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.

  1. I sampled 500 students of 12th class in my state.
  2. Measured the height of each student.
  3. 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.


3 Answers 3


Is it necessary to carry out say 100 more samples?


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.


JB on Sampling distributions




JB on the t-distribution




...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.


It is good that you are paying attention to details in @Dave's nice Answers (+1). But the experimental situation you describe is an easy one. You anticipate having plenty of data and it is hard to imagine that your answer needs to have extraordinary precision.

The population parameter $\mu$ is the mean height in your student population. The only way for you to know its exact value is to measure all the students, which you say (quite reasonably) you cannot do.

Data. Suppose you get data to the nearest cm. that are summarized as shown below. [I am using R statistical software, but other software gives similar summaries.]

summary(x); sd(x)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  151.0   168.0   175.0   174.7   181.0   205.0 
[1] 9.083249

The sample mean $\bar X = \frac 1{500} \sum_{i=1}^{500} X_i = 174.7$ cm and the sample standard deviation $S_x = \sqrt{\frac {1}{499}\sum_{i=1}^{500}(X_i - \bar X)^2} = 9.083.$ With $n = 500$ subjects, we expect the population mean to be $\mu \approx 175$ and the population standard deviation to be $\sigma \approx 9.$ These are first impressions, to be refined presently.

A histogram of the data is shown below.

enter image description here

Assumption of normality. In past experience, people's heights have usually been approximately normally distributed. Also, the fact the the sample mean 174.7 and median 175 are nearly equal and the general shape of the histogram indicate that the data are at least roughly normally distributed.

If you are really worried whether you data are nearly normal, you could do a formal test. For the data shown above, a Shapiro-Wilk test of normality gives the P-value 0.146. A P-value below 0.05 would indicate that the data are not from a normal population.

[1] 0.1461765

Also, the t confidence interval described below is known to perform well even if the data are not perfectly normal. The Central Limit Theorem guarantees that samples as large as $n = 500$ can depart a bit from normality and still give very useful results.

Confidence interval. In order to get an idea how far our estimate $\bar X = 174.7$ might be from the unknown population mean $\mu,$ we can make a 95% confidence interval (CI) of the form $\bar X \pm 1.965 S/\sqrt{n},$ where the numbers $\pm 1.965$ cut off probability 0.025 from the upper and lower tails of Student's t distribution with $n - 1 = 499$ degrees of freedom (which leaves 95% of the probability between these two numbers). For samples as large as $n=500$ this number is roughly $2$ and some people just use 2 when making a 95% confidence interval.

qt(.975, 499)
[1] 1.964729

The procedure t.test in R, makes a 95% confidence interval. (Most other statistical software packages have procedure that do the same.) The resulting 95% CI is $(173.9, 175.5).$

[1] 173.9419 175.5381
 [1] 0.95

At this point, it is OK to round to one decimal place because we usually aren't interested in expressing people's heights more precisely than one mm.

We conclude it is likely that the population mean height $\mu$ is between 173.9 and 175.5. There is a small chance that $\mu$ may be a little bit outside this interval, but for practical purposes it seems good enough to say that $\mu \approx 174.7$ or $175$ cm.--with a margin of error around $8$ mm.

If you wanted to have more than 95% confidence in your interval, you could make a 99% confidence interval $(173.7, 175.8)$, which is a little longer (with a margin of error about $1$ cm).

t.test(x, conf.lev=.99)$conf.int
[1] 173.6896 175.7904
 [1] 0.99

Note: In order to determine whether student's heights decrease between morning and evening, a very careful study was done in India in the mid-1940s. Students were measured as accurately as possible in the AM and PM by two different people. Results were analyzed to make sure the two technicians made consistent height measurements. They tried (almost successfully) to measure student heights to the nearest mm.

They concluded that overall most students lose about a cm in height between morning and evening (gaining it back after a night's sleep).

If you are doing your own study of student heights, you may be interested in details of their work and analysis, reported by Majumbar DN and Rao CR (1958) "Bengal anthropometric survey, 1945," Sankhya, V.19, Parts 3 & 4.

  • $\begingroup$ thanks a lot for the explanation!! After reading all the answers, I under stand that we can get the CI from t distribution but my only contention left is, what if the sample mean happened by chance only and we are creating an interval around it which will be skewed. P.S thanks a lot Dave for the links. I am still going through the videos $\endgroup$ Jul 31, 2019 at 13:46

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