Two Factor Anova You are interested in determining what type

Two Factor Anova You are interested in determining what type of weight training program will produce the most improved times in swimming a 400- meter individual medley event for 1st year high school swimmers. You randomly assign 20 1st year swimmers to one of two 2- hour weight - training programs, one that meets 2 times a week and another that meets 3 times a week ( Factor 1- treatment condition: 2 days a week or 3 days a week). Within each of these two treatments groups, you randomly assign swimmers to a low or high intensity workout program ( Factor 2- Intensity of Treatment: Low or High). The weight- training program lasted a total of 3 months. The dependent or outcome variable is the difference in seconds between the swimmers\' pre- and post- treatment swim times for the 400- meter individual medley. For this question, please use data set below and determine the possible main effects of the treatment condition and intensity of the workout, and the interaction between the two factors. (54 Points) Factor A treatment condition: 2 days a week starts at low 3 amd ends at high 8 Factor B intensity 3 days a week starts at low 5 and ends at high 8

a. what is the appropicate null and alternative hypothesis for the two main effects and interaction effect? ( 6 points) b. what is the critical value for each of the F- ratios for alpha .05? Be sure to SHOW ALL OF YOUR WORK FOR THE DF VALUES ( 7 points) c. What is the calculated value of F? BE SURE TO SHOW ALL OF YOUR WORK ( df, SS, MS, and three f- Ratio calcualtions). ( 26 points) d. Write an APA format sentence to report the results of this study. Please write a sentence for each main effect and interaction effect. ( 9 points) e. Compute the three partial n^2 for this study. ( 6 points)

Solution

Here there are two factors as treatment condition and intensity.

Treatment condition has two levels 2 days a week and 3 days a week.

And intensity has two levels low and high.

First factor (A) :

The null and hypothesis for the main effects is,

H0 : i = 0 for all i = 1, . . . , I

H1 : Not all i are zero

second factor (B) :

The null and hypothesis for the main effects is,

H0 : j = 0 for all i = 1, . . . , I

H1 : Not all j are zero

Interaction effect (AB) :

The null and hypothesis for the main effects is,

H0 : ()ij = 0 i=1,. . .,I; j =1,. . .,J

H1 : Not all ()ij are zero

First enter the data in MINITAB as :

high

Steps in MINITAB for two way ANOVA are,

START --> STAT --> ANOVA --> Two way Analysis Of Variance --> Response : obs --> Row factor : treatment condition --> Column factor : Intensity --> ok

Output :

Two-way ANOVA: obs versus treatment condition, Intensity

Source DF SS MS F P
treatment condit 1 16.2 16.2 16.20 0.001
Intensity 1 80.0 80.0 80.00 0.000
Interaction 1 0.8 0.8 0.80 0.384
Error 16 16.0 1.0
Total 19 113.0

S = 1 R-Sq = 85.84% R-Sq(adj) = 83.19%

Critical values we can find by using EXCEL.

Assume level of significance (alpha) = 0.05

syntax :

=FINV(probability, deg_freedom1,deg_freedom2)

where, probability = alpha

deg_freedom1 = degrees of freedom for treatment condition = 1

deg_freedom2 = degrees of freedom for error = 16

Similarly plug the values for intensity and interaction.

critical value for treatment condition :

CV_treatment = 4.4940

critical value for intensity,

CV_intensity = 4.4940

critical value for interaction,

CV_interaction = 4.4940

degrees of freedom :

df_treatment = number of levels for treatment condition - 1 = 2 - 1 = 1

df_intensity = number of levels for intensity - 1 = 2 - 1 = 1

df_interaction = (number of levels for treatment condition - 1) * (number of levels for intensity - 1)

df_interaction = (2-1)*(2-1) = 1

F-ratio for factor A is ,

F = MSA / MSE = 16.2 / 1.0 = 16.2

F-ratio for factor B is ,

F = MSB / MSE = 80.0 / 1.0 = 80.0

F-ratio for interaction AB is,

F = MSAB / MSE = 0.8 / 1.0 = 0.80

Here we see that for the two factors A and B,

F > CV_treatment

and F > CV_intensity

Reject H0 at 5% level of significance.

Conclusion : Not all i are zero and

Not all j are zero.

But for interaction F < CV_interaction

Accept H0 at 5% level of significance.

Conclusion : All ()_ij are 0.

Compute the three partial n^2 for this study.

n^2 for treatment condition :

n^2_treatment = SS_treatment / SS_total = 16.2 / 113.0 = 0.1434

n^2_intensity = SS_intensity / SS_total = 80.0 / 113.0 =0.7080

n^2_interaction = SS_interaction / SS_total = 0.8 / 113.0 = 0.0071

obs	treatment condition	Intensity
3	2 days a week	low
4	2 days a week	low
3	2 days a week	low
4	2 days a week	low
5	2 days a week	low
4	3 days a week	low
6	3 days a week	low
5	3 days a week	low
5	3 days a week	low
6	3 days a week	low
7	2 days a week	high
6	2 days a week	high
9	2 days a week	high
7	2 days a week	high
8	2 days a week	high
10	3 days a week	high
9	3 days a week	high
11	3 days a week	high
10	3 days a week	high
8	3 days a week	high