Someone goes fishing and catches a bucket of fish of the same species. Their wei

Question

Someone goes fishing and catches a bucket of fish of the same species. Their weights are: W_fish = [1.25, 2, 1.75, 1, 1.3, 1.8, 1.7, 1.2, 1.5] Looking at pictures of fish, the expected weight of the fish in the image of a webpage is stated to be 1.65. What does the degrees of freedom refer to in the Student-T distribution? When would we use the student-T test over the Z score? What is the t-test equation? Using the t-test can you test the weights of the fish caught support or reject that these fish belong to the same species seen in the catalogue?

Explanation / Answer

Given that,
population mean(u)=1.65
sample mean, x =1.5
standard deviation, s =0.3326
number (n)=9
null, Ho: =1.65
alternate, H1: !=1.65
level of significance, = 0.02
from standard normal table, two tailed t /2 =2.896
since our test is two-tailed
reject Ho, if to < -2.896 OR if to > 2.896
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =1.5-1.65/(0.3326/sqrt(9))
to =-1.353
| to | =1.353
critical value
the value of |t | with n-1 = 8 d.f is 2.896
we got |to| =1.353 & | t | =2.896
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != -1.353 ) = 0.2131
hence value of p0.02 < 0.2131,here we do not reject Ho
ANSWERS
---------------
null, Ho: =1.65
alternate, H1: !=1.65
test statistic: -1.353
critical value: -2.896 , 2.896
decision: do not reject Ho
p-value: 0.2131

a.
degrees o freedom is the value of 'n' we check in t-table
for the given level of significance
b.
t - test
c.
(t) = x-u/(s.d/sqrt(n))
d.
Yes

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