Question

A quality control manager at a plant wants to determine if the average width of bolts is different than 4 mm. A sample of 28 bolts yields sample mean,  mm and sample standard deviation, s = 0.6. The decision with α = 0.05 would be:

Answer 1

Assumed data,
samplemean is 3 mm because not given in the data
Given that,
population mean(u)=4
sample mean, x =3
standard deviation, s =0.6
number (n)=28
null, Ho: μ=4
alternate, H1: μ!=4
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =2.052
since our test is two-tailed
reject Ho, if to < -2.052 OR if to > 2.052
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =3-4/(0.6/sqrt(28))
to =-8.8192
| to | =8.8192
critical value
the value of |t α| with n-1 = 27 d.f is 2.052
we got |to| =8.8192 & | t α | =2.052
make decision
hence value of | to | > | t α| and here we reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != -8.8192 ) = 0
hence value of p0.05 > 0,here we reject Ho
ANSWERS
---------------
null, Ho: μ=4
alternate, H1: μ!=4
test statistic: -8.8192
critical value: -2.052 , 2.052
decision: reject Ho
p-value: 0
we have enough evidence to support the claim that if the average width of bolts is different than 4 mm.