Question

Hi guys. Just a bit of statistics (CI) work i'm trying to get my head around:

A Salmon Marketing Board decided to estimate the number of salmon in a lake that had been closed to fishing.

80 salmon were caught, tagged and released back into the lake. A week later another sample of 50 salmon was caught, and 17 of these were salmon that had been tagged the week before.

Using this information, calculate a 95% confidence interval for the number of salmon in the
lake.

Answer 1

If we take our sample to be 50, and 17 of those 50 were tagged, then our sample proportion would be 0.34, giving a (1-p) of 0.66. Then we get a margin of error of 0.1313

Answer 2

So the confidence interval for the proportion of salmon in the lake that were tagged (is this correct?) is

\[0.2087 \le \pi \le 0.4713\]

Answer 3

Question is, how to we get from that confidence interval to the number of fish in the lake CI?

Answer 4

i have forgotten my formulas for proportions :{
how did you calculate margin of error?

if it is correct, then to change CI from proportion to num of fish take 80 divided by proportion
\[\pi = \frac{80}{N} \rightarrow N = \frac{80}{\pi}\]
\[\frac{80}{.4713} \le N \le \frac{80}{.2087}\]

Answer 5

Oh, so the 0.2087 represents the proportion of tagged fish in the entire lake as well... That makes perfect sense.

Margin of error for 95% CI (CLT implies normal distribution and hence z=1.96) is