welcome to chapter 12 mba 541
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Welcome to Chapter 12 MBA 541. B ENEDICTINE U NIVERSITY Two-Sample Tests & Analysis of Variance Analysis of Variance Chapter 12. Chapter 12. Please, Read Chapter 12 in Lind before viewing this presentation. Statistical Techniques in Business & Economics Lind. Goals. - PowerPoint PPT PresentationTRANSCRIPT
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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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Example 2 (Continued)
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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Example 2 (Continued)
Step 1: State the null and alternate hypotheses.H0: μAynor = μLoris = μLander
H1: The means are not all equal.
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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Example 2 (Continued)
• 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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Example 2 (Continued)
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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Example 2 (Continued)
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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Example 2 (Continued)
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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Example 2 (Continued)
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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Example 2 (Continued)
• 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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Example 2 (Continued)
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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Example 2 (Continued)
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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Example 3 (Continued)
• 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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Example 4 (Continued)
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
H1: The means are not all equal.
• 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 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
H1: The means are not all equal.
• 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 3.84 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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Example 4 (Continued)
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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Example 4 (Continued)
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