Question

To test whether students in a higher grade level will be less disruptive in class, a school psychologist records the number of documented interruptions during one day of classes from nine local high schools. The sample consisted of nine (n = 9) freshman, sophomore, junior, and senior high school classes. The data for each high school class are given in the table. High School Class

Freshman      Sophomore     Junior     Senior

2                       3                    1               5

4                       2                        3              2

1                       2                        5              5

0                       0                      5               2

6                       5                       0               4

3                       4                        4              4

1                        0                       6              3

2                      1                       4               2

4                       4                      3               3

(a) Complete the F-table. (Round your answers to two decimal places.)

Source of Variation      SS     df     MS     F

Between groups            ?        ?        ?        ?

Within groups (error)     ?        ?        ?       

Total                              ?        ?

(b) Is it necessary to compute a post hoc test? Explain. (Assume alpha equal to 0.05.)

Yes, post hoc analyses are appropriate because the ANOVA is significant.

OR

No, post hoc analyses are not appropriate because the ANOVA is not significant.

Answer 1

a)

Null and Alternative Hypothesis:

H₀: µ_Freshman = µ_Sophomore = µ_Junior= µ_Senior

H₁: Not all Means are equal

Alpha = 0.05

N = 36

n = 9

Degress of Freedom:

df_Between = a – 1 = 4-1 =3

df_Within = N-a = 36-4 = 32

df_Total = N-1 = 36-1 = 35

Critical Values:

Time (df_Between, df_Within): (3,32) = 2.9

Decision Rule:

If F is greater than 2.9, reject the null hypothesis

Test Statistics:

SS_Betwen = ∑(∑a_i)²/n - T²/N = 8.31

SS_Within = ∑(Y)² - ∑(∑a_i)²/n = 96.44

SS_Total = SS_{Betwen ­­}+ SS_Within = 104.75

MS = SS/df               

F = MS_effect / MS_error

Hence,

F = 2.77/3.01 = 0.92

Result:

Our F = 0.92, we fail to reject the null hypothesis

Conclusion:

All means are equal

b)

No, post analyses are not appropriate because the ANOVA is not significant