confidence interval for population parameters This is my second question based on the understanding from this
suppose I want to estimate the mean height of all the students studying in 12th class in my state. I do not have access to the entire population so I go for a sample and estimate it. 


*

*I sampled 500 students of 12th class in my state.

*Measured the height of each student.

*Calculated the mean height.


Now, I can say my population parameter is this mean height.
But to provide a confidence value, we need to have a sampling distribution as suggested in the answer to the linked question above and One of the articles which I went through. So, it will allow us to say that 90% chances are that the mean height will be the calculated one.
Fair enough. Now,
Is it necessary to carry out say 100 more samples? We know, irrespective of the distribution of population, the sample means will always follow the normal distribution
because of the central limit theorem.
So can I not use, my very first sample of 500 students, I found the mean, I can calculate the variance too and plot the normal curve using these values?
Would that be incorrect? Do we essentially need to carry out such tedious activity to give out confidence intervals?
consider we already do not know if our first sample was from the unlucky 5% or the lucky 95% as specified in the answer to my previous question. So constructing a normal curve around those value, how correct that will be if it happens from the 5% which is purely by chance?
I do not know, but while writing the question I feel like the answer lies in hypothesis testing and not constructing the confidence interval. Will be great if you can provide some good insights for the above questions.
Thanks a lot to the entire community for answering all my queries.
 A: Is it necessary to carry out say 100 more samples?
No!
We happen to know a lot about the distribution of sample means, and we're able to estimate the standard error from just one sample. The standard error is the standard deviation of the sampling distribution (distribution of $\bar{X}$), which is $N(\mu,\sigma^2/n)$. Since we don't know $\mu$ or $\sigma^2$, we estimate them and use a sampling distribution of $N(\bar{x},\sigma^2/n)$. This means that we can calculate the middle 95% of the sampling distribution by going $2 \sqrt{\sigma^2/n}$ above and below $\bar{x}$, since a normal distribution has 95% of its density within two standard deviations of the mean.
Therefore, the confidence interval for $\bar{x}$ is $\bigg[ \bar{x} - 2 \sqrt{\sigma^2/n}, \bar{x} + 2 \sqrt{\sigma^2/n}\bigg]$.
Except that this isn't quite true. We don't know the standard deviation of the sampling distribution. All we did is estimate it. Consequently, instead of going 2 standard deviations in either direction, we go an amount given by something called the t-distribution. We go to the $0.025$ and $0.975$ quantiles of the t-distribution, with degrees of freedom equal to $n-1$. Therefore, the full answer is that the confidence interval is:
$$ \bigg[ \bar{x} + t_{0.025} \sqrt{\sigma^2/n}, \bar{x} + t_{0.975} \sqrt{\sigma^2/n}
\bigg]$$
(The $t_{0.025}$ value will be negative, so we add it rather than subtract it.)
JB Statistics has some videos on YouTube that I very highly recommend. 
Edit:
JB on Sampling distributions
https://www.youtube.com/watch?v=Zbw-YvELsaM
https://www.youtube.com/watch?v=q50GpTdFYyI
https://www.youtube.com/watch?v=V4Rm4UQHij0
JB on the t-distribution
https://www.youtube.com/watch?v=Uv6nGIgZMVw
https://www.youtube.com/watch?v=T0xRanwAIiI
A: 
...so I go for a sample and estimate it.

The most important thing here is that you need to be able to actually take a simple random sample from your population (or sample via some other specified randomisation method).  At minimum, this is going to require you to have a list of the number of students in each 12th form class in your State.  Before you concern yourself with the statistical mechanics of the confidence interval, you should make sure you are able to randomly sample from your population of interest.

But to provide a confidence value, we need to have a sampling distribution...

For this part I will assume that you have a simple random sample from the (large) population of students.  Fortunately, when we are dealing with sample means, we can appeal to a useful statistical theorem (called the central limit theorem) which gives us a very good approximation to the distribution.  We can do this even without specifying the underlying sampling distribution of the height values.  For any distribution of height values where the underlying mean is $\mu$ and the underlying variance is finite,$^\dagger$ for "large" $n$ we have the useful approximating distribution:
$$\frac{\bar{X}_n - \mu}{S_n / \sqrt{n}} \overset{\text{Approx}}{\sim} \text{Student T} (df = n-1).$$
The value $\bar{X}_n$ is your sample mean and the value $S_n$ is the sample standard deviation (upper case because we are considering them here as random variables).  Inversion of this distributional result, and substitution of the observed sample values, gives the standard confidence interval formula:
$$\text{CI}_\mu(1-\alpha) = \Bigg[ \bar{x}_n \pm \frac{t_{n-1, \alpha/2}}{\sqrt{n}} \cdot s_n \Bigg].$$
Your sample size of $n=500$ is more than sufficient to appeal to the approximate distribution above, and therefore to use the standard confidence interval formula.  The accuracy (width) of your confidence interval will depend on the chosen confidence level $1-\alpha$ and the observed sample standard deviation $s_n$.

$^\dagger$ The only condition we require for the CLT is that the distribution is not heavy-tailed (i.e., it has finite variance).  Heights of people are not a heavy-tailed distribution, so the sample mean of randomly sampled height values is subject to the CLT.
