Not sure of my answers
The whole data is the little table at the right upper corner
Given that,
population mean(u)=95
sample mean, x =98
standard deviation, s =5.7619
number (n)=6
null, Ho: μ=95
alternate, H1: μ!=95
level of significance, alpha = 0.1
from standard normal table, two tailed t alpha/2 =2.015
since our test is two-tailed
reject Ho, if to < -2.015 OR if to > 2.015
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =98-95/(5.7619/sqrt(6))
to =1.275
| to | =1.275
critical value
the value of |t alpha| with n-1 = 5 d.f is 2.015
we got |to| =1.275 & | t alpha | =2.015
make decision
hence value of |to | < | t alpha | and here we do not reject
Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != 1.2754 )
= 0.2582
hence value of p0.1 < 0.2582,here we do not reject Ho
ANSWERS
---------------
a.
option:D
b.
null, Ho: μ=95
alternate, H1: μ!=95
c.
test statistic: 1.275
critical value: -2.015 , 2.015
decision: do not reject Ho
d.
p-value: 0.2582
e.
we do not have enough evidence to support the claim that avearge of
trap spacing measurements
for the population of lobster fishermen in the area differ from the
95.
Not sure of my answers The whole data is the little table at the right upper...