A study of computer-assisted learning examined the learning of \"Blissymbols\" b

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? SOLVE: What is the P-value? 0.0025 < P-value < 0.005 P-value < 0.0005 0.001 < P-value < 0.0025 0.0005 < P-value < 0.001

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, = 0.001
from standard normal table,right tailed z /2 =3.09
since our test is right-tailed
reject Ho, if zo > 3.09
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 0.001% is 3.09
we got |zo| =0 & | z | =3.09
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 p0.001 < 0.5,here we do not reject Ho
ANSWERS
---------------
null, Ho: p1 = p2
alternate, H1: p1 > p2
test statistic: 0
critical value: 3.09
decision: do not reject Ho
p-value: 0.5

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, = 0.005
from standard normal table,right tailed z /2 =2.576
since our test is right-tailed
reject Ho, if zo > 2.576
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 0.005% is 2.576
we got |zo| =0 & | z | =2.576
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 p0.005 < 0.5,here we do not reject Ho
ANSWERS
---------------
null, Ho: p1 = p2
alternate, H1: p1 > p2
test statistic: 0
critical value: 2.576
decision: do not reject Ho
p-value: 0.5

we do not have enough evidence that active learning is superior to passive learning
level of significance is 0.005,0.0025,0.001, 0.0005 changes and also critical values changes but P value doesnot change in this problem

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