Question

How can I make a problem like this simpler? and what equation do I use for part D? Thanks!

A researcher wonders if mental simulation will help one more successfully achieve one’s goals. To test this idea, the researcher recruited six students from a general psychology class, several weeks before their first midterm. After their first midterm, the researcher got the students’ scores on this test from their professor. Several weeks later, each of the students was brought into the lab where they completed a mental simulation exercise. In this exercise, students were asked to imagine themselves preparing to study by turning off the radio and TV, then sitting at one’s desk with no distractions around, and studying for several hours. They were asked to imagine themselves going through this process several times. Finally, they were asked to simulate taking the test. The students were asked to repeat this simulation exercise at least once daily until their next test. After the students took their second test, the researcher compared their scores to those they received on the first test. The students’ scores on their tests are presented below:

Student	Test Score(No simulation)	Test(Simulation)
1	80	92
2	79	86
3	73	76
4	91	96
5	90	94
6	85	93

a. What is the null hypothesis?

b. What is the research hypothesis?

c. What is the critical value of your test statistic? Be sure to specify df.

d. What is the obtained value of your test statistic?

e. What is your statistical conclusion?

f. Provide a substantive conclusion.

Answer 1

Given that,
null, H0: Ud = 0
alternate, H1: Ud != 0
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =2.571
since our test is two-tailed
reject Ho, if to < -2.571 OR if to > 2.571
we use Test Statistic
to= d/ (S/√n)
where
value of S^2 = [ ∑ di^2 – ( ∑ di )^2 / n ] / ( n-1 ) )
d = ( Xi-Yi)/n) = -6.5
We have d = -6.5
pooled variance = calculate value of Sd= √S^2 = sqrt [ 307-(-39^2/6 ] / 5 = 3.271
to = d/ (S/√n) = -4.867
critical Value
the value of |t α| with n-1 = 5 d.f is 2.571
we got |t o| = 4.867 & |t α| =2.571
make Decision
hence Value of | to | > | t α| and here we reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != -4.8674 ) = 0.0046
hence value of p0.05 > 0.0046,here we reject Ho
ANSWERS
---------------
a.
null, H0: Ud = 0
alternate, H1: Ud != 0
b.
the research hypothesis
T test for paired difference
d.
test statistic: -4.867
c.
critical value: reject Ho, if to < -2.571 OR if to > 2.571
decision: Reject Ho
p-value: 0.0046
e.
at level of significance =0.05,there is statistically significant differently.
f.
we have enough evidence to support the claim that difference of means between test score( no simulation) and test(simulation).