21.
Given that,
population mean(u)=15.5
sample mean, x =15.42
standard deviation, s =0.16
number (n)=17
null, Ho: μ=15.5
alternate, H1: μ!=15.5
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =2.12
since our test is two-tailed
reject Ho, if to < -2.12 OR if to > 2.12
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =15.42-15.5/(0.16/sqrt(17))
to =-2.062
| to | =2.062
critical value
the value of |t α| with n-1 = 16 d.f is 2.12
we got |to| =2.062 & | t α | =2.12
make decision
hence value of |to | < | t α | and here we do not reject
Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != -2.0616 )
= 0.0559
hence value of p0.05 < 0.0559,here we do not reject Ho
ANSWERS
---------------
i.
null, Ho: μ=15.5
alternate, H1: μ!=15.5
test statistic: -2.062
critical value: -2.12 , 2.12
decision: do not reject Ho
p-value: 0.0559
we donot have enough evidence to support the claim that if true
mean number of parts produced each hour by the overhaul isnot equal
to 15.5.
ii.
Given that,
population mean(u)=15.5
sample mean, x =15.42
standard deviation, s =0.16
number (n)=17
null, Ho: μ=15.5
alternate, H1: μ<15.5
level of significance, α = 0.05
from standard normal table,left tailed t α/2 =1.746
since our test is left-tailed
reject Ho, if to < -1.746
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =15.42-15.5/(0.16/sqrt(17))
to =-2.062
| to | =2.062
critical value
the value of |t α| with n-1 = 16 d.f is 1.746
we got |to| =2.062 & | t α | =1.746
make decision
hence value of | to | > | t α| and here we reject Ho
p-value :left tail - Ha : ( p < -2.0616 ) = 0.02794
hence value of p0.05 > 0.02794,here we reject Ho
ANSWERS
---------------
null, Ho: μ=15.5
alternate, H1: μ<15.5
test statistic: -2.062
critical value: -1.746
decision: reject Ho
p-value: 0.02794
we have enough evidence to support the claim that if true mean
number of parts produced each hour by the overhaul is less than
15.5
Answer:
True
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