Example 4.3.4. We are interested in estimating the population of fish in a lake. A sample of 150 fish is caught, tagged, and released back to the lake. A week later, a new sample of 100 fish is...


Estimate the fish population of Example 4.3.4, page 144, with a 68% confidence level.


Example 4.3.4. We are interested in estimating the population of fish in a<br>lake. A sample of 150 fish is caught, tagged, and released back to the lake. A<br>week later, a new sample of 100 fish is caught, and the number of tagged fish<br>is found to be 12. What is the number of fish in the lake?<br>The method is based on the observation that if the tagged fish mix well<br>with the rest of the population, and the second sample is randomly chosen,<br>then the proportion of tagged fish in the sample will be approximately the<br>same as the proportion of tagged fish in the population. Thus we have a<br>proportion of 4 numbers<br>and therefore<br>N'<br>n<br>n·t<br>(4.2)<br>where<br>r is the number of tagged fish in the (second) sample, that is the number of<br>fish recaptured,<br>n is the size of the (second) sample,<br>t is the number of tagged fish in the whole population. That is, the size of<br>the first sample,<br>N is the size of the population.<br>In our example we haver =<br>12, n = 100, t<br>150 and N is unknown. We<br>conclude that<br>100 · 150<br>= 1250,<br>12<br>i.e. the fish population in the lake is approximately 1,250.<br>

Extracted text: Example 4.3.4. We are interested in estimating the population of fish in a lake. A sample of 150 fish is caught, tagged, and released back to the lake. A week later, a new sample of 100 fish is caught, and the number of tagged fish is found to be 12. What is the number of fish in the lake? The method is based on the observation that if the tagged fish mix well with the rest of the population, and the second sample is randomly chosen, then the proportion of tagged fish in the sample will be approximately the same as the proportion of tagged fish in the population. Thus we have a proportion of 4 numbers and therefore N' n n·t (4.2) where r is the number of tagged fish in the (second) sample, that is the number of fish recaptured, n is the size of the (second) sample, t is the number of tagged fish in the whole population. That is, the size of the first sample, N is the size of the population. In our example we haver = 12, n = 100, t 150 and N is unknown. We conclude that 100 · 150 = 1250, 12 i.e. the fish population in the lake is approximately 1,250.

Jun 09, 2022
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