HW Problem 46. (5 pts) The owner of a high-end fishing camp on Gunflint Lake, wh

Question

HW Problem 46. (5 pts) The owner of a high-end fishing camp on Gunflint Lake, which lies along the Minnesota/Ontario border, has claims that the average Lake Trout in Gunflint Lake are bigger in terms of weight than in any other lake in North America. The owner of a rival fishing camp situated on Cree Lake, located in northern Saskatchewan, claims the Lake Trout in Cree lake are bigger. A friend of the two owners commissions a study to settle this most important question. A random sample of 42 Lake Trout caught in Gunflint Lake results in a sample mean weight of 24.2 pounds with a sample standard deviation of 4.9 pounds. A random sample of 43 Lake Trout caught in Cree Lake results in a sample mean weight of 26.3 pounds with a sample standard deviation of 4.8 pounds. At the .01 level of significance, test to determine if there's a difference in the mean weight of Lake Trout between Gunflint and Cree lakes. List and clearly label all eight steps.

Explanation / Answer

H0: 1 - 2 = 0 i.e. (1 = 2)

H1: 1 - 2 0 i.e. (1 2)

Assuming population variances are equal, we would have to calculate pooled-variance t-Test keeping Lake Trout from Gunflint Lake as population 1 and Lake Trout from Cree Lake as population 2: -

Sp^2 = (n1-1)S1^2+(n2-1)S2^2/(n1-1)+(n2-1)

         = (42-1)*4.9^2+(43-1)*4.8^2/41+42

         = 984.41+967.68/83

         = 23.52

tSTAT=(X1-X2)-(µ1-µ2)/Sp^2(1/n1+1/n2)

       =(24.2-26.3)-0/23.52(1/42+1/43)

       =-2.1/1.05

       =-1.996

tCRIT is +/-2.6363 and hence we cannot reject the null hypothesis (because computed value does not lie in the rejection region). We have evidence to prove that the mean weight of Lake Trout from both the lakes is the same.

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