESS n o Step 01
The population of differences is/ is assumed t o be normally distribut ed. ris a Condition p m Step 02 o Det ermine t he null and C : = : ≤ : ≥ e h alt ernative hypot heses : ≠ : > : < : T ( 8 and) r 0 Step 03
Situation I: When < 30, we use and ⁄√ te
Calculat e t he test stat istic The test st atistic value: p a h value. − C = ss | The crit ical-value √ e approach: sin At the level of significance, u
, the critical value(s): With = − 1 Det ermine the crit ical r B value(s). ± = ± = (,) = −(,) s fo (,)
The -value approach: tistic ta Calculat e t he -value. Situation II: When ⁄ S
≥ 30, we use and √ 2
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International University IU The test st atistic value:
Note: The -value − =
approach is used for
√
The crit ical-value approach: s
At the level of significance, , t he crit ical value(s): n ± = ± / = = − tio la u The p-value approach: p o P If < 0, o w
− = ( < ) × 2
− = ( > )
− = ( < ) f T If > 0, o n
− = ( > ) × 2 o Step 04
Wit h the level of significance ( ) ris a M ake the decision
Situation I: We can reject when p m
The crit ical-value approach: o C [−,] > < e h [−,] > < : T 8 The p-value approach: r 0 te p
− <
− <
− < a h
Situation II: We cannot reject C w hen
The crit ical-value approach: ss | e ∈ [−,] < > sin u ∈ [−,] < > r B The p-value approach: s fo
− >
− >
− > tistic ta S 3
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International University IU CONFIDENCE INTERVALS
Situat ion I: When , w e use and < 30
√ ⁄
= ± s ( ,) n √ tio la
Situat ion II: When , we use and ≥ 30
√ ⁄ u p o P ± o
= w √ f T o n o ris a p m o C e h : T 8 r 0 te p a h C ss | e sin u r B s fo tistic ta S 4
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Example I-01: (Case of the Comparison of Two Population M eans by using Pair-Observation) PROBLEM I-01A:
Recent advances in cell phone screen qualit y have enabled t he showing of movies and commercials on cell phone screens. (Situat ion I)
But according t o t he New York Tim es, advert ising is not as successf ul as movie view ing. Suppose t he follow ing dat a are
numbers of view ers for a movie (M ) and for a com mercial aired w it h t he movie (C). Test for equalit y of movie and s
commercial view ing, on average, using a tw o-t ailed t est at = 0.05 (dat a in thousands): n tio M : 15 17 25 17 14 18 17 16 14 la u C: 10 9 21 16 11 12 13 15 13 p o P o SOLUTION: M C
Score Differences w 15 10 5 f T 17 9 8 o n 25 21 4 o 17 16 1 ris a 14 11 3 p 18 12 6 m o 17 13 4 C e 16 15 1 h 14 13 1 : T 8 r 0 = 9, = 3.67, = 2.45, = 0.05 te p a h
We assum e t hat the population of score dif ferences is norm ally dist ribut ed. C ss | e
: = 0 sin : ≠ 0 u
The test st at istic value: r B − 3.67 − 0 = = ≈ 4.49 2.45 s fo √ √9 tistic
At = 0.05, the critical value(s): ta ± = = ± S = ± (,
( ,.) = ± 2 .306 ) 5
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Thus, at 0.05 level of significance, w e can reject since [− , ] . It means that based on the hypothesis testing we
have significant evidence to prove t he differences bet w een movie and com mercial viewing. PROBLEM I-01B:
A study is undert aken to det erm ine how consum ers react to energy conservation ef for ts. A random group of 60 families is s (Situat ion II)
chosen. Their consum pt ion of electricit y is monit ored in a period bef ore and a period af t er t he families are offered cert ain n
discount s t o reduce t heir energy consum pt ion. Bot h periods are t he same lengt h. The difference in elect ric consumpt ion tio
bet w een t he period before and t he period aft er the offer is recorded for each family. Then the average diff erence in la u
consumpt ion and t he st andard deviat ion of t he difference are comput ed. The result s are
= 0.2 kilowatt and sD = 1.0 p o
kilow at t . At = 0.01, is there evidence t o conclude that conservation effort s reduce consum ption? P o w SOLUTION:
Wit h = − f T = 60, = 0.2, = 1.0, = 0.01 o n o
We assum e t hat the population of score dif ferences is norm ally dist ribut ed. ris a p m : ≤ 0 o
: > 0 C e h
The test st at istic value: : T 8 − 0.2 − 0 = = ≈ 1.5492 r 0 1.0 te √ p √60 a
The critical-value approach: h C
At = 0.01, the critical value: = = 2.33 ss | e
Thus, at 0.01 level of significance, w e cannot reject since ∈ [−,]. It means that with the hypothesis testing we do sin
not have sufficient evidence to prove t hat conservat ion eff ort s reduce consumpt ion. u r B
The p-value approach: s fo
− = ( > 1.5492) ≈ 0.0607 tistic
Since − > , w e cannot reject since ∈ [−, ] . It means that with the hypothesis testing we do not have ta
suff icient evidence to pr ove that conservat ion effort s reduce consumpt ion. S 6
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International University IU PART I
COM PARISON OF TW O POPULATION M EANS M ETHOD 02 s n tio
COM PARISON OF TW O POPULATION M EANS USING INDEPENDENT RANDOM SAM PLES la u p o HYPOTHESIS TESTING P
Two – tailed Test ing
Right – tailed Testing
Left – tailed Test ing o PROCESS w Step 01 f T
The populat ion from which the samples are select ed is/ is assumed to be normally dist ribut ed. o Conditions n o Step 02 ris a Det ermine t he null and
: − = ( − )
: − ≤ ( − )
: − ≥ ( − ) p m alt ernative hypot heses
: − ≠ ( − )
: − > ( −)
: − < ( − ) o C (and) e h Step 03 Situation I: : T 8
Calculat e t he test stat istic r 0 value.
Condition: and are known. te ) ) p
M ethod: We use and ( ⁄ + ( ⁄ a h The crit ical-value C approach: The test st atistic value: ss | Det ermine the crit ical
( − ) − ( − ) e = sin value(s). u + r B s fo
The crit ical-value approach:
At the level of significance, , t he crit ical value(s): tistic ± = ± / = = − ta S 7
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The -value approach: The p-value approach: Calculat e t he -value. If < 0,
− = ( < ) × 2
Note: The -value
− = ( > )
− = ( < ) If > 0,
approach is used for
− = ( > s ) × 2 n
tio Situation II: la u p o Condition: P and are unknow n. o
= ( and are believed to equal (although unknown)) w f T o M ethod:
We use with = ( ) + ( − 1 1 − ) n o and (1 ⁄ + 1⁄) with ris a p ( + ( m − 1) − 1) o =
() ( ( C − 1) + − 1) e h : T The test st atistic value: 8
( − ) − ( − ) r 0 = te 1 p 1 + a h C
At the level of significance, , t he crit ical value(s): With = ( − 1) + ( − 1) , ss | e ± = ± = (,) (,) = −(,) sin u Situation III: r B s fo Condition:
and are unknown. tistic
≠ ( and are believed to unequal (although unknown)) ta S 8
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International University IU M ethod:
We use w ith ( ⁄ ⁄ + ) = (
⁄ )/ ( − 1) + ( ⁄ )/ ( − 1) s n
and ( ⁄) + ( ⁄) tio la u The test st atistic value: p o
( − ) − ( − ) P = o w + f T o n ⁄ ⁄ o
At the level of significance, , t he crit ical values(s): With = ris
⁄ / () ⁄ / () a p m ± o = ± (, = ) (, ) = −(,) C e h : T 8 Situation IV: r 0 te p Condition:
and are unknown. a h
≥ 30and ≥ 30 (Large samples). C ss | M ethod:
We use and ( ⁄ ) + ( ⁄ ) e sin u r B The test st atistic value: s fo ( −
) − ( − ) = tistic + ta S 9
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The crit ical-value approach:
At the level of significance, , t he crit ical values(s): ± = ± / = = − The p-value approach: s n If < 0, tio
− = ( < ) × 2
− = ( > la )
− = ( < ) If u > 0, p o
− = ( > ) × 2 P Step 04
Wit h the level of significance ( ) o w M ake the decision f T
Situation I: We can reject when o n
The crit ical-value approach: o ris [−, ] > < a p [−, ] > < m o The p-value approach: C e h
− <
− <
− < : T 8 r 0 te
Situation II: We cannot reject when p a
The crit ical-value approach: h C
∈ [ −,] < > ss |
∈ [ −,] < > e sin The p-value approach: u r B
− >
− >
− > s fo tistic ta S 10
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International University IU CONFIDENCE INTERVALS Situat ion I:
Condition: and are known. s n
M ethod: We use and ( ⁄ ) + ( ⁄ ) tio la u p o
( − ) = ( − ) ± / + P o w f T Situat ion II: o n o Condition: ris and are unknow n, a p m
= ( and are believed to equal (although unknown)) o C e h : T 8 M ethod:
We use with = ( − 1) + ( − 1) r 0 te p and (1 ⁄ + 1⁄) with a h C (
− 1) + ( − 1) =
() ss | ( ( e − 1) + − 1) sin u r B ( − ) = (
− ) ± ,/ + s fo tistic ta S 11
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International University IU Situat ion III: Condition:
and are unknown.
≠ ( and are believed to unequal (although unknown)) sn 2 2 1 1+ 2 2 tio M ethod:
We use − with = 2 2 la 1 1
/ ( 1−1) 2 2 / (2−1) u p o P
And, = ( ⁄ ) + ( ⁄ ) o w f T o n
( − ) = ( − ) ± ,/ + o ris a p m Situat ion IV: o C e Condition: h
and are unknown. : T 8
≥ 30 and ≥ 30 (Large samples). r 0 te p M ethod:
We use and ) + ( ) a ( ⁄ ⁄ h C ss | e
( − ) = ( − ) ± / + sin u r B s fo tistic ta S 12
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Example I-02: (Case of the Comparison of Two Population M eans by using Independent Random Samples) PROBLEM I-02A:
Suppose t hat t he makers of Duracell bat t eries w ant t o dem onstrat e t hat t heir size AA bat t ery lasts an average of at least (Situation I)
45 minut es longer t han Duracell’s main compet it or, t he Energizer. Two independent random samples of 100 bat t eries of
each kind are select ed, and t he bat t eries ar e run cont inuously unt il t hey are no longer operational. The sample average
life f or Duracell is found to be s
= 308 minutes. The result for the Energizer batteries is = 254 minutes. Assume = n
84 minutes and = 67 minutes
. Is t here evidence to substant iate Dur acell’s claim t hat it s bat t eries last , on average, at tio
least 45 minut es longer than Energizer bat t eries of t he sam e size? la u p o SOLUTION: Duracell bat t eries Energizer batt eries P (1) (2) o w = 100 = 100 f T = 308 = 254 o n = 84 = 67 o
: − ≤ 45 ris a
: − > 45 p
The test st at istic value: m ( − ( 308 − 254) − 45 o
) − ( − ) = ≈ 0.8376 C = e 84 67 h + + 100 100 : T 8
The critical-value approach: r 0
At = 0.05, the critical value: te p a
= = . = 1.645 h C
Thus, at 0.05 level of significance, w e cannot reject since < . It means that with the hypothesis testing we do not ss |
have suf ficient evidence t o prove t hat Duracell bat t eries last , on average, at least 45 m inut es longer t han Energizer e
bat t eries of t he sam e size. sin u
The p-value approach: r B
− = ( > 0.8376) ≈ 0.2011 s fo
Since − > . we cannot reject . It means that with the hypothesis testing w e do not have sufficient evidence
t o prove that Duracell bat t eries last , on average, at least 45 minut es longer t han Energizer bat t eries of t he same size. tistic ta S 13
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International University IU PROBLEM I-02B:
The pow er of supercomputers derives from t he idea of parallel processing. Engineers at Cr ay Research are int erest ed in (Situation II)
det erm ining whet her one of tw o par allel processing designs produces faster average com puting t ime, or whet her t he tw o
designs are equally fast. The follow ing are t he result s, in seconds, of independent random comput at ion t imes using t he t w o designs. Design 1 Design 2 2.1, 2.2, 1.9, 2.0, 1.8, 2.4, 2.6, 2.5, 2.0, 2.1, 2.6, 3.0, s n 2.0, 1.7, 2.3, 2.8, 1.9, 3.0,
2.3, 2.0, 2.4, 2.8, 3.1, 2.7, tio 2.5, 1.8, 2.2 2.6 la u p
Assume that the tw o populat ions of computing t ime are normally distributed a nd that t he tw o population variances o P
are equal. Is there evidence t hat one parallel processing design allow s for faster average comput at ion than t he ot her? o w f T SOLUTION: Design 1 Design 2 o (1) (2) n o = 15 = 13 ris = 2.173 = 2.515 a p = 0.375 = 0.351 m
: − = 0 o C
: − ≠ 0 e h ( + ( ( : T − 1) − 1)
15 − 1) 0.375 + (13 − 1)0.351 8 = = ≈ 0.1326
( − 1) + ( − 1) ( 15 − 1) + (13 − 1) r 0 te p
The test st at istic value: a h
( − ) − ( − ) ( 2.515 − 2.173) − 0 C = = ≈ 2.4785 1 1 1 + 0.1326 + ss | 1 15 13 e sin u
At = 0.05,with = ( − 1) + ( − 1) = (15 − 1) + (13 − 1) = 26, the critical value(s): r B ± = ± = ± ( ( , )
,.) = ± 2.056 s fo
Thus, at 0.05 level of significance, we can reject since [−,]. It means that based on the hypothesis testing we tistic
have suf ficient evidence t o prove t hat one parallel processing design allow s for faster average comput ation t han t he ta S ot her. 14
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International University IU PROBLEM I-02C:
Air Transport W orld recent ly named t he Dut ch airline KLM “ Airline of t he Year.” One measure of t he airline’s excellent (Situation III)
managem ent is it s research effort in developing new rout es and improving service on existing rout es. The airline w anted
t o t est t he profitability of a certain transatlantic flight route and offered daily flights from Europe to the United States
over a period of 6 w eeks on t he new proposed rout e. Then, over a period of 9 w eeks, daily flights were offered from
Europe t o an alt ernat ive airport in t he Unit ed St at es. Weekly profitability data for the two samples were collected, under
t he assumpt ion t hat t hese may be view ed as independent random samples of w eekly profits from the two populations s n
(one populat ion is flights to the proposed airport , and t he ot her populat ion is flights to an alternative airport). Data are tio
as follow s. For t he proposed rout e, = $96,540 per week and
= $12,522. For the alternative route, = $85,991 and la u
= $19,548. Test the hypothesis that the proposed route is more profitable than the alternative route. Use a p o
significance level of your choice. P o w SOLUTION: Proposed Rout e Alt ernat ive Rout e (1) (2) f T o = 6 = 9 n o = 96,540 = 85,991 ris = 12,522 = 19,548 a p m
We assum e t hat tw o populat ions are normally dist ribut ed. o C
: − ≤ 0 e h
: − > 0
The test st at istic value: : T 8
( − ) − ( − ) ( 96,540 − 85,991) − 0 r 0 = = ≈ 1.2737 ( 19,548) te (12,522)+ p + 6 9 a h C ss |
At = 0.05 and e ( ⁄ + ⁄ ) ( 12,522 6 ⁄ + 19,548 9 ⁄ ) sin = = ≈ 12.9993 ≈ 12 u
( ⁄)/ ( − 1) + ( ⁄ ⁄ ⁄ )/ ( − 1) ( 12,522 6) ( 19,548 9 ) + ( 6 ( r B − 1) 9 − 1)
We have t he crit ical value: = (,) = (,.) = 1.782 s fo
Thus, at 0.05 level of significance, w e cannot reject since < . It means that with the hypothesis testing we do not tistic
have sufficient evidence to prove that t he proposed rout e is more profitable t han t he alt ernat ive rout e. ta S 15
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International University IU PROBLEM I-02D:
The phot ography depart ment of a fashion magazine needs t o choose a cam era. Of t he t w o m odels t he depart ment is (Situation IV)
considering, one is made by Nikon and one by M inolt a. The depart ment cont ract s w it h an agency t o det ermine if one of
t he t w o models get s a higher average perfor mance rat ing by professional phot ographers, or whet her the average
perf orm ance ratings of t hese tw o cameras are not st at istically different . The agency asks 60 different professional
phot ographers t o rat e one of t he cameras (30 photographers rat e each m odel). The rat ings are on a scale of 1 t o 10. The
average sample rating for Nikon is 8.5, and t he sample standard deviat ion is 2.1. For t he M inolt a sample, t he average s n
sample rat ing is 7.8, and the st andard deviat ion is 1.8. Is t here a difference betw een t he average population r at ings of tio
t he tw o cam eras? If so, w hich one is rat ed higher? la u p o SOLUTION: Nikon M inolta P (1) (2) o w = 30 = 30 f T = 8.5 = 7 .8 o = 2 .1 = 1.8 n o ris
We assum e t hat tw o populat ions ar e norm ally dist ribut ed a p
: − = 0 m
: − ≠ 0 o
The test st at istic value: C e ( − ( 8.5 − 7.8) − 0 h
) − ( − ) = = ≈ 1.3862 : T ( 1.8) 8 (2.1) + + 30 30 r 0 te p a h
The critical-value approach: C
At = 0.05, the critical value(s): ± = ± / = ± 1.96 ss | e
Thus, at 0.05 level of significance, we cannot r eject since ∈ [−,]. It means that with the hypothesis testing we sin
do not have sufficient evidence t o prove the diff erence bet w een the average populat ion ratings of t w o cameras. u r B
The p-value approach: − = 2( > 1.3862) ≈ 0.1657 s fo
Since − > , w e cannot reject . It means that w ith the hypothesis testing w e do not have sufficient evidence tistic
t o prove the dif ference bet w een the average populat ion rat ings of t w o cameras. ta S 16
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International University IU PART II
COM PARISON OF TW O POPULATION PROPORTIONS HYPOTHESIS TESTING s n
Two – tailed Test ing
Right – tailed Testing
Left – tailed Test ing PROCESS tio la
The populat ion from which the samples are selected is/ is assumed to be normally dist ribut ed. Step 01 u p o Conditions
:/ ℎ ≥ 5
≥ 5 P (1 − ) ≥ 5 ( 1 − ) ≥ 5 o Step 02 w Det ermine t he null and
: − = ( − )
: − ≤ ( − )
: − ≥ ( − ) f T o alt ernative hypot heses n
: − ≠ ( − )
: − > ( − )
: − < ( − ) o (and) ris a Step 03
In order t o perform a test of hypot hesis about the comparison of tw o populat ion proportions, p m
Calculat e t he test stat istic () o
w e use and () + C value. e h
Situation I: If ( − ) = 0, the test statistic value: : T The crit ical-value ( − 8 ) − 0 + = ℎ = r 0 approach: 1 + ( 1 ) + te − 1 Det ermine the crit ical p a h value(s). C
Situation II: If ( − ) = , the test statistic value: ss | The
( − ) − e -value approach: = sin Calculat e t he -value.
(1 − ) + (1 − ) u r B
Note: The -value s fo
The crit ical-value approach: approach is used for
At the level of significance, , t he crit ical value(s): tistic = ± / = = − ta S 17
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International University IU The p-value approach: If < 0,
− = ( < ) × 2
− = ( > )
− = ( < ) If > 0,
− = ( > s ) × 2 n Step 04
Wit h the level of significance ( ), tio M ake the decision
Situation I: We can reject la w hen u
The crit ical-value approach: p o [ −,] > < P o w The p-value approach: f T o
− <
− <
− < n o
Situation II: We cannot reject when ris a
The crit ical-value approach: p m ∈ [−,] < > o C The p-value approach: e h : T
− >
− >
− > 8 r 0 te p a h CONFIDENCE INTERVALS C ss | () e
For all instances, w e alw ays use − / and = () + sin u r B
Situation I: If ( − ) = 0, s fo + tistic
( − ) = ( − ) ±
/ ( − ) + = ta + S 18
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Situation II: If ( − ) = , ( − ) ( − ) ( − ) = (
− ) ± + / sn
Example II: (Case of the Comparison of Tw o Population Proportions) tio la u p PROBLEM 01:
A physicians’ group is int erest ed in t esting t o det ermine w het her more people in small t ow ns choose a physician by word of o
mout h in comparison wit h people in large met ropolit an ar eas. A random sam ple of 1,000 people in small t ow ns reveals t hat P (Situat ion I) o
850 chose their physicians by w ord of mout h; a random sample of 2,500 people living in large met ropolit an areas reveals t hat w
1,950 chose a physician by w ord of mout h. Conduct a one-tailed t est aimed at proving t hat t he percent age of popular f T
recommendat ion of physicians is larger in small tow ns t han in large met ropolitan ar eas. Use = 0.01. o n o ris SOLUTION: Small Towns
Large met ropolit an areas a (1) (2) p m = 850 = 1,950 o C = 1,000 = 2,500 e h : T
:/ ℎ ≥ 5 ≥ 5 8 (1 − ) ≥ 5 (1 − ) ≥ 5 r 0 te p
: − ≤ 0 a h
: − > 0 C We have: ss | + 850 + 1,950 4 e = = = = 0.8 + 1,000 + 2,500 5 sin u r B
The test st at istic value: s fo ( − ( ) − ) − 0 0.85 − 0.78 0 = = ≈ 4.6771 tistic 1 1 ( 1 − ) 1 +
0.8(1 − 0.8) 1 + ta 1,000 2,500 S 19
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The critical- value approach:
At = 0.01, the critical value: = = . = 2.33
Thus, at 0.01 level of significance, w e can reject since < . It means that based on the hypothesis testing we have
suff icient evidence to prove t hat the percent age of popular recom mendation of physicians is larger t han in small t ow ns rather
t han in large met ropolit an areas. s n tio
The p-value approach: − = ( > 4.6771) ≈ 1.4566 × 10 la u p
Since − < , we can reject o
. It means that based on the hypothesis testing we have sufficient evidence to prove P
t hat t he percent age of popular recommendat ion of physicians is larger t han in small t owns rat her t han in large m et ropolit an o areas. w f T o PROBLEM 02:
A random sam ple of 2,060 consum ers show s t hat 13% pr efer California w ines. Over t he next t hr ee m ont hs, an advert ising n o (Situat ion II)
campaign is undert aken t o show that California w ines receive aw ards and win t ast e t est s. The organizers of t he cam paign ris
w ant t o prove t hat t he t hree-mont h campaign raised t he proport ion of people w ho prefer California w ines by at least 5%. At a p
t he end of t he campaign, a random sam ple of 5,000 consumers show s that 19% of t hem now prefer California w ines. Conduct m o
t he test at = 0.05. C e h SOLUTION: Before campaign Aft er cam paign : T (1) (2) 8 r 0 = 0.13 = 0.19 te = 2,060 = 5,000 p a h C
:/ ℎ ≥ 5
≥ 5 (1 − ) ≥ 5 (1 − ) ≥ 5 ss | e
: − ≤ 0.05 sin u
: − > 0.05 r B s fo
The test st at istic value: ( − ( ) − ) − 0.19 − 0.13 0.05 = = ≈ 1.0803 tistic (1 −) 0.13 ( 1 − 0.13) ta (1 − ) + 0.19(1 − 0.19) + S 5,000 2,060 20
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International University IU
The critical- value approach:
At = 0.05, the critical value: = = . = 1.645( = 0.5 − 0.05 = 0.45 )
Thus, at 0.05 level of significance, w e cannot reject since > . It means that there is no evidence that the three-month
campaign raised t he proport ion of people w ho pr efer California w ines by at least 5%. s n
The p-value approach: tio
− = ( > 1.0803) ≈ 0.14 la u p o
Since − > , we cannot reject . It means that there is no evidence that the three-month campaign raised the P
proport ion of people who prefer California wines by at least 5%. o w f T o n PART III o ris a p
COM PARISON OF TW O POPULATION VARIANCES m o C e HYPOTHESIS TESTING h
Two – tailed Test ing
Right – tailed Testing : T PROCESS 8 Step 01 r 0
The populat ion from which the samples are selected is/ is assumed to be normally distribut ed. te Condition. p a Step 02 h = ≤ C Det ermine t he null and : : ss | alt ernative hypot heses : ≠ : > e (and) sin u Step 03 r B
Comput e t he test st at ist ic In order to perform a test of hypothesis about the comparison of two population variances, s fo
value(s) ( ) and the we use − . crit ical value(s) ( ) tistic ta S 21
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International University IU The test st at ist ic = < ) = < ) value(s) ( ( ( ) () ( ) The crit ical value(s) = ( ) ( , ) ( = ( )
(,) s ) n tio Step 04
Wit h the level of significance ( la ), u p M ake the decision. o P
Situation I: We can reject when o > > w ( ) () () () f T o n o
Situation II: We cannot reject when ris < < a ( ) () () () p m o C e h
Example III: (Case of the Comparison of Two Population Variances) : T 8 r 0 PROBLEM :
The f ollow ing dat a are independent random samples of sales of t he Nissan Pulsar m odel made in a joint vent ure of Nissan te
and Alfa Romeo. The dat a represent sales at dealerships bef ore and aft er t he announcement t hat t he Pulsar m odel w ill no p a
longer be made in It aly. Sales num bers are monthly. h C
Before: 329, 234, 423, 328, 400, 399, 326, 452, 541, 680, 456, 220 ss | e
Aft er: 212, 630, 276, 112, 872, 788, 345, 544, 110, 129, 776 sin u
Do you believe t hat t he variance of t he number of car s sold per m ont h before t he announcement is equal to t he variance of r B
t he number of cars sold per mont h aft er t he announcement ? s fo tistic ta S 22
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International University IU SOLUTION: Before Aft er (1) (2) = 12 = 11 = 128.03 = 294.70 = 16 ,384 = 86,849.09 s n
We assum e t hat tw o populat ions ar e norm ally dist ribut ed tio
: = la u ≠ p : o
The test st at istic value: P o 86,849.09 w = ≈ 5.3 ( ) = 16,384 f T
At = 0.05, the critical value o n = (
) = ( ,) = 3.53 o ( ) , ris a
Thus, at 0.05 level of significance, w e can reject p
since > . It means that based on the hypothesis testing we ( ) () m
have suf ficient evidence to prove t hat t he variance of t he number of cars sold per month before t he announcement is o C
differ ent from t he variance of t he number of cars sold per mont h af t er t he announcement . e h : T 8 r 0 te p a h C ss | e sin u r B s fo tistic ----- THE END ------ ta S 23
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