Question

There is no question that both nature (genetics) and nurture (environment) play a role in human characteristics and behavior. One way to tease apart the effect of nature and the effect of nurture is to study children who have been adopted at birth. These children are unique in that they received the genetic influence from their birth parents and the environmental influence from their adoptive parents. For example, a research study can be conducted to see if the amount of time a child spends watching TV is more correlated with the TV-watching time of his/her birth parent or adoptive parent. If the TV-watching time of the child is more correlated with the birth parent’s TV-watching time, it means that the propensity to watch TV might be genetically determined to a larger extent compared to environmental influence. If it’s more correlated with the adoptive parent’s TV-watching time, then it means that the propensity to watch TV might be influenced by the behavior of the adoptive parent to a larger extent.

Use the following dataset to answer the questions below:

	Amount of Time Spent Watching Television (# of hours per day)
Subject ID #	Child	Child’s Birth Parent	Child’s Adoptive Parent
1	4	3	6
2	3	1	5
3	5	4	7
4	2	6	3
5	3	2	3
6	1	2	1
7	5	3	6
8	2	2	1
9	3	3	4
10	2	4	4

Calculate the Z scores for all the data points for the children and their birth parents. (please show work for at least a few so I can understand how to do them)

Explain the direction and strength of the relationship based on the r.

Calculate the Z scores for all the data points for the adoptive parents. (please show work for at least a few so I can understand how to do them)

Explain the direction and strength of the relationship based on the r.

Answer 1

Subject ID #	Child	Child mean	Child sd	Zcore=X-mean/sd	Child’s Birth Parent	Chid' Birth parent mean	Child's Birth Parent sd	Child'Birth Parent Z score	Child’s Adoptive Parent	Childs Adopive parents mean	Childs Adpotive parent sd	Z score Childs Adoptive Parent
1	4	3	1.333333	0.75000019	3	3	1.41421356	0	6	4	2.054805	0.973328
2	3	3	1.333333	0	1	3	1.41421356	-1.414214	5	4	2.054805	0.486664
3	5	3	1.333333	1.50000038	4	3	1.41421356	0.707107	7	4	2.054805	1.459993
4	2	3	1.333333	-0.75000019	6	3	1.41421356	2.12132	3	4	2.054805	-0.486664
5	3	3	1.333333	0	2	3	1.41421356	-0.707107	3	4	2.054805	-0.486664
6	1	3	1.333333	-1.50000038	2	3	1.41421356	-0.707107	1	4	2.054805	-1.459993
7	5	3	1.333333	1.50000038	3	3	1.41421356	0	6	4	2.054805	0.973328
8	2	3	1.333333	-0.75000019	2	3	1.41421356	-0.707107	1	4	2.054805	-1.459993
9	3	3	1.333333	0	3	3	1.41421356	0	4	4	2.054805	0
10	2	3	1.333333	-0.75000019	4	3	1.41421356	0.707107	4	4	2.054805	0

To get Z score

find mean

find standard deviation

which be done using Excel function

=average(datarane)

=stddev.s(datarange)

Find Z score as

Z=x-mean/sd

for subject id 1

Child=4

mean=3

stddev=1.33

Z=x-mean/sd

=4-3/1.33

=1/1.33

=0.7500

similarly calculated others and tabulated above

.Using Data >Data analysis >Correlation in Excel

Correlation matrix can be found as

	Child	Child’s Birth Parent
Child	1
Child’s Birth Parent	0.058926	1

there exists a weak positive relation ship between child and childs birth parent

r=0.0589

Direction :Positive

Form Linear

STrength Weak

	Child	Child’s Adoptive Parent
Child	1
Child’s Adoptive Parent	0.892217816	1

r=0.8922

there exists a strong positive relationship btween child and Childs Adoptive parent

Form :Linear

Strength: Strong

Direction: positive