In: Math
A child development specialist is interested in learning if a new learning program increases students’ memory. 15 Subjects learned a list of 50 words. Learning performance was measured using a recall test. Students were initially tested and then tested again after using the new program. Below is the number of words remembered by each student.
Student # Score 1 Score 2
1 24 26
2 17 24
3 32 31
4 14 17
5 16 17
6 22 25
7 26 25
8 19 24
9 19 22
10 22 23
11 21 26
12 25 28
13 16 19
14 24 23
15 18 22
Did the learning program significantly improve the student’s ability to recall words? Report standard error of means, df, obtained and critical t, and whether you would accept or reject the null hypothesis.
Given that,
mean(x)=21
standard deviation , s.d1=4.7359
number(n1)=15
y(mean)=23.4667
standard deviation, s.d2 =3.8148
number(n2)=15
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =2.145
since our test is two-tailed
reject Ho, if to < -2.145 OR if to > 2.145
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =21-23.4667/sqrt((22.42875/15)+(14.5527/15))
to =-1.57
| to | =1.57
critical value
the value of |t α| with min (n1-1, n2-1) i.e 14 d.f is 2.145
we got |to| = 1.57098 & | t α | = 2.145
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 != -1.571 )
= 0.139
hence value of p0.05 < 0.139,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: -1.57
critical value: -2.145 , 2.145
standard error of means = s.d1/sqrt(n) = 4.7359/sqrt (15) =
1.222
standard error of means = s.d2/sqrt(n) = 3.8148/sqrt(15)
=0.9849
decision: do not reject Ho
p-value: 0.139
we do not have enough evidence to support the claim that the
learning program significantly improve the student’s ability to
recall words