A psychologist is interested whether or not students will learn more effectively

Question

A psychologist is interested whether or not students will learn more effectively with a constant background sound, as opposed to a soothing music or no sound at all. She randomly divides twenty-four students into three groups of eight. All students study a passage of text for 30 minutes. Those in group 1 study with background sound at a constant volume in the background. Those in group 2 study with noise that changes volume periodically. Those in group 3 study with no sound at all. After studying, all students take a 10 point multiple choice test over the material. Their test scores follow:

Group 1: constant sound: 7, 4, 6, 8, 6, 6, 2, 9

Group 2: Soothing music: 5, 5, 3, 4, 4, 7, 2, 2

Group 3: No sound: 2, 4, 7, 1, 2, 1, 5, 5

a. Test the null hypothesis (is background noise a function of the number of the amount of material learned? At Alpha .05). Use appropriate IV and DV terms.

b, Report the three group means and SD

c. Present the ANOVA source table with all relevant statistics.

Note. This is a One-Way ANOVA (i.e., 1 IV). The one IV (factor) has three levels and one ther is one DV (score on the test).

Explanation / Answer

Step 1                              
Null Hypothesis Ho :           µ1 =µ2 =µ3               
Alternative Hypothesis :           µ1 µ2 µ3               
Step 2                              
Degrees of freedom between = k - 1 = 3 - 1 = 2                          
Degrees of freedom Within = n - k = 24 - 3 = 21                          
Degrees of freedom Total F( k-1,n - k,) at 0.05 is = F Crit = 3.467                          
Step 3                              
Grand Mean = G / N = 6+4+3.375 / 3 = 4.458                          
SST = ( Xi - GrandMean)^2 = (7-4.458)^2 + (4-4.458)^2 + (6-4.458)^2 + ……..& so on = 117.958                          
SS Within = (Xi - Mean of Xi ) ^2 =,(7-6)^2 + (4-6)^2 + (6-6)^2 + ……..& so on = 49.594                          
SS Between = SST - SS Within = 117.958 - 49.594 = 68.364                          
Step 4                              
Mean Square Between = SS Between / df Between = 68.364/2 = 34.182                          
Mean Square Within = SS Within / df Within = 49.594/21 = 2.362                          
Step 5                              
F Cal = MS Between / Ms Within = 34.182/2.362 = 14.474                          
We got |F cal| = 14.474 & |F Crit| =3.467                          
MAKE DECISION                              
Hence Value of |F cal| > |F Crit|and Here We Reject Ho                          

ONE WAY ANOVA Treatments Mean = X /n Yellow 7 4 6 8 6 6 2 9 6 White 5 5 3 4 4 7 2 2 4 Green 2 4 7 1 2 1 5 5 3.375

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