researcher is interested in comparing student exam scores in a class wheret earn
ID: 3322758 • Letter: R
Question
researcher is interested in comparing student exam scores in a class wheret earn from a traditional textbook and one where they learn using only a CD-ROM. Two classes consisting of randomly selected students are used in the same semester. The researcher is curious whether the scores differ, but that these data are normal and have equal variances. Select the appropriate statistica null and alternative hypotheses that are appropriate, test the data, and draw a conclusion. 1Do methods differ? Hand calculate this test. as to does not have any prior idea of which method is better. You may assume l test, state the Final exam scores (in percents): Standard textbook: 95, 87, 86, 85, 82, 80, 80, 75, 72, 71, 70, 70, 69, 63 CD-ROM: 100, 85, 82, 81, 80, 80, 79, 75, 74, 73, 72, 65, 63, 62, 58Explanation / Answer
Given that,
mean(x)=123.9286
standard deviation , s.d1=172.0695
number(n1)=14
y(mean)=75.2667
standard deviation, s.d2 =10.6601
number(n2)=15
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, = 0.05
from standard normal table, two tailed t /2 =2.05
since our test is two-tailed
reject Ho, if to < -2.05 OR if to > 2.05
calculate pooled variance s^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
s^2 = (13*29607.9128 + 14*113.6377) / (29- 2 )
s^2 = 14314.585
we use test statistic (t) = (x-y)/sqrt(s^2(1/n1+1/n2))
to=123.9286-75.2667/sqrt((14314.585( 1 /14+ 1/15 ))
to=48.6619/44.4609
to=1.0945
| to | =1.0945
critical value
the value of |t | with (n1+n2-2) i.e 27 d.f is 2.05
we got |to| = 1.0945 & | t | = 2.05
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - ha : ( p != 1.0945 ) = 0.2831
hence value of p0.05 < 0.2831,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: 1.0945
critical value: -2.05 , 2.05
decision: do not reject Ho
p-value: 0.2831
the methods are don't differ
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