BTM8107-8 Week: 4 Analyze t Test and ANOVA Correct Answers

Solutions for t-tests and ANOVA statistical analysis.

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BTM8107-8 Week: 4 Analyze t Test and ANOVA Correct Answers
Part A. Dependent t test
For this assignment, we are interested in finding out whether participation in a creative
writing course results in increased scores of a creativity assessment. For this part of the
activity, you will be using the data file “Activity 4a.sav”. In this file, “Participant” is the
numeric student identifier, “CreativityPre” contains creativity pre-test scores, and
“CreativityPost” contains creativity post-test scores. A total of 40 students completed the
pre-test, took the creativity course, and then took the post-test.
Answer:
Here the null hypothesis H0: the participation in a creative writing course results is not
increased in scores of creative assessment against the alternative H1: the participation in
a creative writing course results is increased in scores of creative assessment. In other
words we can say the null hypothesis H0: d=0 against the alternative hypothesis H1: d>0
where d is mean of the sample differences of the two sample.
This is a paired sample t test. And the test statistics is
t = Σ(differences of the observation) / ((NΣ(differences of observation)2 - (Σ(differences
of observation))2) / (N-1)). Where N is the sample size, here N=40.
The SPSS result:
Paired Samples Statistics
Mean N Std. Deviation Std. Error Mean
Pair 1 Creativity pre-test 40.15 40 8.304 1.313
Creativity post-test 43.35 40 9.598 1.518
The means of Creativity pre-test is 40.15, standard deviation is 8.304 and the mean of
Creativity post-test is 43.35 and standard deviation is 9.598.
Paired Samples Correlations
N Correlation Sig.
Pair 1 Creativity pre-test &
Creativity post-test
40 .650 .000
From the correlation matrix of these two variables we see that the correlation between them is
0.650 and also from the scatter plot we observe that the correlation between them is high and
they are highly correlated.
From the result we can see that the t value is -2.671, corresponding to the t value, p
value is 0.011 which is less than 0.05 at 5% level of significance therefore we reject the
null hypothesis and accept the alternative hypothesis that the participation in a creative
writing course results is increased in scores of creative assessment.
Lower Upper
Pair 1 Creativity
pre-test -
Creativity
post-test
-3.200 7.576 1.198 -5.623 -.777 -2.671 39 .011
Paired Samples Test
Paired Differences
t df
Sig. (2-
tailed)Mean
Std.
Deviation
Std. Error
Mean
95% Confidence
Interval of the
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Document Details

University
Northcentral University
Subject
Statistics

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