1. Weinstein, McDermott, and Roediger (2010) report that students who were given

Question

1.   Weinstein, McDermott, and Roediger (2010) report that students who were given questions to be answered while studying new material had better scores when tested on the material compared to students who were simply given an opportunity to reread the material. In a similar study, an instructor in a psychology class gave one group of students (n = 15) questions to be answered while studying for the final exam. The remaining students (n = 15) were instructed to reread the materials in the chapter. A t-value of 3.46 was computed.

A.   State the null and the alternative hypothesis (using words and symbols)
B.   Set the decision criteria
C.   Use the t-value provided (you don’t need to do calculations).
D.   Make the decision (fail to reject or reject the null hypothesis at either or both the .01 or .05 alpha level)

Explanation / Answer

Q1.
PART A.
When LOS = 0.01
null, Ho: u1 = u2
alternate, ew material had better scores compared to students to who given an opportunity to reread the material, H1: u1 > u2
level of significance, = 0.01
from standard normal table,right tailed t /2 =2.624
since our test is right-tailed
reject Ho, if to > 2.624
Critical Value
The Value of t at 0.01 LOS with n-1 = 14 is +2.624          
P-Value : Right Tail - Ha :( P > 3.46) = 0.0019          
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 > u2
test statistic: 2.624
critical value: 1.761
decision: do not reject Ho
p-value: 0.0019  

PART B.
When LOS = 0.05
null, Ho: u1 = u2
alternate, ew material had better scores compared to students to who given an opportunity to reread the material, H1: u1 > u2
level of significance, = 0.05
from standard normal table,right tailed t /2 =1.761
since our test is right-tailed
reject Ho, if to > 1.761
Critical Value
The Value of t at 0.05 LOS with n-1 = 14 is +1.761                      
P-Value : Right Tail - Ha :( P > 3.46) = 0.0019          
hence value of p0.05 < 0.1033,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 > u2
test statistic: 3.46
critical value: 1.761
decision: do not reject Ho
p-value: 0.0019  

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