session 8b. decision models -- prof. juran2 overview hypothesis testing review of the basics...

Session 8b

Decision Models -- Prof. Juran

2

OverviewHypothesis Testing • Review of the Basics

– Single Parameter– Differences Between Two Parameters

• Independent Samples• Matched Pairs

– Goodness of Fit• Simulation Methods


3

Basic Hypothesis Testing Method

1. Formulate Two Hypotheses

2. Select a Test Statistic

3. Derive a Decision Rule

4. Calculate the Value of the Test Statistic; Invoke the Decision Rule in light of the Test Statistic


4

H0 is true HA is true Do Not Reject H0 Correct Decision ERROR (Type II)

Reject H0 ERROR (Type I) Correct Decision

Type I: Reject H0 when H0 is in fact true (reject a true hypothesis).

Type II: Do Not Reject H0 when HA is in fact true ("accept" a false hypothesis).

Let = probability of a Type I Error = P(reject H0 | H0 is true),

= probability of a Type II Error = P("accept" H0 | HA is true).

Applied Regression -- Prof. Juran

5

Hypothesis Testing: Gardening Analogy

Rocks Dirt


6



7



8



9


Screened out stuff:Correct decision or Type I

Error?

Stuff that fell through:Correct decision or Type II

Error?


10

The p-value of a test is the probability of observing a sample at least as “unlikely” as ours.

In other words, it is the “minimum level” of significance that would allow us to reject H0.

Small p-value = unlikely H0


11

Example: Buying a Laundromat

A potential entrepreneur is considering the purchase of a coin-operated laundry. The present owner claims that over the past 5 years the average daily revenue has been $675. The buyer would like to find out if the true average daily revenue is different from $675.

A sample of 30 selected days reveals a daily average revenue of $625 with a standard deviation of $75.


12

H0: 675 HA: 675


13

Test Statistic:

Decision Rule, based on alpha of 1%: Reject H0 if the test statistic is greater than 2.575 or less than -2.575.

ns

Xz 0

0


14

0z ns

X 0

3075

675625

65.3


15

We reject H0. There is sufficiently strong evidence against H0 to reject it at the 0.01 level. We conclude that the true mean is different from $675.


16

Example: Reliability Analysis

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10111213

A B C D E F G H IStart Rank Fail Rank Battery Life Start Time Fail Time

Distribution of battery lifetimes (lognormal) 1 2 Battery A 25.28 0.0 25.3Mean 20 2 1 Battery B 12.02 0.0 12.0Stdev 5 3 3 Battery C 17.00 12.0 29.0

4 5 Battery D 33.15 25.3 58.45 4 Battery E 23.53 29.0 52.5

SimulationFailure# Insert Battery Current time In Position 1 In Position 2

0 Battery A Battery B1 Battery C 12.017 Battery A Battery C 12.022 Battery D 25.275 Battery D Battery C 25.283 Battery E 29.016 Battery D Battery E 29.024 (none) 52.542 Battery D (none) 52.54 <---- Device Fails


17

The brand manager wants to begin advertising for this product, and would like to claim a mean time between failures (MTBF) of 45 hours.

The product is only in the prototype phase, so the design engineer uses Crystal Ball simulation to estimate the product’s reliability characteristics. Extracted data:

1234567891011121314

A BStatistics Fail Time

Trials 1000Mean 43.82Median 43.40Mode ---Standard Deviation 6.09Variance 37.11Skewness 0.47Kurtosis 3.13Coeff. of Variability 0.14Range Minimum 29.12Range Maximum 64.44Range Width 35.32Mean Std. Error 0.19


18

123456789101112131415

A B C D E F GStatistics Fail Time

Trials 1000 Hypothesized MTBF 45Mean 43.82 Test Statistic -6.103Median 43.40 p-ValuesMode --- Upper Tail N/ AStandard Deviation 6.09 Two-Tail 0.0000Variance 37.11 Lower Tail 0.0000Skewness 0.47Kurtosis 3.13Coeff. of Variability 0.14Range Minimum 29.12Range Maximum 64.44Range Width 35.32Mean Std. Error 0.19

=(B3-E2)/(B14)

=IF(E3>0,1-NORMSDIST(E3),"N/A")=(1-NORMSDIST(ABS(E3)))*2=IF(E3<0,NORMSDIST(E3),"N/A")

H0 here is that the product lasts 45 hours (on the average).

There is sufficiently strong evidence against H0 to reject it at any reasonable significance level. We conclude that the true MTBF is less than 45 hours.


19

Example: Effects of Sales Campaigns

In order to measure the effect of a storewide sales campaign on nonsale items, the research director of a national supermarket chain took a random sample of 13 pairs of stores that were matched according to average weekly sales volume.

One store of each pair (the experimental group) was exposed to the sales campaign, and the other member of the pair (the control group) was not.


20

The following data indicate the results over a weekly period:

STORE WITH SALES CAMPAIGN WITHOUT SALES CAMPAIGN 1 67.2 65.3 2 59.4 54.7 3 80.1 81.3 4 47.6 39.8 5 97.8 92.5 6 38.4 37.9 7 57.3 52.4 8 75.2 69.9 9 94.7 89.0 10 64.3 58.4 11 31.7 33.0 12 49.3 41.7 13 54.0 53.6


21

Is the campaign effective?

Basically this is asking: Is there a difference between the average sales from these two populations (with and without the campaign)?


22

Two Methods• Independent Samples

– General Method• Matched Pairs

– Useful Only in Specific Circumstances– More Powerful Statistically– Requires Logical One-to-One

Correspondence between Pairs


23

Independent Samples Method

Let the true population mean sales with the campaign be represented by X, and the population mean sales w ithout the campaign be represented by Y .

H 0: 0 YX

H A : 0 YX

We have a small-sample test, so we’ll get our critical value from the t-table. We have one tail, alpha = 0.05, and 12 degrees of freedom, so the critical t-value w ill be 1.782. If the value of our test statistic is greater than 1.782, then we will reject the null hypothesis.


24

0t

Y

Y

X

X

n

s

n

s

YX22

0

13

49.19

13

03.20

019.5985.6222

4714.0

We do not reject the null hypothesis. The campaign made no significant difference in sales.


25

123456789

1011121314

A B C D E F G H I J K

STOREWITH SALES CAMPAIGN

WITHOUT SALES CAMPAIGN

1 67.2 65.3 Sample 1 mean 62.852 59.4 54.7 stdev 20.033 80.1 81.3 n 134 47.6 39.8 Sample 2 mean 59.195 97.8 92.5 stdev 19.496 38.4 37.9 n 137 57.3 52.4 H0 08 75.2 69.99 94.7 89 test stat 0.471410 64.3 58.4 tails 111 31.7 33 p-value 0.322912 49.3 41.713 54 53.6

=AVERAGE($B$2:$B$14)=STDEV($B$2:$B$14)=COUNT($B$2:$B$14)=AVERAGE($C$2:$C$14)=STDEV($C$2:$C$14)=COUNT($C$2:$C$14)

=((G2-G5)-G8)/SQRT(((G3^2)/G4)+((G6^2)/G7))

=TDIST(G10,G4-1,G11)


26

Matched-Pairs Method

Let the true difference between the population means be represented by D. H0: 0D

HA: 0D

The critical t-value will still be 1.782. If the value of our test statistic is greater than 1.782, then we will reject the null hypothesis.


27

0t

n

sD

D

0

13

1855.306538.3

14.4

This time we do reject the null hypothesis, and conclude that the campaign actually did have a significant positive effect on sales.


28

123456789

10111213141516

A B C D E F G H I J K L M

STORE

WITH SALES

CAMPAIGN

WITHOUT SALES CAMPAIGN Difference

1 67.2 65.3 1.90 tails 12 59.4 54.7 4.70 alpha 0.053 80.1 81.3 -1.20 critical t 1.7824 47.6 39.8 7.80 Ho 05 97.8 92.5 5.30 n 136 38.4 37.9 0.50 d-bar 3.65387 57.3 52.4 4.90 stdev 3.18558 75.2 69.9 5.309 94.7 89 5.70 test statistic 4.135610 64.3 58.4 5.90 p-value 0.000711 31.7 33 -1.3012 49.3 41.7 7.6013 54 53.6 0.40

3.653.19

=AVERAGE(D2:D14)=STDEV(D2:D14)

=IF(G2=2,TINV(G3,G6-1),TINV(G3*2,G6-1))

=COUNT(D2:D14)=AVERAGE(D2:D14)=STDEV(D2:D14)

=(G7-G5)/(G8/SQRT(G6))=TDIST(ABS(J10),G6-1,G2)


29

TSB Problem RevisitedTSB Simulation Analysis Results

$32,500

$32,600

$32,700

$32,800

$32,900

$33,000

$33,100

$33,200

$33,300

$33,400

$33,500

$1000 $1250 $1500 $1750 $2000 $2250 $2500 $2750 $3000 $3250

Amount Put Into TSB Account

Me

an

Ne

t In

co

me


30

Independent Samples

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1011121314151617

A B C D E F G H I J KTSB Amount (Decision Variable) 1,000$ 1,250$ 1,500$ 1,750$ 2,000$ 2,250$ 2,500$ 2,750$ 3,000$ 3,250$

Annual Salary 50,000$ 50,000$ 50,000$ 50,000$ 50,000$ 50,000$ 50,000$ 50,000$ 50,000$ 50,000$ Tax Rate 30% 30% 30% 30% 30% 30% 30% 30% 30% 30%After TSB Income 49,000$ 48,750$ 48,500$ 48,250$ 48,000$ 47,750$ 47,500$ 47,250$ 47,000$ 46,750$ Taxes Owed 14,700$ 14,625$ 14,550$ 14,475$ 14,400$ 14,325$ 14,250$ 14,175$ 14,100$ 14,025$ Net Income Before Medical Expenses 34,300$ 34,125$ 33,950$ 33,775$ 33,600$ 33,425$ 33,250$ 33,075$ 32,900$ 32,725$

Total Medical Expenses 2,675.05$ Amount in TSB 1,000.00$ 1,250.00$ 1,500.00$ 1,750.00$ 2,000.00$ 2,250.00$ 2,500.00$ 2,750.00$ 3,000.00$ 3,250.00$ Expenses Not Covered (Must Be Paid Out-Of-Pocket) 1,675.05$ 1,425.05$ 1,175.05$ 925.05$ 675.05$ 425.05$ 175.05$ -$ -$ -$ Money Left Over in TSB (Lost) -$ -$ -$ -$ -$ -$ -$ 74.95$ 324.95$ 574.95$

Net Income After Medical Expenses (Objective) 32,624.95$ 32,699.95$ 32,774.95$ 32,849.95$ 32,924.95$ 32,999.95$ 33,074.95$ 33,075.00$ 32,900.00$ 32,725.00$

Mean 2,000.00$ Standard Deviation 500.00$


31

1234567891011121314

A B C D E F G H I J KStatistics $1000 $1250 $1500 $1750 $2000 $2250 $2500 $2750 $3000 $3250

Trials 10000 10000 10000 10000 10000 10000 10000 10000 10000 10000Mean $33,298.70 $33,362.99 $33,409.96 $33,427.00 $33,401.28 $33,326.98 $33,209.51 $33,061.48 $32,895.94 $32,723.82 Median $33,302.16 $33,377.16 $33,452.16 $33,527.16 $33,600.00 $33,425.00 $33,250.00 $33,075.00 $32,900.00 $32,725.00 Mode $34,300.00 $34,125.00 $33,950.00 $33,775.00 $33,600.00 $33,425.00 $33,250.00 $33,075.00 $32,900.00 $32,725.00 Standard Deviation $490.65 $471.50 $432.23 $369.89 $289.49 $203.64 $128.08 $73.17 $39.85 $20.85 Variance $240,739.17 $222,316.85 $186,824.96 $136,817.78 $83,805.39 $41,470.22 $16,404.53 $5,353.44 $1,587.82 $434.85 Skewness -0.11 -0.28 -0.57 -1.01 -1.64 -2.64 -4.39 -7.81 -13.77 -23.28Kurtosis 2.75 2.64 2.77 3.51 5.58 11.08 27.13 79.33 232.03 623.10Coeff. of Variability 0.01 0.01 0.01 0.01 0.01 0.01 0.00 0.00 0.00 0.00Minimum $31,268.67 $31,343.67 $31,418.67 $31,493.67 $31,568.67 $31,643.67 $31,718.67 $31,793.67 $31,868.67 $31,943.67 Maximum $34,300.00 $34,125.00 $33,950.00 $33,775.00 $33,600.00 $33,425.00 $33,250.00 $33,075.00 $32,900.00 $32,725.00 Range Width $3,031.33 $2,781.33 $2,531.33 $2,281.33 $2,031.33 $1,781.33 $1,531.33 $1,281.33 $1,031.33 $781.33 Mean Std. Error $4.91 $4.72 $4.32 $3.70 $2.89 $2.04 $1.28 $0.73 $0.40 $0.21


32

123456789

10111213141516171819

A B C D E F G H IStatistics $1000 $1250 $1500 $1750 $2000 $2250 $2500 $2750

Trials 10000 10000 10000 10000 10000 10000 10000 10000Mean $33,298.70 $33,362.99 $33,409.96 $33,427.00 $33,401.28 $33,326.98 $33,209.51 $33,061.48 Median $33,302.16 $33,377.16 $33,452.16 $33,527.16 $33,600.00 $33,425.00 $33,250.00 $33,075.00 Mode $34,300.00 $34,125.00 $33,950.00 $33,775.00 $33,600.00 $33,425.00 $33,250.00 $33,075.00 Standard Deviation $490.65 $471.50 $432.23 $369.89 $289.49 $203.64 $128.08 $73.17 Variance $240,739.17 $222,316.85 $186,824.96 $136,817.78 $83,805.39 $41,470.22 $16,404.53 $5,353.44 Skewness -0.11 -0.28 -0.57 -1.01 -1.64 -2.64 -4.39 -7.81Kurtosis 2.75 2.64 2.77 3.51 5.58 11.08 27.13 79.33Coeff. of Variability 0.01 0.01 0.01 0.01 0.01 0.01 0.00 0.00Minimum $31,268.67 $31,343.67 $31,418.67 $31,493.67 $31,568.67 $31,643.67 $31,718.67 $31,793.67 Maximum $34,300.00 $34,125.00 $33,950.00 $33,775.00 $33,600.00 $33,425.00 $33,250.00 $33,075.00 Range Width $3,031.33 $2,781.33 $2,531.33 $2,281.33 $2,031.33 $1,781.33 $1,531.33 $1,281.33 Mean Std. Error $4.91 $4.72 $4.32 $3.70 $2.89 $2.04 $1.28 $0.73

Test Stat 2.996p-Values 1-Tail 0.00136872870

2-Tail 0.00273745741

=(E3-D3)/SQRT(((E7)/(E2))+((D7)/(D2)))=1-NORMSDIST(E17)=2*(1-(NORMSDIST(E17)))


33

Matched Pairs

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1011121314151617181920

A B C D E FTSB Amount (Decision Variable) 1,000$ 1,250$ 1,500$ 1,750$ 2,000$

Annual Salary 50,000$ 50,000$ 50,000$ 50,000$ 50,000$ Tax Rate 30% 30% 30% 30% 30%After TSB Income 49,000$ 48,750$ 48,500$ 48,250$ 48,000$ Taxes Owed 14,700$ 14,625$ 14,550$ 14,475$ 14,400$ Net Income Before Medical Expenses 34,300$ 34,125$ 33,950$ 33,775$ 33,600$

Total Medical Expenses 2,675.05$ Amount in TSB 1,000.00$ 1,250.00$ 1,500.00$ 1,750.00$ 2,000.00$ Expenses Not Covered (Must Be Paid Out-Of-Pocket) 1,675.05$ 1,425.05$ 1,175.05$ 925.05$ 675.05$ Money Left Over in TSB (Lost) -$ -$ -$ -$ -$

Net Income After Medical Expenses (Objective) 32,624.95$ 32,699.95$ 32,774.95$ 32,849.95$ 32,924.95$

Mean 2,000.00$ Standard Deviation 500.00$

Difference 75.00$ (1750 vs 1500)

=E14-D14


34


35

12345678910111213141516171819

A B C D E F GStatistics $1500 $1750 Difference

Trials 10000 10000 10000Base Case $32,774.95 $32,849.95 $75.00Mean $33,408.09 $33,426.42 $16.68Median $33,447.04 $33,522.04 $75.00Mode $33,950.00 $33,775.00 $75.00Standard Deviation $432.25 $371.37 $98.23Variance $186,836.67 $137,917.72 $9,648.54Skewness -0.5887 -1.02 -1.25Kurtosis 2.78 3.49 2.74Coeff. of Variation 0.0129 0.0111 5.89Minimum $31,506.74 $31,581.74 $(175.00)Maximum $33,950.00 $33,775.00 $75.00Range Width $2,443.26 $2,193.26 $250.00Mean Std. Error $4.32 $3.71 $0.98

Test Stat 16.978056970.00000000.0000000

=D4/D15=1-NORMSDIST(D17)=2*(1-NORMSDIST(D17))


36

Goodness-of-Fit Tests

• Determine whether a set of sample data have been drawn from a hypothetical population

• Same four basic steps as other hypothesis tests we have learned

• An important tool for simulation modeling; used in defining random variable inputs

37

Example: Barkevious Mingo

Financial analyst Barkevious Mingo wants to run a simulation model that includes the assumption that the daily volume of a specific type of futures contract traded at U.S. commodities exchanges (represented by the random variable X) is normally distributed with a mean of 152 million contracts and a standard deviation of 32 million contracts. (This assumption is based on the conclusion of a study conducted in 2013.) Barkevious wants to determine whether this assumption is still valid.


38

He studies the trading volume of these contracts for 50 days, and observes the following results (in millions of contracts traded):

142.4 207.5 129.9 84.2 149.3 105.8 152.9 141.5 135.6 205.2 111.1 82.1 97.9 133.8 135.2 124.9 141.7 140.2 215.1 100.4 159.8 144.5 92.9 139.1 173.6 103.3 222.2 195.0 179.7 169.2 192.8 187.0 120.7 156.3 139.8 140.4 96.2 149.3 228.0 180.9 190.3 117.2 127.2 140.3 176.2 151.0 128.4 146.0 131.0 213.4


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Bin Observed Frequency

z-value at Bin Upper Limit

Area under Standard Normal Curve

Expected Frequency out of 50 Observations

0-25 0 -3.969 0.0000 0.00 25-50 0 -3.188 0.0007 0.03 50-75 0 -2.406 0.0073 0.37 75-100 5 -1.625 0.0440 2.20 100-125 7 -0.844 0.1473 7.37 125-150 19 -0.063 0.2757 13.78 150-175 6 0.719 0.2888 14.44 175-200 7 1.500 0.1693 8.47 200-225 5 2.281 0.0555 2.78 225-250 1 3.063 0.0102 0.51 250-275 0 3.844 0.0010 0.05 275-300 0 4.625 0.0001 0.00 300-325 0 5.406 0.0000 0.00


40

Here is a histogram showing the theoretical distribution of 50 observations drawn from a normal distribution with μ = 152 and σ = 32, together with a histogram of Mingo’s sample data:

"Eyeball" Hypothesis Test: Expected Distribution

0

5

10

15

20

0-25 25-50 50-75 75-100 100-125 125-150 150-175 175-200 200-225 225-250 250-275 275-300 300-325

Number of Contracts Traded

Fr

eq

ue

nc

y

"Eyeball" Hypothesis Test: Observed Distribution

0

5

10

15

20

0-25 25-50 50-75 75-100 100-125 125-150 150-175 175-200 200-225 225-250 250-275 275-300 300-325

Number of Contracts Traded

Fr

eq

ue

nc

y


41

2

e

eo

f

ff 2

of = the observed frequency of data in a specific range

ef = the expected frequency of data in a specific range

The Chi-Square Statistic


42

Essentially, this statistic allows us to compare the distribution of a sample with some expected distribution, in standardized terms. It is a measure of how much a sample differs from some proposed distribution.

A large value of chi-square suggests that the two distributions are not very similar; a small value suggests that they “fit” each other quite well.


43

Like Student’s t, the distribution of chi-square depends on degrees of freedom.

In the case of chi-square, the number of degrees of freedom is equal to the number of classes (a.k.a. “bins” into which the data have been grouped) minus one, minus the number of estimated parameters.


44

H ere are grap h s sh ow in g th e ch i -sq u are d istrib u tion for sev eral d iff eren t n u m b ers of d egrees of f reed om :

0 .0 0 0

0 .0 0 5

0 .0 1 0

0 .0 1 5

0 .0 2 0

0 5 1 0 1 5 2 0 2 5 3 0 3 5 4 0 4 5 5 0

C h i - S q u a r e S t a t i s t i c

Proba

bility

0 .0 0 0

0 .0 0 5

0 .0 1 0

0 .0 1 5

0 .0 2 0

0 5 1 0 1 5 2 0 2 5 3 0 3 5 4 0 4 5 5 0


Proba

bility

C h i -S q u are D i strib u tion , d.f. = 5 C h i -S q u are D i strib u tion , d.f. = 10

0 .0 0 0

0 .0 0 5

0 .0 1 0

0 .0 1 5

0 .0 2 0

0 5 1 0 1 5 2 0 2 5 3 0 3 5 4 0 4 5 5 0


Proba

bility

0 .0 0 0

0 .0 0 5

0 .0 1 0

0 .0 1 5

0 .0 2 0

0 5 1 0 1 5 2 0 2 5 3 0 3 5 4 0 4 5 5 0


Proba

bility

C h i -S q u are D i strib u tion , d.f. = 15 C h i -S q u are D i strib u tion , d.f. = 20


45


z-value at Bin Upper Limit

Area under Standard Normal Curve

Expected Frequency out of 50 Observations

0-25 0 -3.969 0.0000 0.00 25-50 0 -3.188 0.0007 0.03 50-75 0 -2.406 0.0073 0.37 75-100 5 -1.625 0.0440 2.20 100-125 7 -0.844 0.1473 7.37 125-150 19 -0.063 0.2757 13.78 150-175 6 0.719 0.2888 14.44 175-200 7 1.500 0.1693 8.47 200-225 5 2.281 0.0555 2.78 225-250 1 3.063 0.0102 0.51 250-275 0 3.844 0.0010 0.05 275-300 0 4.625 0.0001 0.00 300-325 0 5.406 0.0000 0.00


46

Note: It is necessary to have a sufficiently large sample so that each class has an expected frequency of at least 5. We need to make sure that the expected frequency in each bin is at least 5, so we “collapse” some of the bins, as shown here.


z-value at Bin Upper

Limit

Area under Standard

Normal Curve

Expected Frequency out of 50

Observations 0-125 12 -0.844 0.1994 9.97

125-150 19 -0.063 0.2757 13.78 150-175 6 0.719 0.2888 14.44 175-325 13 5.406 0.2361 11.81


47

The number of degrees of freedom is equal to the number of bins minus one, minus the number of estimated parameters. We have not estimated any parameters, so we have d.f. = 4 – 1 – 0 = 3.

The critical chi-square value can be found either by using a chi-square table or by using the Excel function:

=CHIINV(alpha, d.f.) = CHIINV(0.05, 3) = 7.815

We will reject the null hypothesis if the test statistic is greater than 7.815.


48

Bin Observed Frequency Expected Frequency out of 50 Observations e

eo

f

ff 2

0-125 12 9.97 0.413 125-150 19 13.78 1.974 150-175 6 14.44 4.932 175-325 13 11.81 0.120 Chi-Square = 7.439

Our test statistic is not greater than the critical value; we cannot reject the null hypothesis at the 0.05 level of significance.

It would appear that Barkevious is justified in using the normal distribution with μ = 152 and σ = 32 to model futures contract trading volume in his simulation.


49

0.000

0.005

0.010

0.015

0.020

0.025

0.030

0 2 4 6 8 10 12 14 16 18 20

Critical Value of Chi-Square = 7.815Test Statistic = 7.439

The p-value of this test has the same interpretation as in any other hypothesis test, namely that it is the smallest level of alpha at which H0 could be rejected. In this case, we calculate the p-value using the Excel function:

= CHIDIST(test stat, d.f.) = CHIDIST(7.439,3) = 0.0591


50

Example: Catalog Company

If we want to simulate the queueing system at this company, what distributions should we use for the arrival and service processes?


51

Arrivals2425262728293031323334353637383940414243444546

H I J K L M N O PObserved Expected

0-2 0 20 58.43 80.00 21.572-4 2 17 42.68 58.43 15.754-6 4 10 31.18 42.68 11.516-8 6 12 22.77 31.18 8.408-10 8 2 16.63 22.77 6.1410-12 10 6 12.15 16.63 4.4812-14 12 5 8.87 12.15 3.2814-16 14 1 6.48 8.87 2.3916-18 16 2 4.73 6.48 1.7518-20 18 2 3.46 4.73 1.2820-22 20 1 2.53 3.46 0.9322-24 22 1 1.84 2.53 0.6824-26 24 0 1.35 1.84 0.5026-28 26 0 0.98 1.35 0.3628-30 28 0 0.72 0.98 0.2730-32 30 1 0.53 0.72 0.1932-34 32 0 0.38 0.53 0.1434-36 34 0 0.28 0.38 0.1036-38 36 0 0.20 0.28 0.0838-40 38 0 0.15 0.20 0.0640-42 40 0 0.11 0.15 0.04

42

=80*EXP(-$J$5*G28)=80*EXP(-$J$5*G28)

=M30-L30


52

Arrival Rate Analysis

0

5

10

15

20

25

0-2 2-4 4-6 6-8 8-10 10-12 12-14 14-16 16-18 18-20 20-22 22-24 24-26 26-28 28-30 30-32

Interarrival Times (Minutes)

Fre

qu

ency

(80

Arr

ival

s)

Observed

Expected


53

25262728293031323334

Q R S T U V W X Y ZObserved Expected

0-2 20 58.43374 80 21.56626 0.1137512-4 17 42.68127 58.43374 15.75247 0.09884-6 10 31.17533 42.68127 11.50594 0.1971046-8 12 22.77113 31.17533 8.404191 1.5384998-10 2 16.63253 22.77113 6.138604 2.79021810-14 11 12.14876 16.63253 7.758812 1.35398314-32 8 6.481557 8.873719 8.76454 0.066692

6.159046

=(R26-V26)^2/V26

=SUM(W26:W32)


54

2526272829303132333435363738

Q R S T U V WObserved Expected

0-2 20 58.43374 80 21.56626 0.1137512-4 17 42.68127 58.43374 15.75247 0.09884-6 10 31.17533 42.68127 11.50594 0.1971046-8 12 22.77113 31.17533 8.404191 1.5384998-10 2 16.63253 22.77113 6.138604 2.79021810-14 11 12.14876 16.63253 7.758812 1.35398314-32 8 6.481557 8.873719 8.76454 0.066692

6.159046d.f. 6alpha 0.05critical value 12.5916test stat 6.1590p-value 0.4056

=CHIINV(S35,S34)=W33=CHIDIST(S37,S34)


55

Goodness of Fit Test for Arrivals

0.0000

0.0020

0.0040

0.0060

0.0080

0.0100

0.0120

0.0140

0.0160

0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 42 44 46 48 50

Chi Square

Pro

bab

ilit

y

Test Statistic = 6.159

Critical Value = 12.59

Area Under the Curve > 6.159 = 0.4056


56

ServicesService Rate Analysis

0

2

4

6

8

10

12

14

16

18

0-2 2-4 4-6 6-8 8-10 10-12 12-14 14-16 16-18 18-20 20-22 22-24 24-26 26-28 28-30 30-32 32-34 34-36 36-38 38-40 40-42

Interarrival Times (Minutes)

Fre

qu

ency

(80

Arr

ival

s)

Observed

Expected


57

Goodness of Fit Test for Services

0.0000

0.0020

0.0040

0.0060

0.0080

0.0100

0.0120

0.0140

0.0160

0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 42 44 46 48 50

Chi Square

Pro

bab

ilit

y

Test Statistic = 47.79

Critical Value = 11.07

Area Under the Curve > 47.79 = 0.0000


142.4 207.5 129.9 84.2 149.3 105.8 152.9 141.5 135.6 205.2

111.1 82.1 97.9 133.8 135.2 124.9 141.7 140.2 215.1 100.4

159.8 144.5 92.9 139.1 173.6 103.3 222.2 195.0 179.7 169.2

192.8 187.0 120.7 156.3 139.8 140.4 96.2 149.3 228.0 180.9

190.3 117.2 127.2 140.3 176.2 151.0 128.4 146.0 131.0 213.4



59


60


61


62

1

2

3

4

5

6

7

8

9

A B C D

142.4

111.1

159.8

192.8

190.3 0

207.5

82.1

144.5

187.0

63

Other uses for the Chi-Square statistic

• Tests of the independence of two qualitative population variables.

• Tests of the equality or inequality of more than two population proportions.

• Inferences about a population variance, including the estimation of a confidence interval for a population variance from sample data.

The chi-square technique can often be employed for purposes of estimation or hypothesis testing when the z or t statistics are not appropriate. In addition to the goodness-of-fit application described above, there are at least three other important uses for chi-square:



64

SummaryHypothesis Testing • Review of the Basics

– Single Parameter– Differences Between Two Parameters

• Independent Samples• Matched Pairs

– Goodness of Fit• Simulation Methods

session 8b. decision models -- prof. juran2 overview hypothesis testing review of the basics...

Documents

decision models

decision rule

correct decision

test statistic slide

juran4 slide

juran12 slide

juran14 slide

b slide