Given that,
population mean(u)=3
standard deviation, sigma =0.2803
sample mean, x =3.0598
number (n)=40
null, Ho: μ=3
alternate, H1: μ!=3
level of significance, alpha = 0.1
from standard normal table, two tailed z alpha/2 =1.645
since our test is two-tailed
reject Ho, if zo < -1.645 OR if zo > 1.645
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 3.0598-3/(0.2803/sqrt(40)
zo = 1.35
| zo | = 1.35
critical value
the value of |z alpha| at los 10% is 1.645
we got |zo| =1.35 & | z alpha | = 1.645
make decision
hence value of |zo | < | z alpha | and here we do not reject
Ho
p-value : two tailed ( double the one tail ) - ha : ( p != 1.35 ) =
0.18
hence value of p0.1 < 0.18, here we do not reject Ho
ANSWERS
---------------
null, Ho: μ=3
alternate, H1: μ!=3
test statistic: 1.35
critical value: -1.645 , 1.645
decision: do not reject Ho
p-value: 0.18
we do not have enough evidence to support the claim that average
tensile strength is 3 lbs/mm.
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