Exercise 33

Question 1

Having that p=0.455 and the variance were not significant, it can be deduced that the data in question meets the criteria for homogeneity of variance since variation is used to test for correlation.

Test of Homogeneity of Variances
Hours Worked
Levene Statistic df1 df2 Sig.
0.840 2 12 0.455


Question 2

The result from the SPSS analysis depicts a roughly normal distribution of the data. A keen study identifies some slight right skewness. The result recognized that Shapiro-Wilk has p of 0.314 thus considered healthy.


Tests of Normality
  Kolmogorov-Smirnova Shapiro-Wilk
Statistic df Sig. Statistic df Sig.
Hours Worked 0.206 15 0.085 0.934 15 0.314

·         Lilliefors Significance Correction


Question 3

The three groups include:

  • TAU observational 15.2
  • Supported M=14.6 and
  • TAU randomized M=26.6


Hours Worked
  N Mean Std. Deviation Std. Error 95% Confidence Interval for Mean Minimum Maximum
Lower Bound Upper Bound
Supported 5 14.60 6.504 2.909 6.52 22.68 8 24
TAU Observational 5 15.20 3.701 1.655 10.60 19.80 9 18
TAU Randomized 5 26.60 7.436 3.326 17.37 35.83 15 35
Total 15 18.80 8.029 2.073 14.35 23.25 8 35

Question 4

The group df=12, F=6.162, and error df=2



Hours Worked
  Sum of Squares df Mean Square F Sig.
Between Groups 457.200 2 228.600 6.162 .014
Within Groups 445.200 12 37.100    
Total 902.400 14      


Question 5

From the statistical analysis conducted, it can be identified that p=0.014 thus alpha is equal to 0.05 something that can be used to recognize that there exists a type 1 error in the analysis.

Question 6

Under normal circumstance, the probability of obtaining the F value when we have a high level of type 1 error occurs is equal to the P value which in this case is equal to 1.4%

Question 7

Treatment: Randomized group working for most hours’ post-treatment

Null Hypothesis: there exist no a significant difference between the treatment groups, i.e. the post-treatment and the veterans and the spinal code. Having that the null hypothesis suggests that there was no treatment nor support to the injured people who worked for more hours, it must be rejected.

Question 8

From the data we have and computed, it can be identified that the ANOVA test was calculated on post-treatment hours worked of the treatment group indicated that the veterans who were subjected to the treatment as usual worked effectively for significant more hours’ post-treatment than the veterans who observed with TAU and those who had supported employment F(14) = 6.162, =p=0.014, and X=26.6, comparedto 14.60 and 15.20 respectively. It can be concluded from the above that randomization of how the support TAU in vocation rehab approach implementation to increase the work activity of the spinal cord.

Question 9

Comparing the Tukey HSD and LSD post hoc result, there exists no significant difference between them. They both show significant values that TAU has higher mean numbers of hours worked compared to the other two groups under study.

Question 10

A t-test is the most effective technique that can be used in this case since the study involves only two groups.  From research, a t-test is used primarily and relevant when handling means of two groups. ANOVA could be useful if more than two groups were under study. The best way to make t-test effective when we have more than two groups is combining the groups under investigation until will have two main groups to be analyzed.

Do you need high quality Custom Essay Writing Services?

Custom Essay writing Service