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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.
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.

1. Table 1 presents the descriptive statistics for the Pre-test and Post-test 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 pre-test score of creative writing is 40.15 and 8.30.  The average and standard deviation of post-test score of creative writing is 43.35 and 9.60 respectively.

The comparative distribution of Creativity in Pre-test and Creativity in Post-test indicates that overall values are greater for Creativity in post-test scores. The score is higher for maximum and minimum values for post-test score.

Null Hypothesis (H0): Participation in the creative writing course does not produce any difference in the mean pre and post-test scores.

Alternate Hypothesis (HA): Participation in the creative writing course does produce difference in the mean pre and post-test scores (Anderson et al., 2014).

1. Table 2 presents the results of the paired samples t-test (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. (2-tailed) .011

The paired sample t-test is applicable when we would like to test the averages of two similar types of variables having so significant relevance between them.

The t-statistic for paired two samples is –,

Where, X1bar = average of first sample, X2bar = average of second sample, S2 = sample variance, n = sample size, t= Student t-statistic with (n-1) degrees of freedom (Francis, 2014).

The confidence interval of the t-statistic is found to be –

(X1bar – X2bar) ± t *    or, equivalently, [(X1bar – X2bar) ± t * SE(X1bar – X2bar)]

The paired difference is found to be (-3.2) and t-statistic is calculated as (-2.671) with degrees of freedom 39. The p-value of significant t-statistic 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 pre-test and post-test 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 t-test was done.

The paired-samples t-test shows that the mean of Creativity Pre-test (M = 40.15, SD = 8.30) and Creativity Post-test (M = 43.35, SD = 9.60), t(39) = -2.671, p = .011. On an average Creativity Pre-test was about 3.200 points lower than Creativity Post-test score (Berenson, Levine & Krehbiel, 2012).

A data set was created with the scores.

1. The paired samples t-test is used to judge the impact of creative writing course. The pre-test and post-test scores of individual participants are compared. The mean test scores of the participants are compared.

On the other hand, the independent sample t-test 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 T-test Results

Like mean and standard deviation, standard error of 40 samples is also greater for Post-test (1.52) than Pre-test (1.31).

The Null and Alternate Hypothesis of independent samples t-test may be written as:

Null Hypothesis (H0): Participation in the creative writing course does not produce any difference in the mean pre and post-test scores.

Alternate Hypothesis (HA): Participation in the creative writing course does produce difference in the mean pre and post-test 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 t-test for Equality of Means t -1.595 -1.595 df 78 76.418 Sig. (2-tailed) .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 F-test is applicable for relating test score of equal variances assumed and not assumed. The F-test is calculated as 0.632 with p-value 0.429. The p-value 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 t-statistic for independent sample t-test is given as –

Where,  =

The t-test for equality of means is given as (-1.595) with degrees of freedom 78 when equal variances is assumed. The significant t-statistic 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 t-tests that are two sample paired t-test and two samples paired t-test refer the same conclusion. In this analysis, both types of tests, it was recorded that the averages of pre-test and post-test are equal.

When identification number mismatches pre-test and post-test scores, then we applied independent samples t-test with the help of 40 pre-test scores and 40 post-test scores. For comparing pre-test and post-test scores, we applied between subject’s design rather than within subject’s design. Both the samples are imputed vertically and incorporated independent sample t-test. The between–subjects rather than within-subjects 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.

1. 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 T-test 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 (H0): The averages of systolic blood pressures in all the three settings are equal that is µ1 = µ2 =µ3.

Alternative Hypothesis (HA): There exists at least one equality in the averages of all the three settings are systolic blood pressures.

Null Hypothesis (H0): The averages of diastolic blood pressures in all the three settings are equal that is µ1 = µ2 =µ3.

Alternative Hypothesis (HA): There exists at least one equality in the averages of all the three settings are diastolic blood pressures.

1. 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 F-statistic is 9.964 with significant p-value 0.001. The p-value of F-statistic 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 F-statistic is 0.105 with significant p-value 0.900. The p-value of F-statistic 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 (I-J) 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 p-value between Home and classroom setting is 0.623 (>0.05). Therefore, the systolic BP between home and classroom setting are equal. The p-value 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 p-value 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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