Question

Homser, Lake Oregon has an Atlantic salmon catch-and-release program that has been very successful. The average Fisherman's catch has been on average 8.8 Atlantic salmon per day (Source: National Symposium on catch and release fishing, Humboldt State University). Suppose that a new quota system restricting the number of fishermen has been put into effect this season. A random sample of 16 fishermen yielded a mean and standard deviation of and s = 3.45 catches per day. The histogram to the right shows the sample distribution. Assuming the catch per day has an approximate normal distribution, use a significance level of 0.05 to test whether the average is different than 8.8 now that the restriction has been put in place. ()

1) a. What is the p-value for the test? Is it one or two sided?

b. Calculate a 95% confidence interval for µ. Show all work

c. Interpret the results from the hypothesis test and confidence interval in the context of the problem.

Answer 1

Let , the sample mean be .

To calculate the P value, we need to compute the test statistic first. Since the population standard deviation is unknown to us and the sample size is not sufficiently large we will be using single sample t test for mean to test the given hypothesis.

The formula for the test statistic is given by :

where, n = sample size = 16 ; s = sample standard deviation = 3.45 ; ;

So, (I am unable to compute the value of T as the sample mean is not mentioned in the question.Please mention it in the comments section  so that I can do it for you, and I will definitely edit/upade the final answer here. Otherwise you can do it by yourself by putting the sample mean value in the formula given above). Let the computed value of T is t.

Then, P-value is : 2*{P( T> t)} . Since T follows a student's t distribution with degrees of freedom (n-1)=(16-1)=15, this probabilty 2*{P(T>t)}  can be calculated using excel's TDIST(x,deg_freedom, tails) where x = (put the computed T value) , deg_freedom = 15 , tails = 2 (since this is a two tailed test) . By enttering this you will get the p-value.

This is a two sided p-value, because we are testing the hypothesis that against    .

b) The formula for the 95% confidence interval is :

(please put the value of sample mean to get the final answer)

c) If the computed P-value in above exceeds 0.05 and the confidence interval contains the value 8.8, conclude that there is sufficient evidence to warrant the rejection of the claim that  the average is different than 8.8 now aftert the restriction has been put in place.

  If the computed P-value in above is less than 0.05 and the confidence interval does not contain the value 8.8, conclude that , there is insufficient evidence to reject the claim that  the average is different than 8.8 now aftert the restriction has been put in place.

If you are having any trouble doing the above by yourself , please mention the sample mean value in the comment, I will do it for you and update my answer. Thank you so much!