Question 821537: The State Fish and Game Division claims that 75% of the fish in Homestead Creek are rainbow trout. However, the local fishing club caught 200 fish one weekend and found that 130 were rainbow trout. For this problem, assume that no single fish was caught more than once. Does this indicate that the percentage of rainbow trout in Homestead Creek is less than 75%? Use α = 0.01.
(a) What is the level of significance? State the null and alternate hypotheses. Will you use a left-tailed, right-tailed, or two-tailed test?
(b) Identify the sampling distribution you will use: the standard normal or the Student’s t.
Explain the rationale for your choice. What is the value of the sample test statistic?
(c) Find (or estimate) the P value. Sketch the sampling distribution and show the area corresponding to the P value.
(d) Find the critical value(s).
(e) Based on your answers for parts (a) to (d), will you reject or fail to reject the null hypothesis?
Interpret your decision in the context of the application.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! The State Fish and Game Division claims that 75% of the fish in Homestead Creek are rainbow trout.
However, the local fishing club caught 200 fish one weekend and found that 130 were rainbow trout.
Does this indicate that the percentage of rainbow trout in Homestead Creek is less than 75%? Use α = 0.01.
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(a) What is the level of significance?:: 99%
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State the null and alternate hypotheses.
Ho: p >= 0.75
Ha: p < 0.75 (claim)
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Will you use a left-tailed, right-tailed, or two-tailed test?
Since Ha is "less than", it call for a left-tailed test.
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(b) Identify the sampling distribution you will use: the standard normal or the Student’s t. :: standard normal for testing proportions.
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Explain the rationale for your choice.
What is the value of the sample test statistic?
z(130/200) = (0.65-0.75)/sqrt(0.75*0.25/sqrt(200)] = -3.2660
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(c) Find (or estimate) the P value. Sketch the sampling distribution and show the area corresponding to the P value.
p-value = P(z < -3.2660) = normalcdf(-100,-3.266) = 0.00054548..
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(d) Find the critical value(s).
invNorm(0.01) = -2.3263
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(e) Based on your answers for parts (a) to (d), will you reject or fail to reject the null hypothesis?
Since the p-value is less than 1%, reject Ho.
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Interpret your decision in the context of the application.
The test supports Ha; So the percentage of rainbow trout is less than 75.
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Cheers,
Stan H.
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