Topic Anova

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MTE3105 Statistics

Topic 5Analysis of variance (ANOVA)

1.1 Synopsis

In this course, students will revisit the concepts of probability and explore inferential statistics

such as analysis variance (ANOVA) in hypothesis testing.The important of using the appropriate

statistical methods in solving real life problems is emphasized.

1.2 Learning Outcomes

1. Understand the theoretical and empirical (concept) of ANOVA2. Use inferential statistics such as ANOVA in hypothesis testing

3. Calculating ANOVA by hand

4. Calculating ANOVA using EXCEL

1.3 Conceptual Framework

TESTING

HYPHOTESIS

ONE WAY ANOVATWO WAYANOVA CHI-SQUARE

LINEAR

REGRESSION


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1.4 ANOVA

Analysis of variance compares two or more populations of interval data. Specifically, we

are interested in determining whether the differences exist between the population

means. The procedure works by analyzing the sample variance.

1.4.1. Definitions

F-distribution

The ratio of two independent chi-square variables divided by their respective degrees of

freedom. If the population variances are equal, this simplifies to be the ratio of the

sample variances.

Analysis of Variance (ANOVA)A technique used to test a hypothesis concerning the means of three or mor populations.

One-Way Analysis of Variance

Analysis of Variance when there is only one independent variable. The null hypothesis

will be that all population means are equal, the alternative hypothesis is that at least one

mean is different.

Between Group VariationThe variation due to the interaction between the samples, denoted SS(B) for Sum of

Squares Between groups. If the sample means are close to each other (and therefore

the Grand Mean) this will be small. There are k samples involved with one data value for

each sample (the sample mean), so there are k-1 degrees of freedom.

Between Group Variance

The variance due to the interaction between the samples, denoted MS(B) for Mean

Square Between groups. This is the between group variation divided by its degrees offreedom.

Within Group Variation

The variation due to differences within individual samples, denoted SS(W) for Sum of

Squares Within groups. Each sample is considered independently, no interaction


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Interaction Effect

The effect one factor has on the other factor

Main Effect

The effects of the independent variables.

1.4.2.ONE WAY ANOVA

In the analysis of variance, the approach is conceptually similar to the t-test, although

the method differs.When you want to compare more than two means, the ONEWAY

Analysis of Variance (ANOVA) is used. Say, for example you conducted an experiment

in which you compared the effectiveness of three teaching methods in enhancing

reading comprehension.

A One-Way Analysis of Variance is a way to test the equality of three or more means at

one time by using variance.

Assumptions

The populations from which the samples were obtained must be normally or

approximately normally distributed.

The samples must be independent.

The variences of the populations must be equal.

1.4.3 How ANOVA works

ANOVA measures two sources of variation in the data and compares their relative sizes

variation BETWEEN groups

for each data value look at the difference between its group mean and

the overall mean

variation WITHIN groups

for each data value we look at the difference between that value and the

mean of its group

2xxi

2iij xx


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The ANOVA F-statistic is a ratio of the Between Group Variaton divided by the Within

Group Variation

A large F is evidence againstH0, since it indicates that there is more difference between

groups than within groups.

We want to measure the amount of variation due to BETWEEN group variation and

WITHIN group variation

For each data value, we calculate its contribution to:

BETWEEN group variation:

WITHIN group variation:

1.4.4. Example problem using One-way Analysis of Variance

Three groups of students, 5 in each group, were receiving therapy for severe test

anxiety. Group 1 received 5 hours of therapy, group 2 - 10 hours and group 3 - 15

hours. At the end of therapy each subject completed an evaluation of test anxiety (the

dependent variable in the study). Did the amount of therapy have an effect on the level

of test anxiety?

The three groups of students received the following scores on the Test Anxiety Index

(TAI) at the end of treatment.

TAI Scores for Three Groups of Students

Group 1 - 5 hours Group 2 - 10 hours Group 3 - 15 hours

48 55 51

50 52 52

53 53 50

52 55 53

50 53 50

MSE

MSG

Within

BetweenF

2

xxi

2iij xx


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The degrees of freedom between groups is:

dfB = K - 1 = 3 - 1 = 2

Where K is the number of groups.

Next we calculate SSW, the sum of squares within groups.

The degrees of freedom within groups is:

dfW = NT - K = 15 - 3 = 12

Where NT is the total number of subjects.


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Finally, we will calculate SST, the total sum of squares.

As a check SST = SSB + SSW

54.4 = 25.2 + 29.2

We can now calculate MSB, the mean square between groups, MSW, the mean square

within groups, and F, the F ratio.

To test the significance of the F value we obtained, we need to compare it with the

critical F value with an alpha level of .05, 2 degrees of freedom between groups (or

degrees of freedom in the numerator of the F ratio), and 12 degrees of freedom within

groups (or degrees of freedom in the denominator of the F ratio). We can look up the

critical value of F in Appendix Table D of the text book (The 5 percent (Lightface Type)

and 1 percent (Boldface Type) points for the Distribution of F), pages 319-326. Look in

the table under column 2 (2 degrees of freedom for the numerator) and row 12 (12

degrees of freedom for the denominator) and read the non-boldfaced entry (for .05 level)

of 3.88 - this is the critical value for F.


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One way of indicating this critical value of F at the .05 level, with 2 degrees of freedom

between groups and 12 degrees of freedom within groups is

F.05(2,12) = 3.88

When using analysis of variance, it is a common practice to present the results of the

analysis in an analysis of variance table. This table which shows the source of variation,

the sum of squares, the degrees of freedom, the mean squares, and the probability is

sometimes presented in a research article. The analysis of variance table for our

problem would appear as follows:

Analysis of Variance Table

Source of

Variation

Sum of

Squares

Degrees of

Freedom

Mean

SquareF Ratio p

Between

Groups25.20 2 12.60 5.178


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Text Book A Text Book B Text Book C

54 53 4949 56 53

52 57 47

55 51 5048 59 54

With = .05, test if the means of the three populations are equal.

1. State the independent variable and the dependent variable in this study

2. State the assumptions for using a one-way ANOVA

3. State the null hypothesis and the alternative hypothesis

4. Compute SSB, SSw and SST

5. Compute the between and within samples variances

6. Indicate the value of Fcritical.

7. Compute the F value

8. Create and ANOVA table and fill in the above information

9. Describe the conclusion.

Solution:

Text Book A Text Book B Text Book C

54 53 49

49 56 5352 57 47

55 51 5048 59 54

T1 = 258 T2 = 276 T3 = 253

X21 = 13350 X

22 = 15276 X

23 = 12835

n1 = 5 n2 = 5 n3 = 5

1 = 51.6 2 = 55.2 3 = 50.6

1) Independent variable : Text book with three different text books

Dependent variable : scores of mathematics achievement

2) The assumption using one-way ANOVA:

1. The distribution of the populations are normal,

2. The variances of the populations are equal

3. Scores are independent


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4. Samples are independent

5. Samples are random

3) Null Hypothesis, H0 = (the three group mean are equal)

Alternative Hyphotesis, Ha : ( at least one of the means are

unequal)

4) a) Sum of Squares Between Group (SSB)

SSB =

()

()

SSB =()

()

()

()

= 58.5333

b) Sum of Square Within Groups (SSw)

SSw = - =

= 41,461 -()

()

()

= 111.2

c) Sum of Squares Total (SST)

SST = SSB + SSw = 58.5333 + 111.2 = 169.7333

5) Between Group Variance

MSB =

Within Group Variance

MSw =

=

= 9.2667

6) The value of Fcritical

Fcritical = F (0.05,2,12) = 3.89

Decision Rules: Reject Ho if F> 3.89

7) The value of F

F =

=


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8) One-Way ANOVA Table

Sources of Variation Sum ofSquares(SS)

DegreesofFreedom

(df)

MeanSquare(MS)

TestStatisticValue (F)

F critical

Between 2 58.5333 29.26673.16 3.89

Within 12 112.2000 9.2667

Total 14 169.7333

9) Conclusions

F = 3.16, Fcritical = 3.89. Therefore we fail to reject the Ho. The data indicate

that the means

of populations are equal ( F(2,12) = 3.16, = 0.05). The differences

of the three sample means are simply due to sampling errors.

4.6 Using the Excel Spreadsheet Program to Calculate One-Way Analysis of

Variance

The Excel spreadsheet program has a tool to calculate One-Way Analysis of

Variance, which simplifies our computational task considerably. Let's use the same

research problem we already considered, but use the spreadsheet program to do the

calculations.

Research Problem:

Three groups of students, 5 in each group, were receiving therapy for severe test

anxiety. Group 1 received 5 hours of therapy, group 2 - 10 hours and group 3 - 15hours. At the end of therapy each subject completed an evaluation of test anxiety (the

dependent variable in the study). Did the amount of therapy have an effect on the

level of test anxiety?


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In the Excel Worksheet select Data Analysis under the Tools menu. If Data Analysis is

not available you must install the Data Analysis Tools.

If you need to you can install the data analysis tools as follows:

1. Select Add-Ins from the Tools menu.

2. In the Add-Ins window click on the box next to Analysis Tool Pak to select it.

3. Click OK. You have now installed the Tool Pak.

With the Data Analysis Tools installed, select Data Analysis under the Tools menu.

In the Data Analysis window scroll down and select Anova: Single Factor. Complete

the Anova: Single Factorwindow as follows:

1. Enter$A$2:$C$7 in the Input Range: box (or you can enter that value

automatically by clicking in the box and then selecting the range of cells A2

through C7). Note that we have included the labels, Group 1, Group 2, and

Group 3, in the range of cells we selected.

2. Click the Columns button so that we indicate we our data is grouped by

columns.

3. Click the Labels in first row box so that we indicate we are using labels (Group

1, Group 2, and Group 3)

4. Enter.05 in the Alpha: box.

5. UnderOutput Options click the button forOutput range: and enter$A$9 in the

Output range: box (or click in the box and then click on the cell A9 to cause it to

appear in the box).

6. Click OK.


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Your spreadsheet should now appear as follows:

The results of the one-way analysis of variance can be seen in the resultant tables. The

means for the three groups (as well as the count, sum, and variance for each group) can

be seen in the SUMMARYtable.

The ANOVA table shows the same results as we put in the Analysis of Variance table

when we calculated the results ourselves. The value of F is shown to be 5.178082192,

which rounded to 5.18 is the same value as we received when we calculated F. The P-

Value is shown as .02391684 which indicates that the result is significant at the .02 level.

We have set our alpha level as .05 so we will simply indicate that p < .05. There is an

additional entry to the table showing the critical value of F at the .05 level (F Crit) which

is 3.88529031 which is similar to the result (2.88) we looked up in Appendix Table D in

the textbook.

Unfortunately, the spreadsheet program does not have a program to calculate the

Scheffe test, so we will have too calculate those the way we did before. The results of

our Scheffe tests were:


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Summary of Scheffe Test Results

Group One versus Group Two 4.62

Group One versus Group Three 0.18

Group Two versus Group Three 2.96

We now have all the information we need to complete the six step statistical inference

process:

1. State the null hypothesis and the alternative hypothesis based on your

research question.

Note: Our null hypothesis, for the F test, states that there are no differences

among the three means. The alternate hypothesis states that there are significant

differences among some or all of the individual means. An unequivocal way of

stating this is not H0.

2. Set the alpha level.

Note: As usual we will set our alpha level at .05, we have 5 chances in 100 of

making a type I error.

3. Calculate the value of the appropriate statistic. Also indicate the degrees of

freedom for the statistical test if necessary and the results of any post hoc

test, if they were conducted.

F(2,12) = 5.178, value of the F ratio

F.05(2,12) = 3.88, critical value of F

F12 = 4.630, Scheffe test value for comparing means 1 and 2



4. Write the decision rule for rejecting the null hypothesis.

Reject H0 if F is >= 3.88

Note: To write the decision rule we had to know the critical value for F, with an

alpha level of .05, 2 degrees of freedom in the numerator (df between groups)

and 12 degrees of freedom in the denominator (df within groups). We can do this


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by looking at Appendix Table D and noting the tabled value for the .05 level in the

column for 2 df and the row for 12 df.

5. Write a summary statement based on the decision.

Reject H0, p < .05

Note: Since our calculated value of F (5.178) is greater than 3.88, we reject the

null hypothesis and accept the alternative hypothesis.

6. Write a statement of results in standard English.

There is a significant difference among the scores the three groups of students

received on the Test Anxiety Index.

Group 1 (the five hour therapy group) has a significantly lower score on the TAI

than does Group 2 (the ten hour therapy group).

We can see that the Excel spreadsheet program gives us an easy way to calculate the F

ratio. It also provides us with an analysis of variance table which shows, among other

things, the critical value of F for the alpha level we specified, and the probability level (p)

of the result.
http://f/ANOVA/ed602lesson13.htm


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Question :

(1) State the Assumptions of ANOVA.(2) Describe the Rationale of ANOVA stating the ANOVA table.(3) Solve the following problem using One-Way ANOVA

Four types of advertising displays were set up in 12 retail outlets, with three

outlets randomly assigned to each of the displays, for the purpose of studyingthe point-of-sale impact of the displays. The relevant information is given in the

following table.

Type of Display Sales

A1 40 44 43

A2 53 54 59A3 48 38 46

A4 48 61 47

Carry out the Analysis of Variance to test the differences among the mean sales

values for the four types of displays, using the 5 percent level of significance.

(i) State the Null Hypothesis and Alternative Hypothesis.

Give Step by Step solution using all the required formulas. Give the

ANOVA table and comment on the conclusion.

(ii) Use excel to solve the above problem using the data given in the abovetable and comment on the conclusion.

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Topic Anova