Question

An educational researcher devised a wooden toy assembly project to test learning in 6-year-olds. The time in seconds to assemble the project was noted, and the toy was disassembled out of the child's sight. Then the child was given the task to repeat. The researcher would conclude that learning occurred if the mean of the second assembly times was less than the mean of the first assembly times. At =α0.05, can it be concluded that learning took place? Use the P-value method and a graphing calculator. Assume the variables are normally distributed. Let μ1 be the average of trial 1and =μD−μ1μ2.

Child	1	2	3	4	5	6	7
Trial 1	128	100	111	112	92	103	143
Trial 2	119	91	111	103	90	99	152

Find the P-value.

Reject or claim the hypothesis.

Is there or is there not enough evidence to support the claim?

Answer 1

Given that,
null, H0: Ud = 0
alternate, H1: Ud < 0
level of significance, α = 0.05
from standard normal table,left tailed t α/2 =1.943
since our test is left-tailed
reject Ho, if to < -1.943
we use Test Statistic
to= d/ (S/√n)
where
value of S^2 = [ ∑ di^2 – ( ∑ di )^2 / n ] / ( n-1 ) )
d = ( Xi-Yi)/n) = -3.429
We have d = -3.429
pooled variance = calculate value of Sd= √S^2 = sqrt [ 344-(-24^2/7 ] / 6 = 6.604
to = d/ (S/√n) = -1.374
critical Value
the value of |t α| with n-1 = 6 d.f is 1.943
we got |t o| = 1.374 & |t α| =1.943
make Decision
hence Value of |to | < | t α | and here we do not reject Ho
p-value :left tail - Ha : ( p < -1.3737 ) = 0.10933
hence value of p0.05 < 0.10933,here we do not reject Ho
ANSWERS
---------------
null, H0: Ud = 0
alternate, H1: Ud < 0
test statistic: -1.374
critical value: reject Ho, if to < -1.943
decision: Do not Reject Ho
p-value: 0.10933
we do not have enough evidence to support the claim that if the mean of the second assembly times was less than the mean of the first assembly times.