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1 Revised 2015

Donnelly Text New Text Page 510 -530 Donnelly Edition 2: 501-521 Class Notes and Computer Assignment for One Way ANOVA – Balanced Design Lecture Dataset Parachute The following problem examines equal average tensile strength fabric from four suppliers. (Parachute) is utilized to demonstrate the six-step process below. Use these techniques to complete the homework problem. Problem Definition for ANOVA: You are the production manager at Perfect Parachute Company. You wish to determine if the synthetic fibers from your four suppliers result in parachutes of equal strength.

H0: The average tensile strength of the fabric from the four suppliers is equal H1: The average tensile strength of the fabric from the four suppliers is not equal

Data Display Row Supplier1 Supplier2 Supplier3 Supplier4

1 18.5 26.3 20.6 25.4

2 24.0 25.3 25.2 19.9

3 17.2 24.0 20.8 22.6

4 19.9 21.2 24.7 17.5

5 18.0 24.5 22.9 20.4

To begin the ANOVA analysis you must determine if the variances of the four factor levels may be assumed equal. Problem Definition: Test to determine if variances may be assumed equal. Hypothesis for equal variances

H0:  2

4

2

3

2

2

2

1 

H1: At least two variances are not equal Decision Rule: If p-value for tests is larger than alpha accept null.

Equal Variance Output

Test for Equal Variances: Supplier1, Supplier2, Supplier3, Supplier4 Tests

Test

Method Statistic P-Value

Bartlett 0.88 0.829

If variances are not significantly different ANOVA randomized may be used. If significant you must used Welchs T Test for two means. NOW THAT WE HAVE SATISFIED THAT THE VARIANCES ARE EQUAL THE ONE WAY ANOVA CAN BE COMPLETED. NOTE THAT THE BOXPLOT NEEDED TO SATISFY NORMALITY IS BUILT INTO THE ANOVA COMMANDS AND WILL HAVE TO BE SHOWN IN THE REPORT AFTER THE EQUAL VARIANCES HYPOTHESIS IS SATISFIED. Decision Rule: If the F Test Stat or critical ratio is greater that the critical value of 3.24 (.05 distribution) reject the null of equal means.

2 Revised 2015

Using Minitab to Test for Equal Variances in ANOVA Equal Variance or Homogeneity of Variances Test Use the pull down to select the correct arrangement of your data (single columns or multiple columns). Select Options and set the correct confidence level and normal distributions Storage>Fill in as shown Results button>Fill in as shown

3 Revised 2015

ANOVA Test with if variance are determined to equal (not significantly different). Problem Definition: You are the production manager at Perfect Parachute Company. You wish to determine if the synthetic fibers from your four suppliers result in parachutes of equal strength.

H0: The average tensile strength of the fabric from the four suppliers is equal H1: The average tensile strength of the fabric from the four suppliers is not equal

Decision Rule: If the F Test Stat or critical ratio is greater that the critical value of 3.24 (.05 distribution) reject the null of equal means.

ANOVA Command String Minitab STAT>ANOVA >Oneway (Unstacked)>Double click on four column names in left window Set Confidence Level >Graphs >Choose Boxplots >Comparisons choose Tukey Populations from which samples were selected are normally distributed. The test for normality is extremely robust to departures from normality and utilizes a boxplot to analyze each distribution. Minitab automatically generates the boxplot, shown in the screen capture, while computing the ANOVA

Commands will produce ANOVA table and Boxplot

4 Revised 2015

Minitab 17: Stat>ANOVA>Oneway These commands will produce ANOVA Table>Tukey Test>Boxplots Each of these; ANOVA, Tukey, and Boxplots output will have to be placed in the appropriate places within the report. Normality of each distribution select Graphs>Fill in as shown

5 Revised 2015

Tukey Test Minitab 17

Resulting Boxplots Be sure to analyze each plot separately from the others. Place these after equal variance test and describe each for normality assumption. (Yours will be for the assigned problem)

Supplier4Supplier3Supplier2Supplier1

26

24

22

20

18

16

D a ta

Boxplot of Supplier1, Supplier2, ...

6 Revised 2015

Restated problem definition and hypothesis for readers’ convenience Problem Definition: You are the production manager at Perfect Parachute Company. You wish to determine if the synthetic fibers from your four suppliers result in parachutes of equal strength.

H0: The average tensile strength of the fabric from the four suppliers is equal H1: The average tensile strength of the fabric from the four suppliers is not equal

Decision Rule: If the F Test Stat or critical ratio is greater that the critical value of 3.24 (.05 distribution) reject the null of equal means.

Analysis of Variance (Used for Hypothesis testing of equal means)

Source DF Adj SS Adj MS F-Value P-Value

Factor 3 63.29 21.095 3.46 0.041

Error 16 97.50 6.094

Total 19 160.79

Model Summary

S R-sq R-sq(adj) R-sq(pred)

2.46860 39.36% 27.99% 5.25%

Means

Factor N Mean StDev 95% CI

Supplier1 5 19.52 2.69 ( 17.18, 21.86)

Supplier2 5 24.260 1.919 (21.920, 26.600)

Supplier3 5 22.840 2.134 (20.500, 25.180)

Supplier4 5 21.16 2.98 ( 18.82, 23.50)

Pooled StDev = 2.46860

Conclusion: The F critical ratio of 3.46 is greater than the critical value of 3.24 so we reject the null with .05 chance of making a T1 error. You can double check with decision with Pvalue .041 Interpretation: We know that at least two of the group means are not equal so at least two average fabric strengths are significantly different. Since ANOVA can only tell us the means are not equal and cannot specify which of the means are different we must use the TUKEY test.

7 Revised 2015

Minitab 17 Session Window Output Problem Definition for Pairwise Comparisons (Tukey): We will conduct a Tukey test to determine the pairs of means that are the most significantly different. Minitab does not differentiate between balanced and unbalanced designs. It uses the Tukey-Kramer version of the test. Hypotheses: Show correct number of interval hypotheses on homework. These must be typed on your homework.

1=2 1=3 1=4 2=3 2=4 3=4

12 13 14 23 24 34

Decision Rule: For the hypotheses to be rejected the interval must not contain zero. For example means one (1) and two (2) are not equal as the interval has two negative signs.

Formula for Tukey intervals.   



  

 

21

11

2

21

nn

MSW Qxx

One-way ANOVA: Supplier1, Supplier2, Supplier3, Supplier4 Significance level α = 0.05 Equal variances were assumed for the analysis.

Factor Information

Factor Levels Values Factor 4 Supplier1, Supplier2, Supplier3, Supplier4

Tukey Pairwise Comparisons Grouping Information Using the Tukey Method and 95% Confidence

Factor N Mean Grouping

Supplier2 5 24.260 A

Supplier3 5 22.840 A B

Supplier4 5 21.16 A B

Supplier1 5 19.52 B

Means that do not share a letter are significantly different.

In this case means one and 2 are most significantly different

Tukey Simultaneous Tests for Differences of Means

Difference SE of Adjusted

Difference of Levels of Means Difference 95% CI T-Value P-Value

Supplier2 - Supplier1 4.74 1.56 ( 0.27, 9.21) 3.04 0.036

Supplier3 - Supplier1 3.32 1.56 (-1.15, 7.79) 2.13 0.187

Supplier4 - Supplier1 1.64 1.56 (-2.83, 6.11) 1.05 0.723

Supplier3 - Supplier2 -1.42 1.56 (-5.89, 3.05) -0.91 0.800

Supplier4 - Supplier2 -3.10 1.56 (-7.57, 1.37) -1.99 0.234

Supplier4 - Supplier3 -1.68 1.56 (-6.15, 2.79) -1.08 0.708

Individual confidence level = 98.87%

8 Revised 2015

Problem Definition: A well-known conglomerate claims that its detergent “whitens and brightens better than all the rest.” In order to compare the cleansing action of the top three brands of detergents 24 swatches of white cloth are soiled with red wine and grass stains and then washed in front-loading machines with the respective detergents. The following whiteness readings were obtained:

Detergent

Tide Gain All

84 78 87

79 74 80

87 81 91

85 86 77

94 86 78

89 89 79

89 69 77

83 79 78

Step 1: Perform an ANOVA for the detergent analysis (You must enter the data into the Minitab spreadsheet from the problem. *Hint: It will make it much easier for you to interpret the output if you label your columns (Tide, Gain, All)

Hypothesis test of equal means (Six step process) Determine if means are significantly different and if a Tukey Test will be appropriate. 5pts If appropriate utilize Tukey test and six step process to determine significant differences between matched pairs (Minitab. For null; state appropriate number of hypotheses. State decision rule as to whether 0 lies in the interval. Test to find non- zero intervals. Conclude by identifying no zero intervals. Interpret which treatment or treatments are best to use for the client. 5pts Step 2: Satisfy Assumptions for ANOVA test: Assumptions you will need to satisfy for your homework. Randomness and independence may be assumed for this test.

Satisfy normality assumption for each of the four distributions Use boxplots and be sure to provide an explanation. 2 points.

Satisfy equal variances assumption –Use six step process and p-values to satisfy assumption of equal variances. 5 pts