(a) We will carry out the study on the
effect of two different treatments in that we choose 30 participants
independently and randomly in that they have similar type of body fat, weight,
height, age and habits in order to avoid the other factors of variability on
impact of treatments of which we assign two treatments randomly to these 30 participants.
(b) We will reduce
all other causes of variation in order to choose 15 pairs of the participants
such that they are matched to the greatest extent based on the body structure,
diet habits, age and general health before the treatment. And once the 15 pairs
are configured the two treatments will be randomly assigned with each pair of
(c) In case we have huge difference in the participants
regarding their body structure, age, health and diet habits before the
treatment and the paired units have strong positive correlation than the paired
procedure would likely to have more effective. On contrary not much difference
in the characteristics of the participants and they are quiet similar and we
have prior information about it then the independent samples procedure would be
more powerful as the test statistic of independent samples procedure has degree
of freedom twice as many as the degree of freedom of paired producer’s test
From the data we give the following table that we help us in
Sample Standard deviation ()
Hypothesis to test
is mean difference.
The test is left tailed so the rejection
region is given by
We use the chart to get
The value of statistic is falling in
acceptance region hence there is no sufficient evidence to reject null hypothesis.
Hence there is no significant decrement in mean SENS value after treatment.
Size of error is given
Thus the confidence interval for mean difference is
We draw box plot for the mean differences using excel
The Box plot is not symmetric hence the
condition for normality is not satisfied.
Therefore, the assumption of using t-test
procedure does not appear to be valid.
Test statistic for mean difference test
p-value for the mean difference test will
is very small for the null hypothesis to be
true so there is sufficient evidence to reject null hypothesis and to accept
the claim that there is significant difference in the mean final grade between
the students in an academically oriented home environment and those in a
non-academically oriented home environment.