Question

An experiment is performed to compare the average reading scores of second grade children taught using four different methods. Students are randomly assigned to the four groups with 20 students per group. An analysis of variance is to be used to analyze the data. The following is a partially completed ANOVA table. Fill in the remaining details (2 pts per blank – 12 possible points):

SS total = ?

df between = ?  

df within = ?

MS between = ?

MS within = ?

F = ?

Source	Sum of Squares	DF	Mean Square	F
Between	136.21	?	?	?
Error/Within	675.83	?	?
Total	?

1. State the null and alternative hypothesis (4pts)
2. Based on the information, would you reject the null hypothesis at alpha=.05 that all four groups of second grade students will have the same average reading scores (1pt)? what is the critical value of F (1pt)?

Answer 1

Hypothesis:

Ho: 1 = 2 = 3 = 4 .........Same average reading scores

Ha: At least one of the mean score is different

SS_Total = SS_Between + SS_Error

SS_Total = 136.21 + 675.83

SS_Total = 812.04

Four different methods, So t = 4, n = 20

Degrees of Freedom (DF):

df_Between = t-1 = 4-1 = 3

df_Total = n-1 = 20-1 = 19

df_Error = df_Total - df_Treatment = 19-3 = 16

Mean square:

Mean square Between = SS_Between / DF_Between

Mean square Between = 136.21/3

Mean square Between = 45.4

Mean square Error = SS_Error / DF_Error

Mean square Error = 675.83 / 16

Mean square Between = 42.2

F = Mean square Between / Mean square Error

F = 45.4 / 42.2

F = 1.07

ANOVA TABLE:

Source	Sum of squares	DF	Mean square	F
Between	136.21	3	45.4	1.07
Error	675.83	16	42.2
Total	812.04	19

Critical value:

F_tabulate = F_,(t-1,n-t) = F_0.05,(3,16)= 3.2389   ..............Using F table

Conclusion:

Test statistuc (F) < Critical value, i.e. 1.07 < 3.2389, That is Fail to Reject Ho at 5% level of significance.

Therefore, All four groups of second grade students will have the same average reading scores