Name: Samples Test Levene’s Test for Equality of

 

 

 

 

 

 

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1.      Do the example data meet the assumptions for the independent
samples t -test? Provide a rationale for your answer.

The
assumptions being the variable measured on a continuous or ordinal scale, the
sample is random, the sample size is big enough, normal distribution is seen
when the data is plotted, and there is homogeneity of variance.

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?

The
shape of the distribution is a standard shape; the results of the Shapiro-Wilk
test is shown below.  The Shapiro-Wilk
significance is .194 shows that it is above the 0.05 alpha level of the null
hypothesis that the distribution is normal.

 

Tests of Normality

 

Kolmogorov-Smirnova

Shapiro-Wilk

Statistic

Df

Sig.

Statistic

df

Sig.

Support

.155

20

.200*

.935

20

.194

*. This is a lower bound of the true significance.

a. Lilliefors Significance Correction

 

3.
What are the means for two group’s wages earned?

 

The
means for the two group are $232.70 for the experimental, and $128.40 for the
control.

Group Statistics

 

Group

N

Mean

Std. Deviation

Std. Error Mean

Support

1

10

232.70

65.325

20.658

2

10

128.40

43.025

13.606

 

 

4.
What is the independent samples t -test value?

 

The t-test value is 4.217.

 

5.
Is the t -test significant at a = 0.05? Specify how you arrived at your answer.

The
t-test is significant at the ?= 0.05, with a t-value of 4.217 and 18 degrees of
freedom, a value chart with show that it is significant to the p = 0.001

 
 
Independent Samples Test

 

Levene’s Test for Equality of Variances

t-test for Equality of Means

F

Sig.

t

df

Sig. (2-tailed)

Mean Difference

Std. Error Difference

95% Confidence Interval of the Difference

Lower

Upper

Support

Equal
variances assumed

2.477

.133

4.217

18

.001

104.300

24.736

52.332

156.268

Equal
variances not assumed

 

 

4.217

15.572

.001

104.300

24.736

51.745

156.855

 

 

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?

 

The
probability of making a type 1 error would be the p- value gotten. This was
0.001, which means there is a 0.1% possibility that the null hypothesis is
true.

 

7.
Which group earned the most money post-treatment?

 

The
group with assistance program earned the most money post treatment.

 

8.
Write your interpretation of the results as you would in an APA-formatted
journal.

Veterans
with disabilities (n = 10) who participated in a work assistance program earned
significantly more money (M = $232.70, SD = $65.32) than those veterans (n =
10) with disability who did not receive work assistance (M = $128.40, SD =
$43.02, t(18) = 4.217, p = 0.001).

9.
What do the results indicate regarding the impact of the supported employment
vocational rehabilitation on wages earned?

An
independent samples t-test demonstrated that employment occupational
rehabilitation positively affects the wages earned by impaired veterans. This
program might be advantageous for a few and further research is justified.

10.
Was the sample size adequate to detect significant differences between the two
groups in this example? Provide a rationale for your answer.

The
sample size was adequate to detect significant differences between these two
groups.  This can be found in the glaring
difference between the mean measure of cash earned, the p value, t value, and
the confidence interval between groups.

 

Exercise
32

1. Do the example data meet the
assumptions for the paired samples t -test? Provide a rationale for your
answer.

Yes, data measured is from only one
group of individuals with a normal distribution of scores. The dependent
variable is measured at an interval level. The only difference between the
groups is that the paired scores are independent.

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?

The shape of the distribution
between the two groups is leptokurtic.

 

Tests
of Normality

 

Kolmogorov-Smirnova

Shapiro-Wilk

Statistic

df

Sig.

Statistic

df

Sig.

PDbaseline

.286

10

.020

.885

10

.149

PDposttx

.224

10

.168

.911

10

.287

 

 

 

 

 

 

 

 

3. What are the means for the
baseline and posttreatment affective distress scores, respectively?

            The baseline mean is 5.3 and the post
treatment mean is 3.3

Paired
Samples Statistics

 

Mean

N

Std.
Deviation

Std.
Error Mean

Pair 1

PDbaseline

5.2000

10

.91894

.29059

PDposttx

3.3000

10

.94868

.30000

 

 

 

 

 

 

 

 

 

 

 

 

4. What is the paired samples t
-test value?

            The paired samples for the t test value is
10.59.

 

Paired
Samples Test

 

Paired
Differences

t

df

Sig.
(2-tailed)

Mean

Std.
Deviation

Std.
Error Mean

95%
Confidence Interval of the Difference

Lower

Upper

Pair 1

PDbaseline – PDposttx

1.90000

.56765

.17951

1.49393

2.30607

10.585

9

.000

 

 

 

 

 

 

 

 

 

 

 

 

5. Is the t -test signi?cant at ? =
0.05? Specify how you arrived at your answer.

Yes, at the point when the alpha is
resolved it relies upon how extreme the reactions are. The lower the alpha, the
more serious the consequences are so you need as precise an investigation as
could be expected under the circumstances. On the off chance that the alpha is
a higher number the less consequences the side effects. The study is conducted
on peridontitis which has side effects of a low severity level. Therefore, an
alpha of 0.05 is appropriate.

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?

Assuming a true null hypothesis, P< 0.001 would be the likelihood of such a t-test being obtained. 7. On average, did the affective distress scores improve or deteriorate over time? Provide a rationale for your answer. Overall the scores got worse as time progressed. There were only two individuals that improved. Only 20% of the individuals were better.  8. Write your interpretation of the results as you would in an APA-formatted journal. After performing t tests for both groups the study concluded that the group treated for peridontitis had a lower probing depth from baseline to post-treatment. The t value was 10.59 and P was <0.001. 9. What do the results indicate regarding the impact of the rehabilitation on emotional distress levels? There is a significant decrease in distress after treatment, indicating the treatment is effective. The t value is 2.89 and P is < 0.05 indicating that the null hypothesis can be rejected. 10. What are the weaknesses of the design in this example? Appears to be a pilot study with a low sample size. Although it will help in the creation of further, larger studies, the results of this study can only be interpreted with caution and are certainly not generalizable.