# STAT-C9

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#### Description

ANALYSIS OF VARIANCE (ANOVA) OR F-TEST

ANALYSIS OF VARIANCE (ANOVA ) or F-TEST The analysis of variance (ANOVA) is used to determine whether there are any significant differences between the means of three or more independent (unrelated) groups. this guide will provide a brief introduction to the one-way ANOVA including the assumptions of the test to interpret the output. The one-way ANOVA compares the means between the groups you are interested in and determines whether any of those means are significantly different from each other.

The Basic ANOVA Situation One Variable: Main Question:

1 Categorical, 1 Quantitative Do the (means of) the quantitative variables depend on which group (given by categorical variable) the individual is in?

If categorical variable Has only two values: 2-sample t-test/z-test ANOVA allows for 3 or more groups

Example of ANOVA situation Subjects: Treatments: Measurement: Data[and means]: Treatment A Treatment B Placebo

25 patients with blisters Treatment A, Treatment B, Placebo Number of days until the blisters heal

: 5,6,6,7,7,8,9,10 : 7,7,8,9,9,10,10,11 : 7,9,9,10,10,10,11,12,13

[7.25] [8.875] [10.11]

One-Way ANOVA There is only one factor being studied as the independent variable. Independent variable may be different experimental conditions, teaching methods, guidance technique, values education approaches, educational attainment, socio economic status or other factors that may have two or more levels.

Steps to follow with One-Way ANOVA 1. State the Hypothesis: 𝑯𝒐: The means of all the groups are equal. 𝑯𝒂: Not all the means are equal. 2. Set the level of significance 𝜶 . 3. Choose the statistical test appropriate to test the hypothesis. 4.Determine the tabular value for the test. 5. Compute the value of the statistical test. 6. Determine the significance of the compound value. 7. Interpret and discuss the result.

Formula in Computing One-Way ANOVA a.

Sum of the Squares 2 𝑥 𝑇𝑆𝑆 = 𝑥2 − 𝑁 1 𝑆𝑆𝑏 = 𝑟 𝑆𝑆𝑏

2

𝑥𝑖𝑗

where: 𝑥𝑖𝑗 = 𝑠𝑢𝑚 𝑜𝑓 𝑒𝑎𝑐ℎ 𝑐𝑜𝑙𝑢𝑚𝑛 𝑥 𝑁

𝑥𝑖1 2 𝑥𝑖2 2 = + +⋯ 𝑛1 𝑛2 2 𝑥𝑖𝑗 𝑥 2 + − 𝑛𝑗 𝑁 𝑆𝑆𝑤 = 𝑇𝑆𝑆 − 𝑆𝑆𝑏

2

𝑥 = Sum of the values of all items 𝑇𝑆𝑆 = 𝑡𝑜𝑡𝑎𝑙 𝑠𝑢𝑚 𝑜𝑓 𝑠𝑞𝑢𝑎𝑟𝑒𝑠 𝑆𝑆𝑏 = 𝑠𝑢𝑚 𝑜𝑓 𝑠𝑞𝑢𝑎𝑟𝑒𝑠 𝑏𝑒𝑡 𝑐𝑜𝑙𝑢𝑚𝑛 𝑆𝑆𝑤 = 𝑠𝑢𝑚 𝑜𝑓 𝑠𝑞𝑢𝑎𝑟𝑒𝑠 𝑤𝑖𝑡ℎ𝑖𝑛 𝑐𝑜𝑙𝑢𝑚𝑛 𝑟 𝑜𝑟 𝑛1 … = 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑖𝑡𝑒𝑚𝑠 𝑝𝑒𝑟 𝑐𝑜𝑙𝑢𝑚𝑛

b. Degrees of Freedom

𝑑𝑓𝑇 = 𝑁 − 1 𝑑𝑓𝑏 = 𝑘 − 1 𝑑𝑓𝑤 = 𝑑𝑓𝑇 − 𝑑𝑓𝑏

where: 𝑑𝑓𝑇 = Total degrees of freedom 𝑑𝑓𝑏 = 𝑑𝑒𝑔𝑟𝑒𝑒𝑠 𝑜𝑓 𝑓𝑟𝑒𝑒𝑑𝑜𝑚 𝑏𝑒𝑡 𝑐𝑜𝑙𝑢𝑚𝑛 𝑑𝑓𝑤 = 𝑑𝑒𝑔𝑟𝑒𝑒𝑠 𝑜𝑓 𝑓𝑟𝑒𝑒𝑑𝑜𝑚 𝑤𝑖𝑡ℎ𝑖𝑛 𝑐𝑜𝑙𝑢𝑚𝑛 𝑘 = 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑐𝑜𝑙𝑢𝑚𝑛𝑠 𝑁 = 𝑡𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑒𝑛𝑡𝑟𝑖𝑒𝑠

c. Mean Sum of Squares 𝑆𝑆𝑏 𝑀𝑆𝑆𝑏 = 𝑑𝑓𝑏 𝑆𝑆𝑤 𝑀𝑆𝑆𝑤 = 𝑑𝑓𝑤 where:

𝑀𝑆𝑆𝑏 = 𝑚𝑒𝑎𝑛 𝑠𝑢𝑚 𝑜𝑓 𝑠𝑞𝑢𝑎𝑟𝑒𝑠 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑐𝑜𝑙𝑢𝑚𝑛 𝑀𝑆𝑆𝑤 = 𝑚𝑒𝑎𝑛 𝑠𝑢𝑚 𝑜𝑓 𝑠𝑞𝑢𝑎𝑟𝑒𝑠 𝑤𝑖𝑡ℎ𝑖𝑛 𝑐𝑜𝑙𝑢𝑚𝑛

d. Locating the Tabular Value and Calculating the Computed Value and Comparing Them 1. Locate the tabular value of 𝐹𝑇 by following the format. 𝑑𝑓𝑏 𝑭𝒕 = → 𝑙𝑜𝑐𝑎𝑡𝑒 (𝑟𝑒𝑓𝑒𝑟 𝑡𝑜 𝑡𝑎𝑏𝑙𝑒) 𝑑𝑓𝑤 2. Calculate 𝐹𝑐 𝑀𝑆𝑆𝑏 𝑭𝒄 = 𝑀𝑆𝑆𝑤 3. Compare the Computed against the tabular value a) Reject Ho is the computed value is greater than or equal to the tabular value. (𝑭𝒄 ≥ 𝑭𝒕 ) b) Do not reject Ho if the computed value is less than the tabular value. (𝑭𝒄 < 𝑭𝒕 )

Table for the One-Way ANOVA Source of Variation

Between Columns

Within Columns

Total

Sum of Squares

Degree of Freedom

Mean Sum of Squares

FT

FC

Example: On the following four groups of teaching attitude, test the null hypothesis that academic performance does not vary due to teaching attitude. Superior

Above Average

Average

Below Average

90 89 88 94 93

85 86 84 83 88

80 82 83 81 80

78 76 75 77 75

Solution: Step 1: Formulate the Hypothesis 𝑯𝒐: The academic performance does not vary due to teaching attitude. 𝑯𝒂: The academic performance vary due to teaching attitude. Step 2: Set Level of Significance 𝛼 = 0.01 𝑜𝑟 0.05 Step 3: Choose the statistical test the apply it ANOVA or F-test

Worksheet table for the One-Way ANOVA Superior 𝑋1

𝑋

90

Above Average 2 1

𝑋2

𝑋

8100

85

89

7921

88

2 2

Average 𝑋3

𝑋

7225

80

86

7396

7744

84

94

8836

93 454

Below Average 2 3

2 4

𝑋4

𝑋

6400

78

6084

82

6724

76

5776

7056

83

6889

75

5625

83

6889

81

6561

77

5929

8649

88

7744

80

6400

75

5625

41250

426

36310

406

32974

381

29039

𝑋𝑇 = 1667

𝑋𝑇 2 = 139573

𝑁𝑇 = 20

Compute Sum of Squares 𝑥2 −

𝑇𝑆𝑆 =

𝑥 𝑁

2

1667 = 139573 − 20

2

= 628.55 2

1 𝑆𝑆𝑏 = 𝑟 1 5

𝑥𝑖𝑗

4542

= = 55.2

+ 4262

𝑥 𝑁

2

+ 4062

+ 3812

1667 2 20

continuation… • Degrees of Freedom 𝑑𝑓𝑇 = 𝑁 − 1 𝑑𝑓𝑏 = 𝑘 − 1 = 20 − 1 =4−1 = 19 =3

𝑑𝑓𝑤 = 𝑑𝑓𝑇 − 𝑑𝑓𝑏 = 19 − 3 = 16

• Mean Sum Squares 𝑀𝑆𝑆𝑏 = =

𝑆𝑆𝑏 𝑑𝑓𝑏 573.35 3

= 191.12

𝑀𝑆𝑆𝑤 = =

𝑆𝑆𝑤 𝑑𝑓𝑤 55.2 16

= 3.45

Step 4: Determine the tabular value

𝑭𝒕 =

𝑑𝑓𝑏 𝑑𝑓𝑤

=

3 ; 16

@0.05 → 3.239; @0.01 → 5.292

Step 5: Compute the value of the statistical test 𝑀𝑆𝑆𝑏 191.2 𝑭𝒄 = = = 55.39 𝑀𝑆𝑆𝑤 3.45 Source of Variation Between Column

Sum of Squares

df

Mean Sum of Squares

573.35

3

191.12

Within Column

55.2

16

3.45

Total

628.55 19

𝑭𝑻

𝑭𝑪

3.239 @0.05 5.292 55.39 @0.01

continuation… 𝑺𝒕𝒆𝒑 𝟔: Determine the significance of the computed value → Since the value of 𝑭𝒄 > 𝑭𝑻 , therefore 𝒓𝒆𝒋𝒆𝒄𝒕 the null hypothesis (𝑯𝒐) accept alternative hypothesis (𝑯𝒂) 𝑺𝒕𝒆𝒑𝟕: Interpret the result Decision: Reject the null hypothesis Interpretation: The academic performance varies due to teaching attitude.

The End! *For Easy and Precise Computation let us use the MS Excel. Answer Exercises 9.1 − 9.5 MAHALO!