The p-value was slightly above conventional threshold, but was described as “rap
ID: 3305353 • Letter: T
Question
The p-value was slightly above conventional threshold, but was described as “rapidly approaching significance” (i.e., p =.06). An independent samples t test was used to determine whether student satisfaction levels in a quantitative reasoning course differed between the traditional classroom and on-line environments. The samples consisted of students in four face-to-face classes at a traditional state university (n = 65) and four online classes offered at the same university (n = 69). Students reported their level of satisfaction on a fivepoint scale, with higher values indicating higher levels of satisfaction. Since the study was exploratory in nature, levels of significance were relaxed to the .10 level. The test was significant t(132) = 1.8, p = .074, wherein students in the face-to-face class reported lower levels of satisfaction (M = 3.39, SD = 1.8) than did those in the online sections (M = 3.89, SD = 1.4). We therefore conclude that on average, students in online quantitative reasoning classes have higher levels of satisfaction. The results of this study are significant because they provide educators with evidence of what medium works better in producing quantitatively knowledgeable practitioners.
Explanation / Answer
Given that,
mean(x)=3.39
standard deviation , s.d1=1.8
number(n1)=65
y(mean)=3.89
standard deviation, s.d2 =1.4
number(n2)=69
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, = 0.1
from standard normal table, two tailed t /2 =1.669
since our test is two-tailed
reject Ho, if to < -1.669 OR if to > 1.669
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =3.39-3.89/sqrt((3.24/65)+(1.96/69))
to =-1.787
| to | =1.787
critical value
the value of |t | with min (n1-1, n2-1) i.e 64 d.f is 1.669
we got |to| = 1.7874 & | t | = 1.669
make decision
hence value of | to | > | t | and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != -1.7874 ) = 0.079
hence value of p0.1 > 0.079,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: -1.787
critical value: -1.669 , 1.669
decision: reject Ho
p-value: 0.079
we have enough evidence to support the claim
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