Find the P value

Question

Homser Lake, Oregon, has an Atlantic salmon catch and release program that has been very successful. The average fisherman's catch has been 𝜇 = 5.7 Atlantic salmon per day. Suppose that a new quota system restricting the number of fishermen has been put into effect this season. A random sample of fishermen gave the following catches per day. Assuming the catch per day has an approximately normal distribution, use a 1% level of significance to test the claim that the population average catch per day is now different from 5.7.

12	6	11	12	5	0	2
7	8	7	6	3	12	12

A) Find the P-Value. (Express to 4 decimal places)

B)State your conclusion in the context of the application.

--Fail to reject the null hypothesis, there is sufficient evidence that the catch is different from 5.7 fish per day.

--Reject the null hypothesis, there is insufficient evidence that the catch is different from 5.7 fish per day.    

--Reject the null hypothesis, there is sufficient evidence that the catch is different from 5.7 fish per day.

--Fail to reject the null hypothesis, there is insufficient evidence that the catch is different from 5.7 fish per day.

Answer 1

A. Based on the given information, a one-sample t test is best because you have a population mean (𝜇 ) of 5.7 to which you want to compare your sample mean. A z-test was also considered, but we are not given σ. However, we can calculate standard deviation (s) because we have a sample. As such, a one-sample t-test is most appropriate. The formula for t = (M-𝜇)/s_M. Based on this equation, you are given 𝜇= 5.7. You can calculate M, which is the mean number of fish caught this season. You calculate that by summing up all the data and divide by the number of fishermen (14).

M = (12+7+6+8+11+7+12+6+5+3+0+12+2+12)/14 = 103/14 = 7.357142857 (I tend not to round numbers until I have finished my calculations.)

s_M = s/

=14

s= standard deviation = = you take each X value and subtract it from the mean then square it and add up all the squared differences. Then, you divide by 13 (which is n-1) and take the square root of the whole thing.

(The calculation is pretty complicated, so I won't show it all.) You should end up with s = 4.030788105.

s_M = 4.030788105/ =1.077273435

t= (7.357142857-5.7)/1.077273435 = 1.538

Now, we look up 1.538 in the t-distribution table for two-tailed test with n-1 = 13 degrees of freedom. For alpha = 0.01, critical t = 3.0123, and p = 0.1480.

B. Because p > 0.01 and calculated t of 1.538 < critical t of 3.0123, we fail to reject the null hypothesis, there is insufficient evidence that the catch is different from 5.7 fish per day.