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 pretest scores, and “CreativityPost” contains creativity posttest scores. A total of 40 students completed the pretest, took the creativity course, and then took the posttest.
1.Exploratory Data Analysis/Hypotheses
a.Perform exploratory data analysis on CreativityPre and CreativityPost. Using SPSS, calculate the mean and standard deviation of these two variables.
b.Construct an appropriate chart/graph that displays the relevant information for these two variables.
c.Write the null and alternative hypotheses used to test the question above (e.g., whether participation in the course affects writing scores).
2.Comparison of Means
a.Perform a dependent t test to assess your hypotheses above (note that many versions of SPSS use the term “paired samples t test” rather than “dependent t test”; the test itself is the same.
In APA style, write one or two paragraphs that describe the dataset, gives your hypothesis, and presents the results of the dependent sample t test.
 Table 1 presents the descriptive statistics for the Pretest and Posttest scores of the creative writing course (Christodoulides & Christodoulides, 2017).
Table 1: Descriptive Statistics 

N 
Mean 
Std. Deviation 

CreativePre 
40 
40.15 
8.30 
CreativePost 
40 
43.35 
9.60 
Valid N (listwise) 
40 
Descriptive Statistics
The average and standard deviation of pretest score of creative writing is 40.15 and 8.30. The average and standard deviation of posttest score of creative writing is 43.35 and 9.60 respectively.
The comparative distribution of Creativity in Pretest and Creativity in Posttest indicates that overall values are greater for Creativity in posttest scores. The score is higher for maximum and minimum values for posttest score.
Null Hypothesis (H_{0}): Participation in the creative writing course does not produce any difference in the mean pre and posttest scores.
Alternate Hypothesis (H_{A}): Participation in the creative writing course does produce difference in the mean pre and posttest scores (Anderson et al., 2014).
 Table 2 presents the results of the paired samples ttest (Black, 2016)
Table 2: Paired Samples Test 

Pair 1 

CreativePre  CreativePost 

Paired Differences 
Mean 
3.20000 

Std. Deviation 
7.57594 

Std. Error Mean 
1.19786 

95% Confidence Interval of the Difference 
Lower 
5.62290 

Upper 
.77710 

t 
2.671 

df 
39 

Sig. (2tailed) 
.011 
The paired sample ttest is applicable when we would like to test the averages of two similar types of variables having so significant relevance between them.
The tstatistic for paired two samples is –,
Where, X_{1}bar = average of first sample, X_{2}bar = average of second sample, S^{2} = sample variance, n = sample size, t= Student tstatistic with (n1) degrees of freedom (Francis, 2014).
The confidence interval of the tstatistic is found to be –
(X_{1}bar – X_{2}bar) ± t * or, equivalently, [(X_{1}bar – X_{2}bar) ± t * SE(X_{1}bar – X_{2}bar)]
The paired difference is found to be (3.2) and tstatistic is calculated as (2.671) with degrees of freedom 39. The pvalue of significant tstatistic is found to be 0.11. It is greater than 0.05. Therefore, null hypothesis of equality of means is accepted at 5% level of significance.
Hence, we can infer that we are 95% confident that average scores of pretest and posttest Creativity are equal.
40 students were asked to participate in a creative writing course. Prior to the start of the course they were administered a test whose score was stored as “CreativePre.” At the end of the course the students were again administered a test, the scores was stored as “CreativePost.” The scores of the participants was stored individually. To assess the importance of the creative writing course a paired sample ttest was done.
The pairedsamples ttest shows that the mean of Creativity Pretest (M = 40.15, SD = 8.30) and Creativity Posttest (M = 43.35, SD = 9.60), t(39) = 2.671, p = .011. On an average Creativity Pretest was about 3.200 points lower than Creativity Posttest score (Berenson, Levine & Krehbiel, 2012).
A data set was created with the scores.
 The paired samples ttest is used to judge the impact of creative writing course. The pretest and posttest scores of individual participants are compared. The mean test scores of the participants are compared.
On the other hand, the independent sample ttest is used to judge the impact of creative writing of unrelated groups. The groups are test scores prior and post taking the course.
Table 3: Group Statistics 

Grouping 
N 
Mean 
Std. Deviation 
Std. Error Mean 

TestScore 
Pre 
40 
40.15 
8.30 
1.31 
Post 
40 
43.35 
9.60 
1.52 
Paired Sample Ttest Results
Like mean and standard deviation, standard error of 40 samples is also greater for Posttest (1.52) than Pretest (1.31).
The Null and Alternate Hypothesis of independent samples ttest may be written as:
Null Hypothesis (H_{0}): Participation in the creative writing course does not produce any difference in the mean pre and posttest scores.
Alternate Hypothesis (H_{A}): Participation in the creative writing course does produce difference in the mean pre and posttest scores.
Table 4: Independent Samples Test 

TestScore 

Equal variances assumed 
Equal variances not assumed 

Levene's Test for Equality of Variances 
F 
.632 

Sig. 
.429 

ttest for Equality of Means 
t 
1.595 
1.595 

df 
78 
76.418 

Sig. (2tailed) 
.115 
.115 

Mean Difference 
3.20000 
3.20000 

Std. Error Difference 
2.00675 
2.00675 

95% Confidence Interval of the Difference 
Lower 
7.19514 
7.19644 

Upper 
.79514 
.79644 
Part 3
The Levene’s Ftest is applicable for relating test score of equal variances assumed and not assumed. The Ftest is calculated as 0.632 with pvalue 0.429. The pvalue is greater than 0.05. Therefore, we cannot reject the null hypothesis of equality of variance at 5% level of significance (George & Mallery, 2016).
The tstatistic for independent sample ttest is given as –
Where, =
The ttest for equality of means is given as (1.595) with degrees of freedom 78 when equal variances is assumed. The significant tstatistic is (0.115) which is greater than 0.05. Hence, from this perspective also we accept the null hypothesis.
Therefore, it is 95% evident that the averages of these two variables are equal.
We have used the same data set for analysing both a between and within subject’s design. The both ttests that are two sample paired ttest and two samples paired ttest refer the same conclusion. In this analysis, both types of tests, it was recorded that the averages of pretest and posttest are equal.
When identification number mismatches pretest and posttest scores, then we applied independent samples ttest with the help of 40 pretest scores and 40 posttest scores. For comparing pretest and posttest scores, we applied between subject’s design rather than within subject’s design. Both the samples are imputed vertically and incorporated independent sample ttest. The between–subjects rather than withinsubjects design is useful to take total 80 samples at a time. The within subject design is helpful for commonly in repeated measure analysis. The within subject design helps to measure how much an individual in the sample tends to vary in different observation. In other words, it is the mean of variation for the average individual case in the specified sample.
 The Analysis of variance technique helps to find the differences of mean scores between Systolic blood pressure and Diastolic blood pressure in Home, Doctor’s Office and Classroom setting.
Descriptive Statistics 

Setting 
N 
Mean 
Std. Deviation 

Home 
SystolicBP 
10 
122.9000 
7.09382 
Valid N (listwise) 
10 

Doctor's Office 
SystolicBP 
10 
132.6000 
8.36926 
Valid N (listwise) 
10 

Classroom Setting 
SystolicBP 
10 
118.8000 
5.55378 
Valid N (listwise) 
10 
The descriptive statistic indicates that systolic blood pressure is higher for Doctor’s Office followed by Home. The pressure is least in classroom setting. All the variables show same indication in both average and standard deviation values.
Independent Sample Ttest Results
Descriptive Statistics 

Setting 
N 
Mean 
Std. Deviation 

Home 
DiastolicBP 
10 
82.9000 
2.68535 
Valid N (listwise) 
10 

Doctor's Office 
DiastolicBP 
10 
83.2000 
3.35989 
Valid N (listwise) 
10 

Classroom Setting 
DiastolicBP 
10 
82.6000 
2.67499 
Valid N (listwise) 
10 
The descriptive statistic refers that Diastolic blood pressure has mean and standard deviation values in Doctor’s office followed by Home. The average and standard deviation are least in Classroom setting.
The graph shows that Overall distribution of Systolic blood pressure is highest in Doctor’s office and lowest in Classroom setting.
The distributions of Diastolic blood pressures are almost equal in all the three settings that are Home, Doctor’s Office and Classroom settings.
The Null and Alternate Hypothesis may be written as:
Null Hypothesis (H_{0}): The averages of systolic blood pressures in all the three settings are equal that is µ_{1} = µ_{2} =µ_{3}.
Alternative Hypothesis (H_{A}): There exists at least one equality in the averages of all the three settings are systolic blood pressures.
Null Hypothesis (H_{0}): The averages of diastolic blood pressures in all the three settings are equal that is µ_{1} = µ_{2} =µ_{3}.
Alternative Hypothesis (H_{A}): There exists at least one equality in the averages of all the three settings are diastolic blood pressures.
 We incorporated two ANOVA tables. One is calculated as Systolic BP is dependent variable and other is calculated as Diastolic BP, a dependent variable.
ANOVA 

SystolicBP 

Sum of Squares 
df 
Mean Square 
F 
Sig. 

Between Groups 
1004.467 
2 
502.233 
9.964 
.001 
Within Groups 
1360.900 
27 
50.404 

Total 
2365.367 
29 
For Systolic BP value of Fstatistic is 9.964 with significant pvalue 0.001. The pvalue of Fstatistic is less than 0.05. Therefore, we reject the null hypothesis at 5% level of significance.
Hence, it is 95% evident that the average values of Systolic blood pressures in all the three setting that are Home, Classroom setting and Doctors office are unequal.
ANOVA 

DiastolicBP 

Sum of Squares 
df 
Mean Square 
F 
Sig. 

Between Groups 
1.800 
2 
.900 
.105 
.900 
Within Groups 
230.900 
27 
8.552 

Total 
232.700 
29 
For Diastolic BP value of Fstatistic is 0.105 with significant pvalue 0.900. The pvalue of Fstatistic is greater than 0.05. Therefore, we accept the null hypothesis at 5% level of significance. Hence, it is 95% evident that the average values of Diastolic blood pressures in all the three setting that are Home, Classroom setting and Doctors office are equal.
The Sum of squares within groups and between groups are significantly higher in case of ANOVA test of Systolic BP. On the other hand, sum of squares between groups is insignificant than within group SSE in the ANOVA table of Diastolic blood pressure (Field, 2013).
Multiple Comparisons 

Dependent Variable: SystolicBP 

Bonferroni 

(I) Setting 
(J) Setting 
Mean Difference (IJ) 
Std. Error 
Sig. 
95% Confidence Interval 

Lower Bound 
Upper Bound 

Home 
Doctor's Office 
9.70000^{*} 
3.17502 
.015 
17.8041 
1.5959 
Classroom Setting 
4.10000 
3.17502 
.623 
4.0041 
12.2041 

Doctor's Office 
Home 
9.70000^{*} 
3.17502 
.015 
1.5959 
17.8041 
Classroom Setting 
13.80000^{*} 
3.17502 
.001 
5.6959 
21.9041 

Classroom Setting 
Home 
4.10000 
3.17502 
.623 
12.2041 
4.0041 
Doctor's Office 
13.80000^{*} 
3.17502 
.001 
21.9041 
5.6959 

*. The mean difference is significant at the 0.05 level. 
The table of “multiple comparison” refers that Using Systolic blood pressure as dependent variable; we calculated different types of differences between two types of settings. The first setting involves Home, Doctor’s Office and Classroom setting (McCormick, 2017). The second setting includes paired categorical settings in each of the three cases. The pvalue between Home and classroom setting is 0.623 (>0.05). Therefore, the systolic BP between home and classroom setting are equal. The pvalue between Doctor’s office and classroom setting is 0.001 (<0.005). Therefore, the systolic BP between Doctor’s office and Classroom setting are not equal. Further, the pvalue between Home and Doctor’s Office is 0.015 (<0.05). Hence, the systolic BP between home and Doctor’s office are not equal.
References
Anderson, D. R., Sweeney, D. J., Williams, T. A., Camm, J. D., & Cochran, J. J. (2014). Essentials of statistics for business and economics. Cengage Learning.
Berenson, M. L., Levine, D. M., & Krehbiel, T. C. (2012). Basic business statistics. Upper Saddle River, NJ: Prentice Hall.
Black, K. (2016). Business statistics: Contemporary decision making. John Wiley & Sons.
Christodoulides, C., & Christodoulides, G. (2017). Analysis and Presentation of Experimental Results. Springer
Field, A. (2013). Discovering statistics using IBM SPSS statistics. Sage.
Francis, A. (2014). Business mathematics and statistics. Cengage Learning EMEA.
George, D., & Mallery, P. (2016). IBM SPSS Statistics 23 step by step: A simple guide and reference. Routledge.
McCormick, K., Salcedo, J., Peck, J., & Wheeler, A. (2017). SPSS Statistics for data analysis and visualization. John Wiley & Sons.
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