Question

In Born together—Reared apart: the Landmark Minnesota twin study (2012), Nancy Segal discusses the efforts of research psychologists at the University of Minnesota to understand similarities and differences between twins by studying sets of twins who were raised separately. The Excel Online file below contains critical reading SAT scores for several pairs of identical twins (twins who share all of their genes), one of whom was raised in a family with no other children (no siblings) and one of whom was raised in a family with other children (with siblings). Construct a spreadsheet to answer the following questions.

SAT Score No Siblings	SAT Score With Siblings
400	468
574	568
598	669
416	398
460	419
382	518
496	451
633	544
492	451
659	700
568	467
564	662
526	485
676	607
607	525
650	636
567	526
588	565
459	487
599	591

a. What is the mean difference between the critical reading SAT scores for the twins raised with no siblings and the twins raised with siblings?

(to 2 decimals)

b. Provide a 90% confidence interval estimate of the mean difference between the critical reading SAT scores for the twins raised with no siblings and the twins raised with siblings.

( , ) (to 2 decimals)

c. Conduct a hypothesis test of equality of the critical reading SAT scores for the twins raised with no siblings and the twins raised with siblings.

p-value is blank (to 4 decimals)

Answer 1

a.
mean(x)=545.7
standard deviation , s.d1=87.9522
number(n1)=20
y(mean)=536.85
standard deviation, s.d2 =86.6197
number(n2)=20
b.
TRADITIONAL METHOD
given that,
mean(x)=545.7
standard deviation , s.d1=87.9522
number(n1)=20
y(mean)=536.85
standard deviation, s.d2 =86.6197
number(n2)=20
I.
stanadard error = sqrt(s.d1^2/n1)+(s.d2^2/n2)
where,
sd1, sd2 = standard deviation of both
n1, n2 = sample size
stanadard error = sqrt((7735.589/20)+(7502.972/20))
= 27.603
II.
margin of error = t a/2 * (standard error)
where,
t a/2 = t -table value
level of significance, α = 0.1
from standard normal table, two tailed and
value of |t α| with min (n1-1, n2-1) i.e 19 d.f is 1.729
margin of error = 1.729 * 27.603
= 47.726
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (545.7-536.85) ± 47.726 ]
= [-38.876 , 56.576]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
mean(x)=545.7
standard deviation , s.d1=87.9522
sample size, n1=20
y(mean)=536.85
standard deviation, s.d2 =86.6197
sample size,n2 =20
CI = x1 - x2 ± t a/2 * Sqrt ( sd1 ^2 / n1 + sd2 ^2 /n2 )
where,
x1,x2 = mean of populations
sd1,sd2 = standard deviations
n1,n2 = size of both
a = 1 - (confidence Level/100)
ta/2 = t-table value
CI = confidence interval
CI = [( 545.7-536.85) ± t a/2 * sqrt((7735.589/20)+(7502.972/20)]
= [ (8.85) ± t a/2 * 27.603]
= [-38.876 , 56.576]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 90% sure that the interval [-38.876 , 56.576] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 90% of these intervals will contains the true population proportion
c.
Given that,
mean(x)=545.7
standard deviation , s.d1=87.9522
number(n1)=20
y(mean)=536.85
standard deviation, s.d2 =86.6197
number(n2)=20
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, α = 0.1
from standard normal table, two tailed t α/2 =1.729
since our test is two-tailed
reject Ho, if to < -1.729 OR if to > 1.729
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =545.7-536.85/sqrt((7735.58948/20)+(7502.97243/20))
to =0.321
| to | =0.321
critical value
the value of |t α| with min (n1-1, n2-1) i.e 19 d.f is 1.729
we got |to| = 0.32062 & | t α | = 1.729
make decision
hence value of |to | < | t α | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 0.3206 ) = 0.752
hence value of p0.1 < 0.752,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: 0.321
critical value: -1.729 , 1.729
decision: do not reject Ho
p-value: 0.752
we do not have enough evidence to support the claim that equality of the critical reading SAT scores for the twins raised with no siblings and the twins raised with sibling