I have the following population:

Where the left column is the age of our individuals and the right column is their weight (in kg).

The exercise tells us that we use **Random Sampling with no replacement** to take our sample and we are twice as likely to select an individual whose age is lower than 20.

I have to find 3 things:

- An unbiased estimator for $bar{x}$.
- The probability of obtaining the sample: $S = { 50, 35, 85 } $
- With $S$ as our sample, estimate the total population weight with a 75% confidence interval.

Any help would de appreciated, I have worked on this for hours and gotten nowhere.

Cross Validated Asked by PLanderos33 on November 14, 2021

1 Answers**A small population with twice as many children as adults.**
Suppose there are 4 each of children of ages 5, 10, and 15;
and that there are 3 each of adults of ages 25 and 45.
That means the average weight in the population of $36$ is
$[4(20+35+50) + 3(90+85)]/18 = 52.5$kg.

**What sample size?** Also, suppose
we are take a random sample of $n = 3$ from this population.
(A clue that we should use $n = 3$ is having been
given a sample of size three in the problem.)

The population of 36 weights is `kg`

as follows:

```
kg=c(rep(c(20,35,50), each=4), rep(c(90,85), each=3))
mean(kg)
[1] 52.5
```

**Simulation results.** If we take many samples of size 3 from this population, we
can get a good approximation of the sampling distribution.

```
set.seed(2020)
m = 10^5; n=3; s.3 = wt = numeric(m)
for(i in 1:m) {
x = sort(sample(kg, 3))
s.3[i] = sum(x == c(35,50,85))
wt[i] = mean(x) }
mean(s.3==3) # prob sample has 35,50,85
[1] 0.058995 # aprx 1/17
1/17
[1] 0.05882353 # exact 1/17
mean(wt)
[1] 52.50809 # aprx 52.5
2*sd(wt)/1000
[1] 0.02902334
```

** Probability of specified sample.** With a million samples, one can expect 2 or 3 places of accuracy.
One can show by simple combinatorics that the probability of
getting one each of the weights $35, 50, 85$ (in some order) is $1/7,$ which is consistent with the simulation.

** Unbiased estimator.** Also, the mean weight in the population is $52.5.$ The simulation approximates $E(bar X_3)=
52.508 pm 0.029,$ with a 95% margin of simulation error.

If sampling had been with replacement, it is obvious that the mean of the sample of $n=3$ would be an unbiased estimate of the population weight $52.2.$ It is not hard to show that the same is true for sampling without replacement, and I will leave that to you.

**Confidence intervals.** I don't know what you have studied about confidence intervals.
The sample mean of the *specified sample* of three observations is $bar X_3 = 56.67;$ it should be the center of a CI for the true mean weight of the population. Using it's standard error you should be able to get some style of CI.

Three observations are hardly enough for a good bootstrap CI,
but if you know about bootstrapping this part of the problem
may be a prompt to do whatever kind of bootstrap you may have
studied. A naive percentile 75% nonparametric bootstrap CI
can be found as follows (repeatedly re-sampling *with* replacement from the sample of three). This CI is $(40.0, 73.3),$ which does cover the known population mean.

```
set.seed(721)
re.avg = replicate(10^4, mean(sample(c(35,50,85), 3, rep=T)))
quantile(re.avg, c(.125, .875))
12.5% 87.5%
40.00000 73.33333
```

Answered by BruceET on November 14, 2021

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