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 tTest and ANOVACorrectAnswersPart A. Dependent t testFor this assignment, we are interested in finding out whether participation in a creativewriting course results in increased scores of a creativity assessment. Forthis part of theactivity, you will be using the data file “Activity 4a.sav”. In this file, “Participant” is thenumeric student identifier, “CreativityPre” contains creativity pre-test scores, and“CreativityPost” contains creativity post-test scores. A total of 40 students completed thepre-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 notincreased in scores of creative assessment against the alternative H1: the participation ina creative writing course results is increased in scores of creative assessment. In otherwords we can say the null hypothesis H0: d=0 against the alternative hypothesis H1: d>0where d is mean of the sample differences of the two sample.This is a paired sample t test. And the test statistics ist =Σ(differences of the observation)/((NΣ(differences of observation)2-(Σ(differencesof observation))2)/ (N-1)). Where N is the sample size, here N=40.The SPSS result:Paired Samples StatisticsMeanNStd. DeviationStd. Error MeanPair 1Creativity pre-test40.15408.3041.313Creativity post-test43.35409.5981.518The means of Creativity pre-test is 40.15, standard deviation is 8.304 and the mean ofCreativity post-test is 43.35 and standard deviation is 9.598.Paired Samples CorrelationsNCorrelationSig.Pair 1Creativity pre-test &Creativity post-test40.650.000

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From the correlation matrix of these two variables we see that thecorrelation between them is0.650 and also from the scatter plot we observe that the correlation between them is high andthey are highly correlated.From the result we can see that the t value is-2.671, corresponding to the t value, pvalue is 0.011 which isless than 0.05 at 5% level of significance therefore we reject thenull hypothesis and accept the alternative hypothesis that the participationin a creativewriting course results is increased in scores of creative assessment.LowerUpperPair 1Creativitypre-test -Creativitypost-test-3.2007.5761.198-5.623-.777-2.67139.011Paired Samples TestPaired DifferencestdfSig. (2-tailed)MeanStd.DeviationStd. ErrorMean95% ConfidenceInterval of the
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