Question

A researcher is interested in seeing if different study techniques can influence test scores. An experiment was designed in which students are randomly assigned to one of three groups based on how they study for a quiz. Group 1 simply reads the chapter, Group 2 reads and takes notes, Group 3 reads, takes notes, and uses a study guide. After each group studies, they take a quiz. Their quiz scores are listed below.

Group 1	7	6	4	6	7	4	7	6	8	7	x̅= 6.2
Group 2	7	4	6	7	8	6	7	4	6	8	x̅= 6.3
Group 3	7	7	7	7	8	7	7	7	8	8	x̅= 7.3

a. What type of statistical test should be used to analyze this data (be specific)?

b. Identify the IV

c. Identify the DV

d. state null hypothesis

e. state the alternative hypothesis

f. what are the degrees of freedom?

a. dftotal =

b. dfwithin=

c. dfbetween=

g. What is the grand mean?

h. Calculate SS between

i. Calculate MS between

j. Calculate MS within

k. Calculate F ratio

l. if significant, calculate effect size, SSbetween/SStotal

Answer 1

a:

Since we need to compare three population mean so one way ANOVA will be used.

b:

The independent variable: study technique

c:

The dependent variable: quiz scores

d:

Ho: Different study techniques did not influence test scores.

$H_{0}:\mu_{1}=\mu_{2}=\mu_{3}$

e:

Ha: Different study techniques influence test scores.

Ha: At least one mean is different.

f:

a.

Thete are total 30 data values so

dftotal = 30- 1 =29

b.

There are 3 groups so

dfwithin= 30-3 = 27

c.

dfbetween= 3-1 = 2

g:

The grand mean is

6.6

h:

betweenn1(71 -T 10 . (63-66)2 + 10 . (7.3-6.6)2 = 7.4 22T)

i:

7.43. SS7.4 between M Sbetueendfbetueen between

j:

Following table shows the calculations for SS:

Group 1, x1	Group 2, X2	Group 3, X3	(X1-xbar1)^2	(X2-xbar2)^2	(X3-xbar3)^2
7	7	7	0.64	0.49	0.09
6	4	7	0.04	5.29	0.09
4	6	7	4.84	0.09	0.09
6	7	7	0.04	0.49	0.09
7	8	8	0.64	2.89	0.49
4	6	7	4.84	0.09	0.09
7	7	7	0.64	0.49	0.09
6	4	7	0.04	5.29	0.09
8	6	8	3.24	0.09	0.49
7	8	8	0.64	2.89	0.49
Total			15.6	18.1	2.1

SSwithin (n1-1)sT+(n2-1)s2 + (n3-1)s 15.6+18.1+2.1 35.80

Swithin = SSuithin _ 35.8 dfuithin =-27, 35.81326 withindfwithin dfwithin27 1.326

k:

F test statistics is

M Sbetween79 M Swithiun

So p-value of the test is 0.0791. Since P-value is greater than 0.05 so we fail to reject the null hypothesis. That is on the basis of sample evidence we cannot conclude that different study techniques influence test scores.