A capture-recapture experiment designed to estimate the size...

A capture-recapture experiment designed to estimate the size of a fish population in a lake was carried out in two phases. In the first phase, 400 fish were captured, tagged, and released. In the second phase, 157 fish were captured, of which 27 were found to be tagged. Provide answers to the following to the nearest integer. (a) Making the necessary assumptions about independent sampling, give an estimate of the size of the fish population. 0.1720 (b) Give a standard error for your estimate in (a). 0.0301 (c) Using a more accurate method than simply finding the estimate plus/minus 1.96 SE, give a 95% confidence interval for the population size. ( 0.1130 , 0.2310 )

Transcript

00:01 So you have 400 fish that are tagged, caught and tagged, and then they're released.

00:07 And then we have a sample pulled that is 27 out of 157.

00:13 And your answer for a was to estimate what the proportion is.

00:19 So that's going to be a good estimate of the population proportion.

00:23 And i agree with your 0 .1720 to 4 decimal places.

00:30 And then your standard error.

00:31 Now this is the only catch.

00:34 If we have, let's go through and do this, and your numbers agree if we don't use that finite population correction factor and factor, which is that square root of capital n minus n over n minus 1, that we have to use if we have our same.

01:01 Sample size being too big proportionately to the population size.

01:06 So let's estimate what the population size.

01:09 And if i solve this proportion and basically take the 400 times the 157 divided by 27, i would find that this would estimate this to be 2 ,326 fish.

01:27 That's what the estimate is for capital and how many are in that.

01:31 And if we find out what percent that is, 157 out of 2 ,326, divided by, let me get that 157 divided by 2326.

01:47 That comes out to be 0 .067.

01:52 Now, if your rule is that this has to be greater than, if this, as long as this is less than 5%, you're okay to sample, but this is bigger than 5%.

02:04 Some people use a 10 % rule, but if your rule is 5%, then for that confidence interval, you actually need to utilize this proportion.

02:15 So i'm going to guess that's why you're not getting it correct.

02:18 If you don't use this proportion, i agree with your answers.

02:21 If you take this p hat plus or minus that, let's see, i think you had to find in part b or standard air, if you find your standard air, let me erase that, if you find your standard air by taking the square root of that p hat, the one minus the p hat, and then dividing it by the sample size of 157, i agree with you.

02:50 I get 0 .031.

02:52 And if i find the confidence interval by taking p hat plus or minus, this value and times 1 .96 as the z value for 95%.

03:03 I agree with you...

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