A study of computer-assisted learning examined the learning of \"Blissymbols\" b
ID: 3361601 • Letter: A
Question
A study of computer-assisted learning examined the learning of "Blissymbols" by children. Blissymbols are pictographs (think of Egyptian hieroglyphs) that are sometimes used to help learning-impaired children communicate. The researcher designed two computer lessons that taught the same content using the same examples. One lesson required the children to interact with the material, while in the other the children controlled only the pace of the lesson. Call these two styles "Active" and "Passive." Children were assigned at random to Active and Passive groups. After the lesson, the computer presented a quiz that asked the children to identify 56 Blissymbols. Here are the numbers of correct identifications by the 24 children in the Active group: The 24 children in the Passive group had these counts of correct identifications: Is there good evidence that active learning is superior to passive learning? FORMULATE: What hypotheses should be tested? HA:a<P HA:aP HA:a>P
Explanation / Answer
Given that,
sample one, x1 =24, n1 =56, p1= x1/n1=0.429
sample two, x2 =24, n2 =56, p2= x2/n2=0.429
null, Ho: p1 = p2
alternate, H1: p1 > p2
level of significance, = 0.05
from standard normal table,right tailed z /2 =1.645
since our test is right-tailed
reject Ho, if zo > 1.645
we use test statistic (z) = (p1-p2)/(p^q^(1/n1+1/n2))
zo =(0.429-0.429)/sqrt((0.429*0.571(1/56+1/56))
zo =0
| zo | =0
critical value
the value of |z | at los 0.05% is 1.645
we got |zo| =0 & | z | =1.645
make decision
hence value of |zo | < | z | and here we do not reject Ho
p-value: right tail - Ha : ( p > 0 ) = 0.5
hence value of p0.05 < 0.5,here we do not reject Ho
ANSWERS
---------------
null, Ho: p1 = p2
alternate, H1: p1 > p2
test statistic: 0
critical value: 1.645
decision: do not reject Ho
p-value: 0.5
we do not have enough evidence that active learning is superior to passive learning
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