Welcome to Chapter 12 MBA 541

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Welcome to Chapter 12 MBA 541

BENEDICTINE UNIVERSITY

• Two-Sample Tests & Analysis of Variance

• Analysis of Variance

• Chapter 12

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Chapter 12

Please, Read Chapter 12 in

Lind before viewing this presentation.

StatisticalTechniques in

Business &Economics

Lind

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GoalsWhen you have completed this chapter, you will be

able to:

• ONE– List the characteristics of the F distribution.

• TWO– Conduct a test of hypothesis to determine whether the

variances of two populations are equal.• THREE

– Discuss the general idea of analysis of variance.• FOUR

– Organize data into a one-way and a two-way ANOVA table.

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GoalsWhen you have completed this chapter, you will be

able to:

• FIVE– Conduct a test of hypothesis among three or more

treatment means.• SIX

– Develop confidence intervals for the difference in treatment means.

• SEVEN– Define and understand the terms treatments and blocks.

• EIGHT– Conduct a test of hypothesis to determine whether there

is a difference among block means.

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The F Distribution

• There is a “family” of F distributions.– Each member of the family is determined by two

parameters: the numerator degrees of freedom and the denominator degrees of freedom.

• The F distribution is continuous.– Its values range from 0 to ∞.

• The F distribution cannot be negative.

• The F distribution is positively skewed.

• The F distribution is asymptotic. – As the values of x increase, the F curve approaches the

X-axis but never touches it.

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Comparing Two Population Variances

• The test statistic for comparing two variances is given by:

where, s1² and s2² are the sample variances for the two samples. The larger s² is placed in the numerator.

• The degrees of freedom are:n1-1 for the numerator and

n2-1 for the denominator.

• The null hypothesis is rejected if the computed value of the test statistic is greater than the critical value.

21

22

sF

s

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Example 1

• Colin, a stockbroker at Critical Securities, reported that the mean rate of return on a sample of 10 internet stocks was 12.6 percent with a standard deviation of 3.9 percent.

• The mean rate of return on a sample of 8 utility stocks was 10.9 percent with a standard deviation of 3.5 percent.

• At the 0.05 significance level, can Colin conclude that there is more variation in the internet stocks?

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Example 1 (Continued)

• Step 1: State the null and alternate hypotheses.H0: σ1²≤ σ2²

H1: σ1²> σ2²

• Step 2: State the level of significance.The 0.05 significance level is stated in the problem.

• Step 3: Find the appropriate test statistic.The test statistic is the F distribution.

• Step 4: State the decision rule.The null hypothesis is rejected if F is greater than 3.68 or

if p is less than 0.05. There are n1-1 or 9 degrees of freedom in the numerator and n2-1 or 7 degrees of freedom in the denominator.

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Example 1 (Continued)• Step 5: Compute the value of F and make a decision.

• The p(F>1.2416) is 0.3965.

• Because the computed F of 1.2416 is less than the critical F value of 3.68 and the p-value of 0.3965 is greater than the required level of significance of 0.05, the decision is to not reject the null hypothesis.

• There is insufficient evidence to show more variation in the internet stocks.

221

222

3.9sF 1.2416

s 3.5

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The ANOVA Test

• The F distribution is also used for testing whether three or more sample means came from the same or equal populations.

• This technique is called analysis of variance or ANOVA.

• The null and alternate hypotheses for four sample means is given by:

Ho: μ1= μ2= μ3= μ4

H1: The means are not all equal.

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The ANOVA Assumptions

• ANOVA requires the following conditions:

– The sampled populations follow the normal distribution,

– The samples are independent, and

– The populations have equal standard deviations.

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The ANOVA Test

• The F distribution is used as the test statistic for the ANOVA Test via the following equation.

• The degrees of freedom for the F statistic in ANOVA:– If there are k populations being sampled, the numerator

degrees of freedom is k-1, and– If there are a total of n observations, then the denominator

degrees of freedom is n-k.

Estimate of the population var iance based

on the differences among the sample meansF

Estimate of the population var iance based

on the var iation within the samples

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The ANOVA Test

• ANOVA divides the Total Variation into the variation due to the treatment, Treatment Variation, and to the error component, Random Variation.

• In the following table,i stands for the ith observation,

is the overall or grand mean, andk is the number of treatment groups.

Total Variation Treatment Variation Random Variation

GX

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The ANOVA Table

• SST is Treatment Variation• SSE is Random Variation• TSS is Total Variation

Source of Variation

Sum of Squares

Degrees of Freedom

MeanSquare

F

Treatments(k)

SST k-1 SST/(k-1)= MST

Error SSE n-k SSE/(n-k)= MSE

Total TSS n-1

MST

MSE

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Example 2

• Rosenbaum Restaurants specialize in meals for families. Katy Polsby, President, developed a new meat loaf dinner.

• Before making it a part of the regular menu, she decides to test it in several of her restaurants.

• She would like to know if there is a difference in the mean number of dinners sold per day at the Anyor, Loris, and Lander restaurants.

• Use the 0.05 significance level.

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Number of Dinners Sold by RestaurantRestaurant

DayAynor Loris Lander

Day 1 13 10 18

Day 2 12 12 16

Day 3 14 13 17

Day 4 12 11 17

Day 5 17

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Step 1: State the null and alternate hypotheses.H0: μAynor = μLoris = μLander


Step 2: State the level of significance.The 0.05 significance level is stated in the problem.

Step 3: Find the appropriate test statistic.The test statistic is the F distribution.

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• Step 4: State the decision rule.The numerator degrees of freedom, n-k, is equal to

3-1 or 2.The denominator degrees of freedom, n-k, is equal

to 13-3 or 10.The critical value of F at 2 and 10 degrees of

freedom (with α =0.05) is 4.10.The null hypothesis is rejected if F is greater than

4.10 or if p is less than 0.05.

• Step 5: Select the sample, perform the calculations, and make a decision

The ANOVA calculations follow.

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Computation of SSEAynor# sold

SS(Anyor) Loris# sold

SS(Loris) Lander# sold

SS(Lander)

13 (13-12.75)² 10 (10-11.5)² 18 (18-17)²

12 (12-12.75)² 12 (12-11.5)² 16 (16-17)²

14 (14-12.75)² 13 (13-11.5)² 17 (17-17)²

12 (12-12.75)² 11 (11-11.5)² 17 (17-17)²

17 (17-17)²

2.75 5 2

12.75 11.5 17

SSE: 2.75 + 5 + 2 = 9.75 14.00

kX

GX

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Computation of TSSAynor# sold

TSS(Anyor) Loris# sold

TSS(Loris) Lander# sold

TSS(Lander)

13 (13-14)² 10 (10-14)² 18 (18-14)²

12 (12-14)² 12 (12-14)² 16 (16-14)²

14 (14-14)² 13 (13-14)² 17 (17-14)²

12 (12-14)² 11 (11-14)² 17 (17-14)²

17 (17-14)²

9 30 47

TSS: 9 + 30 + 47 = 86 SSE: 2.75 + 5 + 2 = 9.75 14.00GX

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Computation of SSTRestaurant XT SST

Anyor 12.75 4(12.75-14)²

Loris 11.50 4(11.50-14)²

Lander 17.00 4(17.00-14)²

76.25

Shortcut: SST = TSS – SSE

= 86 – 9.75

= 76.25

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ANOVA Table

Source of Variation

Sum of Squares

Degrees of Freedom

Mean Square

F

Treatments

76.25 3 – 1= 2

76.25 / 2= 38.125 38.125

0.975

= 39.103

Error 9.75 13 – 3= 10

9.75 / 10= 0.975

Total 86.00 13 – 1= 12

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• As shown, the computed value for F is 39.103.

• The p(F>39.103) is 0.000018

• Because an F of 39.103 is greater than the critical F value of 4.10, and the p-value of 0.000018 is less than the required level of significance of 0.05, the decision is to reject the null hypothesis and conclude that at least two of the treatment means are not the same.

• The mean number of meals sold at the three locations is not the same.

• The ANOVA tables on the next two slides are from the Minitab and Excel applications.

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Analysis of VarianceSource DF SS MS F PFactor 2 76.250 38.125 39.10 0.000Error 10 9.750 0.975Total 12 86.000 Individual 95% CIs For Mean Based on Pooled StDevLevel N Mean StDev ---------+---------+---------+-------Aynor 4 12.750 0.957 (---*---) Loris 4 11.500 1.291 (---*---) Lander 5 17.000 0.707 (---*---) ---------+---------+---------+-------Pooled StDev = 0.987 12.5 15.0 17.5

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SUMMARY

Groups Count Sum Average Variance

Aynor 4 51 12.75 0.92

Loris 4 46 11.50 1.67

Lander 5 85 17.00 0.50

ANOVA

Source of Variation SS df MS F P-value F crit

Between Groups 76.25 2 38.13 39.10 2E-05 4.10

Within Groups 9.75 10 0.98

Total 86.00 12

Anova: Single Factor

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Inferences About Treatment Means

• When I reject the null hypothesis that the means are equal, I want to know which treatment means differ.

• One of the simplest procedures is through the use of confidence intervals around the difference in treatment means.

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Confidence Interval for theDifference Between Two Means

• The confidence interval is given by the following equation

• Where, t is obtained from the t table with degrees of freedom equal to (n-k), and

• If the confidence interval around the difference in treatment means includes zero, there is not a difference between the treatment means.

1 2

1 2

1 1X X t MSE

n n

MSE SSE / n k

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Example 3

• Determine the 95% confidence interval for the difference in the mean number of meat loaf dinners sold in Lander and Aynor.

• Can Katy conclude that there is a difference between the two restaurants?

1 117 12.75 2.228 0.975

4 5

4.25 1.48 2.77 , 5.73

95% Confidence Interval

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• Because zero is not in the interval, we conclude that this pair of means differs.

• The mean number of meals sold in Aynor is different from Lander.

95% Confidence Interval 2.77 , 5.73

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Two-Factor ANOVA• Sometimes, there are other causes of variation.

• For the two-factor ANOVA, we test whether there is a significant difference between the treatment effect and whether there is a difference in the blocking effect (a second treatment variable).

where r is the number of blocks,Xb is the sample mean of block b, and

XG is the overall or grand mean.

• In the following ANOVA table, all sums of squares are computed as before, with the addition of the SSB.

2

b GSSB r X X

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Two-Factor ANOVA Table

Source of Variation

Sum of Squares Degrees of

Freedom

MeanSquare

F

Treatments(k)

SST k-1 SST/(k-1)= MST

Blocks(b)

SSB b-1 SSB/(b-1)= MSB

Error SSE(TSS-SST-SSB)

(k-1)(b-1)

= MSE

Total TSS n-1

MST

MSE

MSB

MSE SSE

k 1 b 1

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Example 4

• The Bieber Manufacturing Co. operates 24 hours a day, five days a week.

• The workers rotate shifts each week.

• Todd Bieber, the owner, is interested in whether there is a difference in the number of units produced when the employees work on various shifts.

• A sample of five workers is selected and their output recorded on each shift.

• At the 0.05 significance level, can we conclude that there is a difference in the mean production by shift and in the mean production by employee?

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Employee DayOutput

EveningOutput

NightOutput

McCartney 31 25 35

Neary 33 26 33

Schoen 28 24 30

Thompson 30 29 28

Wagner 28 26 27

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Example 4 (Continued)For the Treatment Effect:• Step 1: State the null and alternate hypotheses.

H0: μ1 = μ2 = μ3




• Step 4: State the decision rule.The null hypothesis is rejected if F is greater than 4.46 or if p is less than 0.05. The degrees of freedom are (2,8).

• Step 5: Perform the calculations and make a decision.

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Example 4 (Continued)For the Block Effect:• Step 1: State the null and alternate hypotheses.

H0: μ1 = μ2 = μ3= μ4 = μ5




• Step 4: State the decision rule.The null hypothesis is rejected if F is greater than 3.84 or if p is less than 0.05. The degrees of freedom are (4,8).

• Step 5: Perform the calculations and make a decision.

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Example 4 (Continued)Block Sum of Squares ( Effects of time of day and worker on productivity)

Employee Day Evening Night Employee Average

SSB

McCartney 31 25 35 30.33 3(30.33-28.87)²= 6.42

Neary 33 26 33 30.67 3(30.67-28.87)²= 9.68

Schoen 28 24 30 27.33 3(27.33-28.87)²= 7.08

Thompson 30 29 28 29.00 3(29.00-28.87)²= 0.05

Wagner 28 26 27 27.00 3(27.00-28.87)²= 10.49

SSB = 6.42 + 9.68 + 7.08 + 0.05 + 10.49 = 33.73

Note: XG = 28.87

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Compute the remaining sums of squares as before:

TSS = 139.73SST = 62.53SSE = 43.47 = (139.73 – 62.53 – 33.73)

df (block) = 4 = (b – 1)df (treatment) = 2 = (k – 1)df (error) = 8 = (k – 1 )(b – 1)

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Two-Factor ANOVA Table

Source of Variation

Sum of Squares

Degrees of Freedom

MeanSquare

F

Treatments(k)

62.53 2 62.53 / 2= 31.275

31.27 / 5.43= 5.75

Blocks(b)

33.73 4 33.73 / 4= 8.73

8.43 / 5.43= 1.55

Error 43.47 8 43.47 / 8= 5.43

Total 139.73 14

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Example 4 (Continued)• For the Treatment Effect (time of day):

– Because the computed F of 5.75 is greater than the critical F of 4.10, and the p-value of 0.03 is less than the required level of significance of 0.05, the null hypothesis is rejected.

– There is a significant difference in the mean number of units produced for the different time periods.

• For the Block Effect (different employees):– Because the computed F of 1.55 is less than the critical F

of 3.84, and the p-value of 0.28 is greater than the required level of significance of 0.05, the null hypothesis is not rejected.

– There is no significant difference in the mean number of units produced for the different employees.

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Example 4 (Continued)• Output Using Minitab.

• Two-Way ANOVA: Units versus Worker, Shift

Analysis of Variance for Units

Source DF SS MS F P

Worker 4 33.73 8.43 1.550.276

Shift 2 62.53 31.27 5.750.028

Error 8 43.47 5.43Total 14 139.73

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Example 4 (Continued)• Output Using Excel.

Anova: Two-Factor Without Replication

SUMMARY Count Sum Average VarianceMcCartney 3 91 30.33 25.33Neary 3 92 30.67 16.33Schoen 3 82 27.33 9.33Thompson 3 87 29.00 1.00Wagner 3 81 27.00 1.00

Day 5 150 30.00 4.50Evening 5 130 26.00 3.50Night 5 153 30.60 11.30

ANOVA

Source of Variation SS df MS F P-value F crit

Rows 33.73 4 8.43 1.55 0.28 3.84Columns 62.53 2 31.27 5.75 0.03 4.46Error 43.47 8 5.43

Total 139.73 14

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Welcome to Chapter 12 MBA 541