Name: ________________________ Program&Year: ________________Score: _____

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Concept Notes Laboratory Individual Quiz Formative Summative

Learning Activity Sheet #9

Lesson/Topic: One-way Analysis of Variance

Learning References:
 Sirug, W.S. (2011). Basic Probability and Statistics. A Step by Step
Approach. Mindshapers Co., Inc: Intramuros, Manila.
 Tattao, L.A. (2007). Basic concepts in Statistics (Worktext). Rex Book
Store: Manila Philippines.
 Mendenhall, W., Beaver, R., Beaver, B. (2013). Introduction to Probability
and Statistics (14th Ed). Brooks/Cole, Cengage Learning: Canada.

Learning Target(s):
 To perform analysis of variance.
 To appropriately interpret results in the analysis of variance.
 To conduct a test for one-way analysis of variance.

1. To Engage
You have exposed already to both z test and t test in our previous lesson, this time
you will be exposed to another statistical tool or technique. This technique extends t test
and z test which have engage only on nominal level variable to have two categories. Why
is Anova analysis of variance? Who did develop this technique?
2. To Explore
ANOVA, also known as Fishers Analysis of Variance was first developed by Ronald
Fisher in 1918. when we are conducting ANOVA, we are attempting to determine if
there is a significant difference among the groups.

3. To Explain
 This module will discuss hypothesis testing steps wherein a specific statement is
generated about a population parameter, and sample statistics are used to assess the
likelihood that the hypothesis is true.

4. To Elaborate
Analysis of Variance (ANOVA) allows researchers to compare two or more
populations of interval or ratio data. It is extremely powerful and commonly used
statistical procedure. The ANOVA technique determines whether differences exist
between population means.

One-way Analysis of Variance

The analysis of variance is an extension or generalization of the difference of
means test. The technique is used to test three or several means, however, the test itself
involves working directly with the variances rather than means and standard errors
(Tattao, 2007).

Figure 1 below shows the sampling process for drawing independent samples. For
each population, we draw independent random sample, where we compute the sample
mean and the sample variance.

FIGURE 1. SAMPLING SCHEME FOR INDEPENDENT SAMPLES

(SIRUG, 2011)

Population N1 Population N2 Population Nk

Mean=µ1 Mean=µ2 Mean=µk

------------
Variance= Variance= Variance=

Sample size n1 Sample size n1 Population N1

Mean=µ1
Mean= Mean=
------------
Variance=

FIGURE 1. NOTATION FOR THE ONE-WAY ANALYSIS OF VARIANCE

Treatment
 1 2 … j … c
X1,1 X1,2 … X1,j … X1,c
X2,1 X2,2 … X2,j … X2,c
X3,1 X3,2 … X3,j … X3,c
: : … : … :
Xn1,1 Xn1,2 … Xnj,3 … Xnc,c
Sample Size n1 n2 … nj … nc
Sample Mean …

Table above presents the process of performing the analysis of variance. The
variable X is called the response variable, and its value refers to responses. The unit
that we intend to measure is called an experimental unit. The criterion by which we
classify the population is called a factor and each population is called factor level.

The analysis of variance test whether there is enough statistical evidence to show
that the null hypothesis is false. If the null hypothesis is true, the population means
would be equal or we would expect that the sample means are close to one another.

Assumptions of One-Way Analysis of Variance

1. The samples are randomly selected and independently assigned to groups.
2. Populations should have approximately equal standard deviation (homogeneous)
3. Population distributions are normal.
4. The level of measurement involved is at least an interval scale.

Procedures for One-way Analysis of Variance

1. Set up the hypotheses:

H1: H0 is false ( or At least two means differ).

2. Set the level of Significance.

3. Calculate the means of each group, grand mean, sum of squares, and mean squares.

, where n= n1 + n2 + . . . + nk is the number of populations.

 Calculate the value of the F-test formula

4. Calculate the degrees of freedoms and determine the critical value of F.

5. Statistical decisions for hypothesis testing:

If , do not reject H0.

If , reject H0.

6. State the conclusion.

A. Total Sum of Squares (SST)

The statistic that measures the total variations of all the data is called the total sum
of squares. The total sum of squares is based on the partitioning of the sum of
squares,denoted by SST. (Sirug, 2011)

SST= SSB + SSW or

where SST = total sum of squares (or total variations).

SSB= sum of squares between groups.
SSW= sum of sqaures within groups.

observation in group j.

= number of obswrvationns in group j.

c = number of groups (or treatment)

B. Sum of Squares Between Groups (SSB)

The statistic that measure the variation attitude to the differences between the
treatment means is called sum of squares between groups (or between -treatment
variations), denoted by SSB.

where: SSB = sum of squares betwen groups.

= grand mean.
 = mean of group j.

= number of observations in group j.

c = number of groups (or treatment)

C. Sum of Squares Within Groups (SSW)

The sum of squares within groups (or within- treatment variations/error)
measure the variation within the sample, denoted by SSW.

where: SSW = sum of squares betwen groups.

observation in group j.

= mean of group j.

= number of observations in group j.

c = number of groups (or treatment)

D. Mean Squares Between Groups (MSB)

The mean square between groups (or mean square for treatment) is computed
by dividing SSB by the number of groups minus 1.

where: MSB = mean square between groups.

SSB = sum of squares between groups.

= degrees of freedom between groups.

c = number f groups (or treatment)
E. Mean Square Within Groups (MSW)
The mean square within groups (or mean square for error) is determined by
dividing SSW by the total number of sample size (labeled n) minus the number of
groups.

where MSW = mean square within groups.

SSW = sum of squares within groups.
c = number of groups (or treatment).
F. Test Statistics
The test statistic is defined as the ratio of the two mean squares.

or

where MSB= mean square between groups.

MSW= mean square within groups.

F= (or F ratio).
The degrees of freedom from this F test are df bg =c-1 and dfwg =n-c. the sample sizes
need not be equal in all groups. The F test to compare means is always right-tailed.

The results of the analysis of variance are usually reported in an analysis of variance
table, table below shows the general organization of the One- way ANOVA table.

Source of Sum of Squares Degrees of Mean Square F

Variations Freedom
Between Groups
(Treatment)
Within Groups
(Error)
Total

(This is taken from Sirug, 2011)

Example 1: A concern consumer organization was interested in determining whether
any difference existed in the average life of four different brands of fluorescent bulbs. A
random sample of five fluorescent bulbs of each brand was tested with the following
results (in months):
Brand A Brand B Brand C Brand D
12 11 12 12
13 10 11 15
14 13 15 10
11 15 16 12
15 14 16 11
 At the 0.05 level of significance, is there evidence of significant difference in the
average life of these four brands of fluorescent bulbs?
Solution:
Step 1. State the hypotheses

(All the flourescent bulbs have equal life time means).

(Not all the flourescent bulbs have equal life time means).

Step 2. The level of significance is α=0.05

Step 3. Determine the degrees of freedom and the critical value of F (refer to table
below)

, ,

Table : F DISTRIBUTION

F-test Numerator Degrees of Freedom (c-1)

1 2 3 4
Denominator 1 161 200 216 …
Degrees of
2 18.51 19.00 19.16 …
Freedom (n-
c) 3 10.13 9.55 9.28 Fcritical
…
: : : : …
16 4.49 3.63 3.24 …
: : : : …

Step 4. Compute for the value of F test.

No. Brand A Brand B Brand C Brand D
1 12 11 12 12
2 13 10 11 15
3 14 13 15 10
4 11 15 15 12
5 15 14 16 11
Total 65 63 69 60

We have to compute for the means of all the groups (Brand A,B,C and D) and the
grand mean.
 = mean of group A (Brand A).

= mean of group B (Brand B).

= mean of group C (Brand C).

= mean og group D (Brand D).

= mean of total subjects (Brand A, B, C, D) or grand mean.

After obtaining the means of the data, we have to compute for the SSB, SSW and
SST.

No. XA XB XC XD

1 12 11 12 12 1.00 2.56 3.24 0.00

2 13 10 11 15 0.00 6.76 7.84 9.00
3 14 13 15 10 1.00 0.16 1.44 4.00
4 15 15 15 12 4.00 5.76 1.44 0.00
5 16 14 16 11 4.00 1.96 4.84 1.00
Total 65 63 69 60 10.00 17.20 18.80 14.00
Mea 13 12. 13. 12
n 6 8
 No. XA X XC XD
B

1 12 11 12 12 0.7225 3.4225 0.7225 0.7225

2 13 10 11 15 0.0225 8.1225 3.4225 4.6225
3 14 13 15 10 1.3225 0.0225 4.6225 8.1225
4 15 15 15 12 3.4225 4.6225 4.6225 0.7225
5 16 14 16 11 4.6225 1.3225 9.9225 3.4225
Total 65 63 69 60 10.1125 17.5125 23.3125 17.6125
12.85

or

Source of Variations Sum of Degrees of Mean Square F

Squares Freedom (Variance)
Between Groups 8.55 3 2.85 0.76
WithinGroups 60.00 16 3.75
Total 68.55 19

Step 5. Decision rule.

Since the computed F-value of 0.76 is less than the F critical value of 3.24 at level of
significance of 0.05, the statistical decision is not to reject the null hypothesis.

Step 6. Conclusion
Since the null hypothesis has not been rejected, we can conclude that there is no
evidence that shows significant difference in the average life time of the fluorescent
bulbs.
5. To Evaluate
Read and analyze the problem below. Complete the steps given below by
applying Analysis of Variance (show your solution)
1. A scorporate manager wanted to know whether there was significant difference in the
monthly sales of four representatives. Michael is strictly on salary and small allowance,
Jason is strictly on commission, Mar is on commission and small salary, and John is on
small commission and a salary. Four months were chosen at random. The data represent
monthly sales in units of product.
Michael 18 23 25 29 23
Jason 26 21 23 26 27
Mar 24 29 28 21 26
John 29 24 25 28 22

Using 0.01 level of significance, test the hypothesis that there is no difference in the
mean monthly sales of the sales representatives.
Solution:
Step 1. State the hypotheses.
H0:_______________________________________________________________
______________________________________________________________
H1:_______________________________________________________________
_______________________________________________________________
Step 2.The level of significance is __________.
Step 3. Determine the degrees of freedom and the critical value of F.
dfbg=__________ dfwg=__________ and Fcritical=__________
Step 4. Complete the table and compute for the value of F.
Source of Sum of Degrees of Mean square F
variations squares freedom
Between groups
Within groups
Total

Step 5. Decision rule.

_________________________________________________________________
_________________________________________________________________
Step 6. Conclusion.
_________________________________________________________________
_________________________________________________________________