ECO 251 Final ExamDr. T. Andrews
Summer 2001West Chester University
10 points each (except the bonus)
Show all your work and explain all of your answers.
Name
GENEVA, Switzerland- four former soviet republics on the Caspian Sea are working on an agreement to reduce over fishing of caviar producing sturgeon. According to the convention on international trade in Endangered Species (CITES), Sturgeon numbers have been in “fast and precipitous decline.” The sturgeon’s black eggs are prized as a delicacy in Europe and the U.S. Caviar prices can exceed $2000 a pound, depending on the species of sturgeon.
While species populations are difficult to determine, the current practice is to compare catch rates over time. Catch rates can be measured in a number of ways
a. Over the past ten years 80 percent of all fishing boats returned to port with a full catch of sturgeon, but last year only 50 of 65 boats returned with a full catch. Does this indicate that the catch rate has declined?
P= 50/65=76% .76+/- 1.28sqrt.(.76(1-.76) )/65 =
Z= 1.28 n = 65 .76 +/- 1.28 (.052) =
.76 +/- (.066) =
69.4% to 82.6%
There is an 80% chance that the percentage of boats returned with a full catch is between 69 and 83 %. So we are not sure that it actually declined because 82% is more than the 80% from the past ten years, but 69% is less.
- Again, over the last decade, sturgeon boats caught 800 pounds of sturgeon per trip, with a standard deviation of 100 pounds. If a fleet of 75 boats goes to sea, what should the average catch rate be if populations are stable?
Z = 1.28 800+/- 1.28(100/√75)=
σ = 100 800 +/- 1.28(8.66)=
μ = 800 n = 75 800 +/- 14.78=
785.21 to 814.78
If the populations are stable the average catch rate for 75 boats should be between 798.3 to 814.7 pounds of sturgeon per trip.
- Last year six boats sampled returned with the following catches: 670, 700, 900, 500, 200 and 800 pounds of sturgeon. Use this information (only) to estimate the catch rate for all boats, being sure state any necessary assumptions. Explain whether or not your assumptions are reasonable. Does your estimate support the conclusion that populations are in decline?
Descriptive Statistics
Variable N Mean Median TrMean StDev SEMean
C1 6 628 685 628 249 102
X bar =628 x +/- t (249/√6)
c.l=90% x +/-2.015(102)
t=2.015x +/- 205.53
s= 249 628 +/- 205.53
n = 6 423 to 833
T- Confidence Intervals
80.0 % C.I 90.0 % C.I. 95.0 % C.I.99.0 % C.I.
( 478, 778) ( 423, 833) ( 367, 890)( 218, 1038)
In order to interpret this as an accurate confidence interval, it is necessary to assume that the population is normally distributed. This means that the distribution of catches among boats is symmetric, continuous, and mound shaped with 68% of boats landing within 1 standard deviation of average. In this case, the variable is continuous, and there is no obvious violation of the normal rules (no significant percentage under 0, etc.)
We can be 80 percent sure that the average catch rate is between 478 and 778 pounds of sturgeon, so it appears that the average is less than 800 pounds (barely). We can not be 90 percent sure that populations are declining. Since there were only 6 boats in the sample, we are leaning heavily on the normality of the population.
- How many boats should be sampled to estimate the average catch rate with a reasonable level of confidence to within 50 pounds? (You may use any information from a through c).
n=?
z=80%/2 = .45=1.64 1.642(1002) = 26896 = 10.75 ~ 11
σ = 100502 2500
E = 50oHow
You would need to sample at least 11 boats to be within 50 pounds of the average catch rate.
- How is a sampling distribution different from a population distribution? You are on your own here.