SOLUTIONS

6.9 EXERCISES

6.27

(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

participants.

(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

statistic.

6.28

From the data we give the following table that we help us in

calculations.

SN

Before treatment

After treatment

1

22.86

6.11

16.75

2

7.74

-4.02

11.76

3

15.49

8.04

7.45

4

9.97

3.29

6.68

5

1.44

-0.77

2.21

6

9.39

6.99

2.4

7

11.4

10.19

1.21

8

1.86

2.09

-0.23

9

-6.71

11.4

-18.11

10

6.42

10.7

-4.28

Sample mean()

2.584

Sample Standard deviation ()

9.4907

(a)

Hypothesis to test

Where

is mean difference.

Test statistic

=0.8610

Critical value

The test is left tailed so the rejection

region is given by

We use the chart to get

Rejection region

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.

(b)

Size of error is given

Thus the confidence interval for mean difference is

(c)

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.

6.29

(a)

Test hypothesis

Test statistic for mean difference test

p-value for the mean difference test will

be

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.