Given that,
population mean(u)=82
sample mean, x =73.9167
standard deviation, s =10.0946
number (n)=12
null, Ho: μ=82
alternate, H1: μ<82
level of significance, α = 0.1
from standard normal table,left tailed t α/2 =1.363
since our test is left-tailed
reject Ho, if to < -1.363
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =73.9167-82/(10.0946/sqrt(12))
to =-2.774
| to | =2.774
critical value
the value of |t α| with n-1 = 11 d.f is 1.363
we got |to| =2.774 & | t α | =1.363
make decision
hence value of | to | > | t α| and here we reject Ho
p-value :left tail - Ha : ( p < -2.7739 ) = 0.00905
hence value of p0.1 > 0.00905,here we reject Ho
ANSWERS
---------------
a.
T test for single mean with unknown population standard
deviation
b.
null, Ho: μ=82
alternate, H1: μ<82
c.
test statistic: -2.774
critical value: -1.363
d.
p-value: 0.00905
e.
p value is less than alpha value
f.
decision: reject Ho
g.
we have enough evidence to support the claim that population mean
score for students who are given colored pens at the begining of
class is lower than 82
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