Single Factor Independent Measures ANOVA

Statistical Methods in Psychology

Homework 8

Single Factor Independent Measures ANOVA

1. Why should you use ANOVA instead of several t-tests to evaluate mean differences when an experiment consists of three or more treatment conditions?

With 3 or more treatment conditions, you need three or more t tests to evaluate all the mean differences. Each test involves a risk of a Type I error (which is equal to the alpha level).  The more tests you do, the more risk there is of a Type I error.  The ANOVA performs all of the tests simultaneously with a single, fixed alpha level.

2. Explain what the purpose of post-hoc tests is.

ANOVA tell us that at least one mean difference is significant. It does not tell us which one, or which ones, are. If we reject the null hypothesis, Post Hoc tests are used to determine exactly which treatment conditions are significantly different from each other.

3. When should you use post- hoc tests?

When you have rejected the null hypothesis, and then have concluded that there is at least one significant mean difference (at least two means are different from each other). And when your IV has 3 or more levels or conditions.

Questions 4 to 8

The following data summarize the results from an independent measures study comparing three treatment conditions.

Use an ANOVA with α = .01 to determine whether there are any significant differences among the three treatment means

Null Hypothesis

There are no significant differences among the three treatment means.

H₀:  μ₁ = μ₂ = μ₃

Alternative Hypothesis

H₁: There is at least one significant mean difference: At least two means significantly differ from each other

Critical F-value

N = 15

k = 3

dfbetween = k – 1 = 3 – 1 = 2

dfwithin = N – k = 15 – 3 = 12

In the F Distribution Table (Table B.4, pages 705 to707) we find the F critical associated with α = .01, and df (2,12): F critical = 6.93

If F-statistic (that we calculate in step 3 below) is greater than 6.93, we reject the null hypothesis.

F-Statistic

SS = 16   SS = 12  SS = 20

N = 15   G = 60    ΣX²= 298

1. Calculate: SS_total , SS_between , and, SS_within

SS_total =   ΣX²  –  G²     

                                       N

SS_total = 298 –  60² = 58

15

SS_within = ∑SS_{inside each condition} = 16 + 12 + 20 = 48

SS_between = SS_total – SS_within

SS_between = 58 – 48 = 10

1. Compute the df_between, df_within , and, df_total

df_between = k – 1 = 3 – 1 = 2

df_within = N – k = 15 – 3 = 12

1. Calculate the mean squares and the F

Source        Sum of squares          df             Mean squares                  F

Between             10            ÷            2       =              5              =           1.25

Within               48             ÷           12      =              4

Total                  58                        14

F-statistic = 1.25              

Your decision is: F = 1.25 is not greater than the F critical = 6.93. Fail to reject the null hypothesis and conclude that there are no significant mean differences among the treatments

Questions 9 to 13

A study surveys student to determine the amount of Facebook use during the time they are doing Math homework. Students are classified into three groups: Non- User, Rarely-Use, Regularly-Use and their Math scores are recorded.  The following data summarize the results.

Use an ANOVA with α = .01 to determine whether there are significant differences among the means of the three groups.

Null Hypothesis

There are no significant differences in math scores means between the three groups of Facebook users (non-user, rarely-use, regularly use).

H₀:  μ₁ = μ₂ = μ₃

Alternative Hypothesis

H₁: There is at least one significant mean difference among the three groups: The math score means of at least two of the groups of Facebook users significantly differ from each other

Critical F-Value

N = 22

k = 3

dfbetween = k – 1 = 3 – 1 = 2

dfwithin = N – k = 22 – 3 = 19

With α = .01

In the F Distribution Table (Table B.4, pages 705 to707) we find the F critical associated with α = .01, and df (2,19): F critical = 5.93

If F-statistic (that we will calculate in step 3 below) is greater than 5.93, we reject the null hypothesis.

F-Statistic

SS = 30   SS = 33  SS = 42

N = 22   G = 72    ΣX²_tot = 393

1. Calculate: SS_total , SS_between , and, SS_within

SS_total =    ΣX² –   G²     

                                           N

SS_total = 393 –  72² = 157.36

22

SS_within = ∑SS_{inside each condition} = 30 + 33 + 42 = 105

SS_between = SS_total – Ss_within

SS_between = 157.36 – 105 = 52.36

1. Compute the df_between, df_within , and, df_total

df_between = k – 1 = 3 – 1 = 2

df_within = N – k = 22 – 3 = 19

1. Calculate the mean squares and the F

Source        Sum of squares          df             Mean squares                  F

Between           52.36            ÷            2       =           26.18               =   4.73

Within               105              ÷           19      =            5.53

Total                 157.36                       21

 

F = 4.73

Your decision is: F = 4.73 is not greater than the F critical = 5.93.

Fail to reject the null hypothesis and conclude that there are no significant mean differences among the three groups regarding Math scores

APA Report

There are no significant mean differences among the three groups regarding Math scores, F (2,19) = 4.73, p > .01

Questions 14 to 19

The following data were obtained from an independent-measures research study comparing three treatment conditions.

Use an ANOVA with α = .05 to determine whether there are any significant mean differences among the treatments.

Null Hypothesis

There are no significant differences among the three treatment means.

H₀:  μ₁ = μ₂ = μ₃

Alternative Hypothesis

H₁: There is at least one mean difference among the three treatment means.

In Anova, we do not state the alternative hypothesis in symbols

Critical F-value

N = 14

k = 3

dfbetween = k – 1 = 3 – 1 = 2

dfwithin = N – k = 14 – 3 = 11

In the F Distribution Table (Table B.4, pages 705 to707) we find the F critical associated with α = .05, and df (2,11): F critical = 3.98

If F-statistic (that we will calculate in step 3 below) is greater than 3.98, we reject the null hypothesis.

F-Statistic

SS = 14   SS = 9  SS = 10

N = 14   G = 42    ΣX² = 182

1. Calculate: SS_total , SS_between , and, SS_within

SS_total = ΣX² –   G²     

                                        N

 

SS_total = 182 –  42² = 56

14

SS_within = ∑SS_{inside each condition} = 14 + 9 + 10 = 33

SS_between = SS_total – Ss_within

SS_between = 56 – 33 = 23

1. Compute the df_between, df_within , and, df_total

df_between = k – 1 = 3 – 1 = 2

df_within = N – k = 14 – 3 = 11

1. Calculate the mean squares and the F

Source        Sum of squares          df             Mean squares                  F

Between           23             ÷            2       =              11.5                  =    3.83

Within               33           ÷           11         =               3

Total                  56                         13

 

F = 3.83                  

Your decision is: F = 3.83 is not greater than the F critical = 3.98. Fail to reject the null hypothesis and conclude that there are not significant differences among the conditions

Use η²to measure the effect size for this study.

   η²  = SS_between /SS_total

η²  = 23/56 = .41 (41%)

APA report

The ANOVA indicates that there were not significant mean differences among the three groups, F (2,11) = 3.83, p > .05, η²  =.41.