In: Statistics and Probability
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.
(b) What is the test statistic? Round your answer to 2 decimal places. t-x= ? (c) Use software to get the P-value of the test statistic. Round to 4 decimal places. P-value = ? |
DATA ( n = 15 ) Salmon Weights
|
here, as the population sd is unknown , we will do 1 sample t test for mean.
hypothesis:-
necessary calculations:-
Pounds() | |
25.7 | 3.264165 |
24.8 | 7.326225 |
36 | 72.13614 |
26.6 | 0.822105 |
18.9 | 74.07528 |
22.5 | 25.06704 |
33.2 | 32.41366 |
25.8 | 2.912825 |
23.5 | 16.05364 |
26.8 | 0.499425 |
32.5 | 24.93304 |
27.7 | 0.037365 |
28.8 | 1.672625 |
28.3 | 0.629325 |
31.5 | 15.94644 |
sum= 412.6 | sum = 277.789 |
sample size (n) = 15
test statistic be:-
degrees of freedom = (n-1) = (15-1) = 14
p value = 0.1056
[ using software for, t = 1.31, df = 14, one tailed test ]
decision:-
p value = 0.1056 >0.01 ()
so, we do not have enough evidence to reject the null hypothesis.
conclusion:-
there is not sufficient evidence to support the claim that the mean weight of Columbia River salmon is greater than 26 pounds at 0.01 level of significance.
*** if you face any trouble to understand the answer to the problem please mention it in the comment box.if you are satisfied, please give me a LIKE if possible.