Salmon (Raw Data, Software Required): Assume that the weights of Chinook Salmon

Question

Salmon (Raw Data, Software Required):
Assume that the weights of Chinook Salmon in the Columbia River are normally distributed. You randomly catch and weigh 15 such salmon. The data is found in the table below. Test the claim that the mean weight of Columbia River salmon is greater than 26 pounds. Test this claim at the 0.01 significance level.

(a) What type of test is this?

This is a left-tailed test.

This is a right-tailed test.    

This is a two-tailed test.

(b) What is the test statistic? Round your answer to 2 decimal places.

tx =

(c) Use software to get the P-value of the test statistic. Round to 4 decimal places.

P-value =  

(d) What is the conclusion regarding the null hypothesis?

reject H0

fail to reject H0    

(e) Choose the appropriate concluding statement.

The data supports the claim that the mean weight of Columbia River salmon is greater than 26 pounds.

There is not enough data to support the claim that the mean weight of Columbia River salmon is greater than 26 pounds.    

We reject the claim that the mean weight of Columbia River salmon is greater than 26 pounds.

We have proven that the mean weight of Columbia River salmon is greater than 26 pounds.

DATA ( n = 15 )
Salmon Weights

Pounds    26.2 25.3 36.5 27.1 19.4 23.0 33.7 26.3 24.0 27.3 33.0 28.2 29.3 28.8 32.0

Explanation / Answer

Given that,
population mean(u)=26
sample mean, x =28.0067
standard deviation, s =4.4544
number (n)=15
null, Ho: <26
alternate, H1: >26
level of significance, = 0.01
from standard normal table,right tailed t /2 =2.624
since our test is right-tailed
reject Ho, if to > 2.624
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =28.0067-26/(4.4544/sqrt(15))
to =1.745
| to | =1.745
critical value
the value of |t | with n-1 = 14 d.f is 2.624
we got |to| =1.745 & | t | =2.624
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value :right tail - Ha : ( p > 1.7448 ) = 0.05146
hence value of p0.01 < 0.05146,here we do not reject Ho
ANSWERS
---------------
null, Ho: =26
alternate, H1: >26
test statistic: 1.745
critical value: 2.624
decision: do not reject Ho
p-value: 0.05146

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