3. Walleye Trouble. Walleye is a fish that is native to the northern United Stat

Question

3. Walleye Trouble. Walleye is a fish that is native to the northern United States and Canada and is known for its excellent taste. It is especially popular in Minnesota where it is deemed their state fish. As a result it is often overfished in Minnesota lakes, especially Lake Mille Lacs. One way to measure the health of the population is through the average length of the fish because fish grow their entire lives. The DNR considers the walleye population in a lake to be strong if the mean length of walleye in the lake is more than 17 inches. The DNR conducted a fish survey and netted 660 walleye and calculated an average length of 17.5 inches and a standard deviation of 4.68 inches. Is this sufficient evidence to say the walleye population in Lake Mille Lacs is strong?

(a) Identify µ in this scenario. (Choose One) • Length of a fish in the lake. • Average length of all fish in the lake. • Average length of the fish in the sample. • Length of the fish in the lake.

(b) Conduct a hypothesis test of the above situation, using a significance level of =0.05. i. Identify the hypotheses H0 and Ha. • µ = 17 • µ = 17.5 • x = 17.5 • x > 17.5 • µ > 17.5 • µ > 17 3 ii. Find the test statistic (Round your answer to 2 decimal places).

iii. Using the table below, select the appropriate p-value. P(T < t) 0.991 P(T > t) 0.009 P(T > |t|) 0.018

iv. (FREE RESPONSE) Provide an interpretation of the p-value you chose from the above table within the context of the problem. v. Make a decision about H0. Use a significance level of = 0.05. • Reject the null hypothesis because the p-value . • Reject the null hypothesis because the p-value > . • Fail to reject the null hypothesis because the p-value . • Fail to reject the null hypothesis because the p-value > . vi.

(FREE RESPONSE) Based on your decision, write a conclusion in the context of the problem

Explanation / Answer

Set Up Hypothesis
Null, H0: U=17
Alternate, H1: U>17
Test Statistic
Population Mean(U)=17
Sample X(Mean)=17.5
Standard Deviation(S.D)=4.68
Number (n)=660
we use Test Statistic (t) = x-U/(s.d/Sqrt(n))
to =17.5-17/(4.68/Sqrt(660))
to =2.745
| to | =2.745
Critical Value
The Value of |t | with n-1 = 659 d.f is 1.647
We got |to| =2.745 & | t | =1.647
Make Decision
Hence Value of | to | > | t | and Here we Reject Ho
P-Value :Right Tail - Ha : ( P > 2.7447 ) = 0.00311
Hence Value of P0.05 > 0.00311,Here we Reject Ho

[ANSWERS]
a. Average length of all fish in the lake
b. H0: U=17 ,H1: U>17
c. (T > t) 0.009
d. Reject the null hypothesis because the p-value

Related Questions

Navigate

• Browse (All)
• Browse 3
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.