An educational psychologist is interested in knowing whether the experience of attending preschool is related to subsequent sociability. She identifies two groups of first graders: those who had...

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Answer To: An educational psychologist is interested in knowing whether the experience of attending preschool...

David answered on Dec 21 2021
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9
9. An educational psychologist is interested in knowing whether the experience of
attending preschool is related to subsequent sociability. She identifies two groups of first
graders: those who had attended preschool and thos
e who had not. Then each child is
assigned a sociability score on the basis of observations made in the classroom and on the
playground. The following sociability results are obtained:
Attended preschool Did not attend preschool
n1 = 12, ∑X1 = 204, SS1 = 192 n2 = 16, ∑X2 = 248, SS2 = 154
a. Set up the appropriate statistical hypotheses.
b. Perform the test (α = .05)
Ans: (Note: I assumed SS is sum of squares)
Here the psychologist wants to see if there is a difference between the scores due to
attending preschools. Let μ1 and μ2 are the population mean of scores of the
graders who attended preschool and did not consecutively.
Null Hypothesis:
So the hypothesis of interest is,
Ho: μ1 = μ2 against Ha: μ1 ≠ μ2
Or equivalently,
Ho: μ1 - μ2 =0 against Ha: μ1 - μ2 ≠ 0
Test statistic
Here we don’t know the sample variance thus we will use the t test to test the
hypothesis. Assuming unequal variance for the populations the test statistic can be
given by,
21
21
2
2
2
1
21
n
s
n
s
XX
t











; Under the null hypothesis this t follows a t-distribution
with appropriate degrees of freedom.
The degrees of freedom can be calculated using the formula,
Degrees of freedom=
   
12
2/
11
1/
21
22
2
22
1
2
2
2
2
1











n
ns
n
ns
n
s
n
s
v
Rejection Condition:
Since under the null hypothesis t follows a t distribution so deviation from null
hypothesis implies deviation from t-distribution or vice versa. Again if we look at
the alternative we can see it is both sided thus the considered test is a two tail test
and we will reject the null hypothesis at 100α% level of significance if,
Observed |t| > tv,α/2
Calculation:
From the given values,
Mean of 1
st
sample = 204/12 = 17
Mean of 2
nd
sample =248/16 = 15.5
Sample variance of 1
st
sample = SS1/(12-1) = 17.45
So standard deviation of 1
st
sample = 4.18
Sample variance of 2
nd
sample = 154/15 = 10.27
Standard deviation of 2
nd
sample=...
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