Given that,
population mean(u)=20
sample mean, x =18.2
standard deviation, s =4.3
number (n)=16
null, Ho: μ=20
alternate, H1: μ<20
level of significance, α = 0.05
from standard normal table,left tailed t α/2 =1.753
since our test is left-tailed
reject Ho, if to < -1.753
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =18.2-20/(4.3/sqrt(16))
to =-1.6744
| to | =1.6744
critical value
the value of |t α| with n-1 = 15 d.f is 1.753
we got |to| =1.6744 & | t α | =1.753
make decision
hence value of |to | < | t α | and here we do not reject
Ho
p-value :left tail - Ha : ( p < -1.6744 ) = 0.05738
hence value of p0.05 < 0.05738,here we do not reject Ho
ANSWERS
---------------
a.
null, Ho: μ=20
alternate, H1: μ<20
test statistic: -1.6744
critical value: -1.753
decision: do not reject Ho
b.
option:B
left tail area is shown in the figure.
p-value: 0.05738
option :C
p value is between 0.05 to 0.10
d.
we do not have enough evidence to support the claim that mean is
less than 20.
option:C
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