STATISTICS FOR BUSINESS - CHAP08 - Comparison of Two Populations.pdf

International University IU

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STATISTICS FOR BUSINESS

CHAPTER 08

COMPARISON OF TWO POPULATIONS

STRUCTURE OF PAPER

PART I - COMPARISON OF TWO POPULATION MEANS

METHOD 01

COMPARISON OF TWO POPULATION MEANS USING PAIRED-OBSERVATION

METHOD 02

COMPARISON OF TWO POPULATION MEANS USING INDEPENDENT RANDOM SAMPLES

PART II - COMPARISON OF TWO POPULATION PROPORTIONS

PART III - COMPARISON OF TWO POPULATION VARIANCES


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PART I

COMPARISON OF TWO POPULATION MEANS

METHOD 01

COMPARISON OF TWO POPULATION MEANS USING PAIRED-OBSERVATION

HYPOTHESIS TESTING PROCESS

Two – tailed Testing Right – tailed Testing Left – tailed Testing

Step 01 The population of differences is normally distributed

Step 02 Determine the null and alternative hypotheses (퐻 and 퐻 )

퐻 :휇 = 휇 퐻 :휇 ≠ 휇

퐻 :휇 ≤ 휇 퐻 : 휇 > 휇

퐻 : 휇 ≥ 휇 퐻 :휇 < 휇

Step 03 Compute the test statistic value (푡 /푧 ) and the critical value(s) (푡 /푧 )

Situation I: When 푛 < 30, we use 푡 − 푑푖푠푡푟푖푏푢푡푖표푛 and 푆퐷 = 푠 √푛⁄ The test statistic value:

푡 =퐷 − 휇푠√푛

The critical value(s): With 푑푓 = 푛 − 1

±푡 = ±푡( , ) 푡 = 푡( , ) 푡 = −푡( , )


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Situation II: When 푛 ≥ 30, we use 푧 − 푑푖푠푡푟푖푏푢푡푖표푛 and 푆퐷 = 푠 √푛⁄ The test statistic value:

푧 =퐷 − 휇푠√푛

The critical value(s):

±푧 = ±푧 / 푧 = 푧 푧 = −푧

Step 04 Make the decision

With the level of significance (훼)

Situation I: We can reject 퐻 when 푡 [−푡 , 푡 ] 푡 > 푡 푡 < 푡

푧 [−푧 , 푧 ] 푧 > 푧 푧 < 푧

Situation II: We cannot reject 퐻 when 푡 ∈ [−푡 , 푡 ] 푡 < 푡 푡 > 푡

푧 ∈ [−푧 , 푧 ] 푧 < 푧 푧 > 푧

CONFIDENCE INTERVALS

Situation I: When 푛 < 30, we use 푡 − 푑푖푠푡푟푖푏푢푡푖표푛 and 푆퐷 = 푠 √푛⁄

푪푰 풇풐풓 흁푫 = 푫 ± 풕(풅풇,휶ퟐ)풔푫√풏푫


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Situation II: When 푛 ≥ 30, we use 푧 − 푑푖푠푡푟푖푏푢푡푖표푛 and 푆퐷 = 푠 √푛⁄

푪푰 풇풐풓 흁푫 = 푫 ± 풛휶ퟐ

풔푫√풏푫

Sample:

PROBLEM 8.1: (Situation I)

Recent advances in cell phone screen quality have enabled the showing of movies and commercials on cell phone screens. But according to the New York Times, advertising is not as successful as movie viewing. Suppose the following data are numbers of viewers for a movie (M) and for a commercial aired with the movie (C). Test for equality of movie and commercial viewing, on average, using a two-tailed test at = 0.05 (data in thousands): M: 15 17 25 17 14 18 17 16 14 C: 10 9 21 16 11 12 13 15 13

SOLUTION: M C Score Differences 15 10 5 17 9 8 25 21 4 17 16 1 14 11 3 18 12 6 17 13 4 16 15 1 14 13 1

푛 = 9, 퐷 = 3.67, 푠 = 2.45, 훼 = 0.05


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We assume that the population of score differences is normally distributed.

퐻 : 휇 = 0 퐻 : 휇 ≠ 0

The test statistic value:

푡 =퐷 − 휇푠√푛

=3.67− 0

2.45√9

≈ 4.49

The critical value(s): ±푡 = ±푡( , ) == ±푡( , . ) = ±2.306

Thus, at 0.05 level of significance, we can reject 퐻 since 푡 [−푡 , 푡 ]. It means that based on the hypothesis testing we have significant evidence to prove the differences between movie and commercial viewing.

PROBLEM 8.2: (Situation II)

A study is undertaken to determine how consumers react to energy conservation efforts. A random group of 60 families is chosen. Their consumption of electricity is monitored in a period before and a period after the families are offered certain discounts to reduce their energy consumption. Both periods are the same length. The difference in electric consumption between the period before and the period after the offer is recorded for each family. Then the average difference in consumption and the standard deviation of the difference are computed. The results are 퐷 = 0.2 kilowatt and sD = 1.0 kilowatt. At = 0.01, is there evidence to conclude that conservation efforts reduce consumption?

SOLUTION: 푛 = 60, 퐷 = 0.2, 푠 = 1.0, 훼 = 0.01 We assume that the population of score differences is normally distributed.

퐻 : 휇 ≥ 0 퐻 : 휇 < 0


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푧 =퐷 − 휇푠√푛

=0.2 − 0

1.0√60

≈ 1.5492

The critical value: 푧 = −푧 = −2.33

Thus, at 0.01 level of significance, we cannot reject 퐻 since 푧 ∈ [−푧 , 푧 ]. It means that with the hypothesis testing we do not have sufficient evidence to prove that conservation efforts reduce consumption.


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PART I

COMPARISON OF TWO POPULATION MEANS

METHOD 02

COMPARISON OF TWO POPULATION MEANS USING INDEPENDENT RANDOM SAMPLES



Step 01 Two populations are normally distributed


퐻 :휇 − 휇 = (휇 − 휇 )

퐻 :휇 − 휇 ≠ (휇 − 휇 )

퐻 :휇 − 휇 ≤ (휇 − 휇 )

퐻 : 휇 − 휇 > (휇 − 휇 )

퐻 : 휇 − 휇 ≥ (휇 − 휇 )

퐻 :휇 − 휇 < (휇 − 휇 )

Step 03 Compute the test statistic value (푧 /푡 ) and the critical value(s) (푧 /푡 )

Situation I:

Condition: 휎 and 휎 are known (for all 푛 and 푛 )

Method: We use 푧 − 푑푖푠푡푟푖푏푢푡푖표푛 and 푆퐷 = (휎 푛⁄ ) + (휎 푛⁄ ) The test statistic value:

푧 =(푥̅ − 푥̅ ) − (휇 − 휇 )

휎푛 + 휎

푛


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±푧 = ±푧 / 푧 = 푧 푧 = −푧

Situation II:

Condition: 휎 and 휎 are unknown 휎 = 휎 (pooled variance) 푑푓 = (푛1 − 1) + (푛2 − 1) ≤ 30

Method: We use 푡 − 푑푖푠푡푟푖푏푢푡푖표푛 with 푑푓 = (푛1 − 1) + (푛2 − 1)

And, 푆퐷 = 푆푃2(1 푛1⁄ + 1 푛2⁄ ) with

푆 =(푛 − 1)푆 + (푛 − 1)푆

(푛 − 1) + (푛 − 1) (푝표표푙푒푑 푣푎푟푖푎푛푐푒)


푡 =(푥̅ − 푥̅ )− (휇 − 휇 )

푆 1푛 + 1

푛

The critical value(s): With 푑푓 = (푛1 − 1) + (푛2 − 1),

±푡 = ±푡( , ) 푡 = 푡( , ) 푡 = −푡( , )


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Situation III:

Condition: 휎 and 휎 are unknown 휎 ≠ 휎

푑푓 = ⁄ ⁄

⁄ /( ) ⁄ /( )≤ 30

Method: We use 푡 − 푑푖푠푡푟푖푏푢푡푖표푛 with

푑푓 =(푆 푛⁄ + 푆 푛⁄ )

(푆 푛⁄ ) /(푛 − 1) + (푆 푛⁄ ) /(푛 − 1)

And, 푆퐷 = (푠 푛⁄ ) + (푠 푛⁄ ) The test statistic value:

푡 =(푥̅ − 푥̅ )− (휇 − 휇 )

푠푛 + 푠

푛

The critical values(s): With 푑푓 = ⁄ ⁄

⁄ /( ) ⁄ /( )

±푡 = ±푡( , ) 푡 = 푡( , ) 푡 = −푡( , )


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Situation IV:

Condition: 휎 and 휎 are unknown If 휎 = 휎 , 푑푓 = (푛1 − 1) + (푛2 − 1) > 30

If 휎 ≠ 휎 , 푑푓 = 푆12 푛1+푆2

2 푛2

푆12 푛1 /(푛1−1) 푆2

2 푛2 /(푛2−1)> 30

Method: We use 푧 − 푑푖푠푡푟푖푏푢푡푖표푛 and 푆퐷 = 푠12 푛1⁄ + 푠2

2 푛2⁄


푧 =(푥̅ − 푥̅ ) − (휇 − 휇 )

푠푛 + 푠

푛

The critical values(s):

±푧 = ±푧 / 푧 = 푧 푧 = −푧



Situation I: We can reject 퐻 when 푡 [−푡 , 푡 ] 푡 > 푡 푡 < 푡

푧 [−푧 , 푧 ] 푧 > 푧 푧 < 푧

Situation II: We cannot reject 퐻 when

푡 ∈ [−푡 , 푡 ] 푡 < 푡 푡 > 푡

푧 ∈ [−푧 , 푧 ] 푧 < 푧 푧 > 푧


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Situation I:

Condition: 휎 and 휎 are known (for all 푛 and 푛 )

Method: We use 푧 − 푑푖푠푡푟푖푏푢푡푖표푛 and 푆퐷 = (휎 푛⁄ ) + (휎 푛⁄ )

푪푰 풇풐풓 (흁ퟏ − 흁ퟐ) = (풙ퟏ − 풙ퟐ) ± 풛휶/ퟐ흈ퟏퟐ

풏ퟏ+흈ퟐퟐ

풏ퟐ

Situation II:

Condition: 휎 and 휎 are unknown

휎 = 휎

푑푓 = (푛1 − 1) + (푛2 − 1) ≤ 30

Method: We use 푡 − 푑푖푠푡푟푖푏푢푡푖표푛 with 푑푓 = (푛1 − 1) + (푛2 − 1)

And, 푆퐷 = 푆푃2 (1 푛1⁄ + 1 푛2⁄ ) with

푆 =(푛 − 1)푆 + (푛 − 1)푆

(푛 − 1) + (푛 − 1) (푝표표푙푒푑 푣푎푟푖푎푛푐푒)

푪푰 풇풐풓 (흁ퟏ − 흁ퟐ) = (풙ퟏ − 풙ퟐ) ± 풕풅풇,휶/ퟐ 푺푷ퟐퟏ풏ퟏ

+ퟏ풏ퟐ


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Situation III:


휎 ≠ 휎

푑푓 = 푆12 푛1+푆2

2 푛2

푆12 푛1 /(푛1−1) 푆2

2 푛2 /(푛2−1)≤ 30

Method: We use 푡 − 푑푖푠푡푟푖푏푢푡푖표푛 with 푑푓 = 푆12 푛1+푆2

2 푛2

푆12 푛1 /(푛1−1) 푆2

2 푛2 /(푛2−1)

And, 푆퐷 = (푠 푛⁄ ) + (푠 푛⁄ )

푪푰 풇풐풓 (흁ퟏ − 흁ퟐ) = (풙ퟏ − 풙ퟐ) ± 풕풅풇,휶/ퟐ풔ퟏퟐ

풏ퟏ+풔ퟐퟐ

풏ퟐ

Situation IV


If 휎 = 휎 , 푑푓 = (푛1 − 1) + (푛2 − 1) > 30

If 휎 ≠ 휎 , 푑푓 = 푆12 푛1+푆2

2 푛2

푆12 푛1 /(푛1−1) 푆2

2 푛2 /(푛2−1)> 30

Method: We use 푧 − 푑푖푠푡푟푖푏푢푡푖표푛 and 푆퐷 = 푠12 푛1⁄ + 푠2

2 푛2⁄


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푪푰 풇풐풓 (흁ퟏ − 흁ퟐ) = (풙ퟏ − 풙ퟐ) ± 풛휶/ퟐ풔ퟏퟐ

풏ퟏ+풔ퟐퟐ

풏ퟐ

Sample:

PROBLEM 8.3: (Situation I)

Suppose that the makers of Duracell batteries want to demonstrate that their size AA battery lasts an average of at least 45 minutes longer than Duracell’s main competitor, the Energizer. Two independent random samples of 100 batteries of each kind are selected, and the batteries are run continuously until they are no longer operational. The sample average life for Duracell is found to be 푥̅ = 308 minutes. The result for the Energizer batteries is 푥̅ = 254 minutes. Assume 휎 = 84 minutes and 휎 = 67 minutes. Is there evidence to substantiate Duracell’s claim that its batteries last, on average, at least 45 minutes longer than Energizer batteries of the same size?

SOLUTION: Duracell batteries Energizer batteries (1) (2)

푛 = 100 푛 = 100 푥̅ = 308 푥̅ = 254 휎 = 84 휎 = 67

퐻 : 휇 − 휇 ≤ 45 퐻 :휇 − 휇 > 45


푧 =(푥̅ − 푥̅ )− (휇 − 휇 )

휎푛 + 휎

푛

=(308− 254)− 0

84100 + 67

100

≈ 0.838

At 훼 = 0.05, the critical value: 푧 = 푧 = 푧 . = 1.645

Thus, at 0.05 level of significance, we cannot reject 퐻 since 푧 < 푧 . It means that with the hypothesis testing we do not have sufficient evidence to prove that Duracell batteries last, on average, at least 45 minutes longer than Energizer batteries of the same size.


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PROBLEM 8.4: (Situation II)

The power of supercomputers derives from the idea of parallel processing. Engineers at Cray Research are interested in determining whether one of two parallel processing designs produces faster average computing time, or whether the two designs are equally fast. The following are the results, in seconds, of independent random computation times using the two designs.

Design 1 Design 2 2.1, 2.2, 1.9, 2.0, 1.8, 2.4, 2.0, 1.7, 2.3, 2.8, 1.9, 3.0, 2.5, 1.8, 2.2

2.6, 2.5, 2.0, 2.1, 2.6, 3.0, 2.3, 2.0, 2.4, 2.8, 3.1, 2.7, 2.6

Assume that the two populations of computing time are normally distributed and that the two population variances are equal. Is there evidence that one parallel processing design allows for faster average computation than the other?

SOLUTION: Design 1 Design 2 (1) (2)

푛 = 15 푛 = 13 푥̅ = 2.173 푥̅ = 2.515 휎 = 0.375 휎 = 0.351

퐻 : 휇 − 휇 = 0 퐻 :휇 − 휇 ≠ 0

푆 =(푛 − 1)푆 + (푛 − 1)푆

(푛 − 1) + (푛 − 1) =(15− 1)0.375 + (13− 1)0.351

(15− 1) + (13− 1) ≈ 0.1326


푡 =(푥̅ − 푥̅ )− (휇 − 휇 )

푆 1푛 + 1

푛

=(2.515− 2.173)− 0

0.1326 115 + 1

13

≈ 2.4785

At 훼 = 0.05, 푑푓 = (푛 − 1) + (푛 − 1) = (15 − 1) + (13− 1) = 26, the critical value(s):

±푡 = ±푡( , ) = ±푡( , . ) = ±2.056


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Thus, at 0.05 level of significance, we can reject 퐻 since 푡 [−푡 , 푡 ]. It means that based on the hypothesis testing we have sufficient evidence to prove that one parallel processing design allows for faster average computation than the other.

PROBLEM 8.5: (Situation III)

Air Transport World recently named the Dutch airline KLM “Airline of the Year.” One measure of the airline’s excellent management is its research effort in developing new routes and improving service on existing routes. The airline wanted to test the profitability of a certain transatlantic flight route and offered daily flights from Europe to the United States over a period of 6 weeks on the new proposed route. Then, over a period of 9 weeks, daily flights were offered from Europe to an alternative airport in the United States. Weekly profitability data for the two samples were collected, under the assumption that these may be viewed as independent random samples of weekly profits from the two populations (one population is flights to the proposed airport, and the other population is flights to an alternative airport). Data are as follows. For the proposed route, 푥̅ = $96,540 per week and 푠 = $12,522. For the alternative route, 푥̅ = $85,991 and 푠 = $19,548. Test the hypothesis that the proposed route is more profitable than the alternative route. Use a significance level of your choice.

SOLUTION: Proposed Route Alternative Route (1) (2)

푛 = 6 푛 = 9 푥̅ = 96,540 푥̅ = 85,991 휎 = 12,522 휎 = 19,548

We assume that two populations are normally distributed.

퐻 : 휇 − 휇 ≤ 0 퐻 :휇 − 휇 > 0


푡 =(푥̅ − 푥̅ )− (휇 − 휇 )

푠푛 + 푠

푛

=(96,540− 85,991)− 0(12,522)

6 + (19,548)9

≈ 1.2737


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At 훼 = 0.05

푑푓 =(푆 푛⁄ + 푆 푛⁄ )

(푆 푛⁄ ) /(푛 − 1) + (푆 푛⁄ ) /(푛 − 1)

푑푓 =(12,522 6⁄ + 19,548 9⁄ )

(12,522 6⁄ )(6 − 1) + (19,548 9⁄ )

(9− 1)

≈ 12.9993 ≈ 13

the critical value(s): 푡 = 푡( , ) = 푡( , . ) = 1.771 Thus, at 0.05 level of significance, we cannot reject 퐻 since 푡 < 푡 . It means that with the hypothesis testing we do not have sufficient evidence to prove that the proposed route is more profitable than the alternative route.

PROBLEM 8.6: (Situation IV)

The photography department of a fashion magazine needs to choose a camera. Of the two models the department is considering, one is made by Nikon and one by Minolta. The department contracts with an agency to determine if one of the two models gets a higher average performance rating by professional photographers, or whether the average performance ratings of these two cameras are not statistically different. The agency asks 60 different professional photographers to rate one of the cameras (30 photographers rate each model). The ratings are on a scale of 1 to 10. The average sample rating for Nikon is 8.5, and the sample standard deviation is 2.1. For the Minolta sample, the average sample rating is 7.8, and the standard deviation is 1.8. Is there a difference between the average population ratings of the two cameras? If so, which one is rated higher?

SOLUTION: Nikon Minolta (1) (2)

푛 = 30 푛 = 30 푥̅ = 8.5 푥̅ = 7.8 휎 = 2.1 휎 = 1.8


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We assume that two populations are normally distributed

퐻 : 휇 − 휇 = 0 퐻 :휇 − 휇 ≠ 0


푧 =(푥̅ − 푥̅ )− (휇 − 휇 )

푠푛 + 푠

푛

=(8.5− 7.8)− 0(2.1)

30 + (1.8)30

≈ 1.3862

At 훼 = 0.05, the critical value(s): ±푧 = ±푧 / = ±1.96

Thus, at 0.05 level of significance, we cannot reject 퐻 since 푧 ∈ [−푧 , 푧 ]. It means that with the hypothesis testing we do not have sufficient evidence to prove the difference between the average population ratings of two cameras.


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PART II

COMPARISON OF TWO POPULATION PROPORTIONS





퐻 : 푝 − 푝 = (푝 − 푝 )

퐻 :푝 − 푝 ≠ (휇 − 푝 )

퐻 : 푝 − 푝 ≤ (푝 − 푝 )

퐻 :푝 − 푝 > (푝 − 푝 )

퐻 :푝 − 푝 ≥ (푝 − 푝 )

퐻 : 푝 − 푝 < (푝 − 푝 )

Step 03 Compute the test statistic value (푡 /푧 ) and the critical value(s) (푡 /푧 )

For all instances, we always use 푧 − 푑푖푠푡푟푖푏푢푡푖표푛 푧훼/2 and 푆퐷 = 푝1 1−푝1

푛1+

푝2 1−푝2

푛2


푧 =(푝̂ − 푝̂ ) − (푝 − 푝 )푝̂ (1− 푝̂ )

푛 + 푝̂ (1− 푝̂ )푛


푧 = ±푧 / 푧 = 푧 푧 = −푧



Situation I: We can reject 퐻 when 푧 [−푧 , 푧 ] 푧 > 푧 푧 < 푧

Situation II: We cannot reject 퐻 when 푧 ∈ [−푧 , 푧 ] 푧 < 푧 푧 > 푧


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For all instances, we always use 푧 − 푑푖푠푡푟푖푏푢푡푖표푛 푧훼/2 and 푆퐷 = 푝1 1−푝1

푛1+

푝2 1−푝2

푛2

푪푰 풇풐풓 (풑ퟏ − 풑ퟐ) = (풑ퟏ − 풑ퟐ) ± 풛휶/ퟐ풑ퟏ(ퟏ − 풑ퟏ)

풏ퟏ+풑ퟐ(ퟏ − 풑ퟐ)

풏ퟐ

Sample:

PROBLEM 8.7:

A physicians’ group is interested in testing to determine whether more people in small towns choose a physician by word of mouth in comparison with people in large metropolitan areas. A random sample of 1,000 people in small towns reveals that 850 chose their physicians by word of mouth; a random sample of 2,500 people living in large metropolitan areas reveals that 1,950 chose a physician by word of mouth. Conduct a one-tailed test aimed at proving that the percentage of popular recommendation of physicians is larger in small towns than in large metropolitan areas. Use = 0.01.

SOLUTION: Small Towns Large metropolitan areas (1) (2)

푝̂ = 푥 /푛 = 850/1000 = 0.85 푝̂ = 푥 /푛 = 1,950/2,500 = 0.78 푛 = 1,000 푛 = 2,500

퐻 : 푝 − 푝 ≤ 0

퐻 :푝 − 푝 > 0 The test statistic value:

푧 =(푝̂ − 푝̂ ) − (푝 − 푝 )푝̂ (1 − 푝̂ )

푛 + 푝̂ (1 − 푝̂ )푛

=(0.85− 0.78)− 0

0.85(1− 0.85)1,000 + 0.78(1− 0.78)

2,500

≈ 4.9982


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At 훼 = 0.01, the critical value: 푧 = 푧 = 푧 . = 2.33 Thus, at 0.01 level of significance, we can reject 퐻 since 푧 < 푧 . It means that based on the hypothesis testing we have sufficient evidence to prove that the percentage of popular recommendation of physicians is larger than in small towns rather than in large metropolitan areas.


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PART III

COMPARISON OF TWO POPULATION VARIANCES





퐻 : 휎 = 휎

퐻 : 휎 ≠ 휎

퐻 : 휎 ≤ 휎

퐻 :휎 > 휎

퐻 : 휎 ≥ 휎

퐻 :휎 < 휎

Step 03 Compute the test statistic value (퐹 ) and the critical value(s) (퐹 )

For all instances, we always use 퐹 − 푑푖푠푡푟푖푏푢푡푖표푛

The test statistic value (퐹 )

퐹 ( ) =푠푠

(푠 < 푠 )

퐹 ( ) =푠푠

(푠 < 푠 ) 퐹 ( ) =

푠푠

(푠 < 푠 ) 퐹 ( ) =푠푠

(푠 < 푠 )

The test statistic value (퐹 )

퐹 ( ) =1

퐹( , )

퐹 ( ) = 퐹( , ) 퐹 ( ) = 퐹( , ) 퐹 ( ) =

1퐹( , )


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With the level of significance (훼) Situation I: We can reject 퐻 when

퐹 ( ) 퐹 ( ),퐹 ( )

퐹 ( ) 퐹 ( ),퐹 ( ) 퐹 ( ) > 퐹 ( ) 퐹 ( ) < 퐹 ( )

Situation II: We cannot reject 퐻 when

퐹 ( ) ∈ 퐹 ( ),퐹 ( )

퐹 ( ) ∈ 퐹 ( ),퐹 ( ) 퐹 ( ) < 퐹 ( ) 퐹 ( ) > 퐹 ( )

Sample:

PROBLEM 8.8:

The following data are independent random samples of sales of the Nissan Pulsar model made in a joint venture of Nissan and Alfa Romeo. The data represent sales at dealerships before and after the announcement that the Pulsar model will no longer be made in Italy. Sales numbers are monthly. Before: 329, 234, 423, 328, 400, 399, 326, 452, 541, 680, 456, 220 After: 212, 630, 276, 112, 872, 788, 345, 544, 110, 129, 776 Do you believe that the variance of the number of cars sold per month before the announcement is equal to the variance of the number of cars sold per month after the announcement?

SOLUTION: Before After (1) (2)

푛 = 12 푛 = 11 푠 = 128.03 푠 = 294.70 s = 16,384 s = 86,849.09


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We assume that two populations are normally distributed

퐻 : 휎 = 휎

퐻 :휎 ≠ 휎 The test statistic value:

퐹 ( ) =푠푠

=86,849.09

16,384 ≈ 5.3

At 훼 = 0.05, the critical value(s): 퐹 ( ) = 퐹( , ) = 퐹( , ) = 2.86

Thus, at 0.05 level of significance, we can reject 퐻 since S퐹 ( ) > 퐹 ( ) . It means that based on the hypothesis

testing we have sufficient evidence to prove that the variance of the number of cars sold per month before the announcement is different from the variance of the number of cars sold per month after the announcement.

STATISTICS FOR BUSINESS - CHAP08 - Comparison of Two Populations.pdf

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