Grand Canyon University Workbook Exercises 31 & 32 - 1100 Words

Grand Canyon University: Workbook Exercises 31 & 32 (Statistics Project Sample) Content: Workbook Exercises 31 32NameGrand Canyon University 1 Exercise 31 1 Do the example data meet the assumptions for the independent samples t-test? Provide a rationale for your answer.The data meet the assumption for the independent samples t-test because the participants the first group are different from those in the second group. Take, for instance, the participants in the treatment group runs from 1 to 10 with no number repeated with the participants in the control group run from 11 to 20, again with no number repeated. This makes the two data from the two groups to be statistically different. 2 If calculating by hand, draw the frequency distributions of the dependent variable, wages earned. What is the shape of the distribution? If using SPSS, what is the result of the Shapiro-Wilk test of normality for the dependent variable? (Instructors note: Use SPSS for this question.)The Shapiro-Wilk p-Value for the weekly wages was 0.194. The p-Value indicates that the frequ ency distribution did not significantly deviate from normality. 3 What are the means for two groups wages earned?The mean for the weekly wages earned for the treatment group was 232.7 while the mean for the weekly wages earned for the control group was 128.4. However, the combined mean of the two group was 180.55 4 What is the independent samples t-test value?The independent sample t-test value for the weekly wages earned by the treatment group is t = 4.217. 5 Is the t-test signiï ¬ cant at a = 0.05? Specify how you arrived at your answer.At a = 0.05, the t-values is statistically significance. We arrived at this answer by comparing the t-value of 4.217 with the lower limit of 52.33 and upper limit of 156.27. the value 4.217 lies outside the two limits. 6 If using SPSS, what is the exact likelihood of obtaining a t-test value at least as extreme or as close to the one that was actually observed, assuming that the null hypothesis is true? (Instructors note: Use SPSS for this questi on.)Using the SPSS, we find that there is a 95% chance of obtaining a t-test value at least as an extreme or close to the one observed. 7 Which group earned the most money post-treatment?The total weekly wage earned by treatment group was $2,327 while the total weekly wage earned by control group was 1,284. It therefore, follows that the treatment group earned the most money post treatment 8 Write your interpretation of the results as you would in an APA-formatted journal.With a sample of size n = 20 for the treatment group, a = 0.05 t = 4.217, the result indicated statistical significance for the t-test. It, therefore, follows that there is no difference between participants number and weekly wages earned. 9 What do the results indicate regarding the impact of the supported employment vocational rehabilitation on wages earned?The results indicate that supported vocational rehabilitation has significant impacts on wages earned for both the treatment and control groups. 10 Was the s ample size adequate to detect signiï ¬ cant differences between the two groups in this example? Provide a rationale for your answer.The sample size was adequate to determine the difference between the two groups. This is because our analysis gave us a level of significance at alpha = 0.05.Exercise 32 1 Do the example data meet the assumptions for the paired samples t-test? Provide a rationale for your answer.Yes, the data sample meet the criteria for paired sample t-test. This is because the sample variables have been collected from a single group of participants with similar characteristics for baseline and post-treatment probing. 2 If calculating by hand, draw the frequency distributions of the two variables. What are the shapes of the distributions? If using SPSS, what are the results of the Shapiro-Wilk tests of normality for the two variables? (Instructors note: Use SPSS for this question.)The Shapiro-Wilk p-Value for the weekly wages was 0.716. The p-Value indicates that the frequency distribution did not significantly deviate from normality. 3 What are the means for the baseline and posttreatment affective distress scores, respectively?The mean for the baseline affective distress was 3.030 while the mean for the Post Treatment Distress was 2.040. 4 What is the paired samples t-test value? The paired samples t-test value is 2.865. 5 Is the t-test signiï ¬ cant at = 0.05? Specify how you arrived at your answer.At = 0.05, the t-test is significant. We arrived at this answer by using the calculated p-Value of 0.019, which is less than the = 0.05. 6 If using SPSS, what is the exact likelihood of obtaining a t- test value at least as extreme as or as close to the one that was actually observed, assuming that the null hypothesis is true? (Instructors note: Use SPSS for this question.)the exact likelihood of obtaining a t-test value at least as an extreme as or as close to the one that was actually observed if we assume that the null hypothesis is true w ill be more than 0.099% but less than 0.1%. 7 On average, did the affective distress scores improve or deteriorate over time? Provide a rationale for your answer.On average, the affective scores deteriorated overtime but finally became constant. The constant value in affective distress are represented by a constant Q-Q line for the baseline and post-treatment affective distress. 8 Write your interpretation of the results as you would in an APA-formatted journal.The paired sample t-test calculated from baseline affective distress and post-treatment affective distress revealed deteriorating of distress scores over time, t(9) = 2.865, p 0.001; mean = 3.03 and 2.04 for the two tests. Therefore, the mode of treatment is effective for affective distress. 9 What do the results indicate regarding the impact of the rehabilitation on emotional distress levels?The results indicate that rehabilitation lowers emotional distress levels among patients. 10 What are the weaknesses of the design in t his example?It may be hard to determine whether or not the changes in one sample have been caused by the intervention program. The design does not take into consideration other factors that may have contributed towards the changes.Exercise 31 SPSS OutputCase Processing Summary Group Cases Valid Missing Total N Percent N Percent N Percent WeeklyWage 20 100.0% 0 0.0% 20 100.0% Tests of Normality Group Kolmogorov-Smirnova Shapiro-Wilk Statistic df Sig. Statistic df Sig. WeeklyWage .155 20 .200* .935 20 .194 *. This is a lower bound of the true significance. a. Lilliefors Significance Correction WeeklyWage Stem-and-Leaf Plot forGroup=Frequency Stem Leaf3.00 0 . 7774.00 1 . 23445.00 1 . 556674.00 2 . 00122.00 2 . 592.00 3 . 33Stem width: 100.00Each leaf: 1 case(s)Descriptive Statistics N Minimum Maximum Mean Std. Deviation WeeklyWage 20 70.00 330.00 180.5500 75.90123 Valid N (listwise) 20 Group Statistics Grou p N Mean Std. Deviation Std. Error Mean WeeklyWage Participant#TreatmentGroup 10 232.7000 65.32491 20.65755 Participant#ControlGroup 10 128.4000 43.02506 13.60572 Independent Samples Test Levenes Test for Equality of Variances t-test for Equality of Means F Sig. t df, Sig.(2-tailed) Mean Diff. STD. Error Difference 95% Confidence Interval of the Difference Lower Upper Weekly Wage Equal Variance assumedEqual variance not assumed 2.477 .133 4.2174.217 1815.572 .001.001 104.30000104.30000 24.7356024.73560 52.3324351.74540 156.26757156.85460 Exercise 32 SPSS OutputDescriptive Statistics N Minimum Maximum Mean Std. Deviation BaselineAffectiveDistressScores 10 .3 5.2 3.030 1.6640 PostTreatmentAffectiveDistress 10 .7 3.7 2.040 .9834 Valid N (listwise) 10 ...