Based on the amount of time spent on Facebook, students are classified into 3 groups and their grade point averages are recorded. The following datashow the typical pattern of results.
FACEBOOK USE WHILE STUDYING
NON-USER RARELY USE REGULARLY USE
3.70 3.51 3.02
3.45 3.42 2.84
2.98 3.81 3.42
3.94 3.15 3.10
3.82 3.64 2.74
3.68 3.20 3.22
3.90 2.95 2.58
4.00 3.55 3.07
3.75 3.92 3.31
3.88 3.45 2.80
USE AN ANOVA WITH o=.05 to determine whether there are significant meandifferences among the three groups. Complete the following table by selecting the correct values in each cell (rounding)
SOURCE SS df MS F Fcritical
BETWEEN TREATMENTS ________ _____ _____ _____ _______
WITHIN TREATMENTS ________ _____ _____
TOTAL ________ _____
F DISTRIBUTION
NUMERATOR DEGREES OF FREEDOM=6
DENOMINATOR DEGREES OF FREEDOM=16
*REJECT THE NULL HYPOTHESIS. THERE IS SIGNIFICANT MEAN DIFFERENCE AMOUND THE 3 GROUPS
*FAIL TO REJECT THE NULL HYPOTHESIS. THERE IS A SIGNIFICANT MEAN DIFFERENCE AMOUNG THE THREE GROUPS
* FAIL TO REJECT THE NULL HYPOTHESIS. THERE IS NO SIGNIFICANT MEAN DIFFERENCE AMOUNG THE THREE GROUPS.
N2=_____
THE RESULTS SHOW SIGNIFICANT DIFFERENCE IN MEAN GRADE POINT AVERAGES BETWEEN GROUPS, ________________.
I input the run the data using one factor ANOVA, the result is as follows:
One factor ANOVA
Mean n Std. Dev
3.710 10 0.3013 Non-user
3.460 10 0.2986 Rarely use
3.010 10 0.2677 Regularly …
The solution gives detailed steps on run one factor ANOVA using Excel and then analyzing the results including finding effect size.