Question

2. *(15 points). For these data, you will first need to test for equality of variances, and then apply the appropriate two-sample test to test differences between means. A group of students are randomly assigned to one of two classes. Class 1 uses a traditional lecture- and discussion method of teaching biochemistry. Class 2 uses the "Keller Plan," a self- paced method where students study units of biochemistry independently and then take short tests on each unit. At the end of the semester, both classes take a common final exam. Does the Keller Plan group perform better on the final than the traditional group? Class 1 78 Class 2 68 anriate test to determine if

Answer 1

2.

Given that,
mean(x)=73.5333
standard deviation , s.d1=5.5661
number(n1)=15
y(mean)=75.8
standard deviation, s.d2 =5.7221
number(n2)=15
null, Ho: u1 = u2
alternate, H1: u1 > u2
level of significance, α = 0.05
from standard normal table,right tailed t α/2 =1.7
since our test is right-tailed
reject Ho, if to > 1.7
calculate pooled variance s^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
s^2 = (14*30.9815 + 14*32.7424) / (30- 2 )
s^2 = 31.8619
we use test statistic (t) = (x-y)/sqrt(s^2(1/n1+1/n2))
to=73.5333-75.8/sqrt((31.8619( 1 /15+ 1/15 ))
to=-2.2667/2.0611
to=-1.0997
| to | =1.0997
critical value
the value of |t α| with (n1+n2-2) i.e 28 d.f is 1.7
we got |to| = 1.0997 & | t α | = 1.7
make decision
hence value of |to | < | t α | and here we do not reject Ho
p-value: right tail -ha : ( p > -1.0997 ) = 0.85959
hence value of p0.05 < 0.85959,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 > u2
test statistic: -1.0997
critical value: 1.7
decision: do not reject Ho
p-value: 0.85959
we do not have enough evidence to support the claim that keller plan group perform better
on the final than the traditional group.