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Apologies if this may sounds a dumb question, but the more I'm trying to understand how to interpret and evaluate the results, the more I can't find a proper answer.

I've been trying to understand the confidence level and the standard deviation topics.

Given a single column of values (whether time, votes, points ... you choose), I got the mean, the standard deviation and the 95% confidence level.

I've been doing this in Excel for the sake of practicality

Excel screenshot

Given the above, the CI lays between 3 (5.74-2.74) and 8.48 (5.74+2.74)

Questions

  • In looking at the Confidence interval, I assume the interval has to be taken in consideration against the individual results. Is this the case?
  • The 95% confidence or 5% alpha, and the results next to it, what does exactly tell? And how am I supposed to use that figure?
  • Should I want to find what are the results in my 95%, what should I do? Very simplistically, and perhaps incorrectly, I did evaluate whether the results are in the proposed range, but here's the catch. The CI proposed once evaluated against the dataset shows me only 14 returns are in the range. 14 out of 23 is 60% and not a 95%. What am I doing wrong?

Or should I just take the average of 21.85 (so 22) of those results and compare against the mean previously calculated?

Thanks for your patience and help.

UPDATE: Have been trying to get this right somehow via the help provided, but I'm not getting there.

Whether I calculate the CI adding/subtracting the confidence value to the mean (which I understand it's the correct value) or the value of the standard deviation (incorrect), the people value spanning in between that CI is either 7 or 14. And those are not next to the 95% of the confidence level I have considered. That's the part that I cannot understand. If this number has to give me confidence that the 95% of the people in the subset has to be in a range, I would expect a counter-proof. What am I doing wrong?

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Confidence intervals are hard to explain. Before looking at your data, you can say that --- if you are going to draw a sample from a normal population with unknown mean and unknown standard deviation --- there is a 95% probability that the calculated confidence interval from some future experiment:

$$\left[\overline{X}-t_{n-1;1-\frac{\alpha}{2}}\frac{s_n}{\sqrt{n}},\overline{X}+t_{n-1;1-\frac{\alpha}{2}}\frac{s_n}{\sqrt{n}}\right]$$ encompasses the true value of the mean ($\alpha=0.05$, $t_{n-1;1-\frac{\alpha}{2}}$ is the $1-\alpha/2$ percentile of a $t$ Student variable with $n-1$ degrees of freedom, $s_n=\sqrt{\sum(x_i-\overline{x})^2/(n-1)})$. See here).

An example in R:

> library(mvtnorm)
> set.seed(1234)
> sigma <- diag(rep(4, 5000))                 # true standard deviation = 2
> mean <- rep(5, 5000)                        # true mean = 5
> X <- rmvnorm(23, mean, sigma)               # 5000 columns with samples of size 23
> str(X)
 num [1:23, 1:5000] 2.59 3.35 1.37 8.12 1.62 ...
> X_means <- apply(X, 2, mean)                # 5000 sample means
> X_sds <- apply(X, 2, sd)                    # 5000 sample standard deviations
> t22 <- qt(1-0.05/2, 22)                     # t_{1-alpha/2; n-1}
> CI <- matrix(NA, nrow=5000, ncol=3)         # 5000 CI's
> for (i in 1:5000) {
+     midwid <- t22 * X_sds[i]/sqrt(23)
+     CI[i,1] <- X_means[i] - midwid          # lower bound
+     CI[i,2] <- X_means[i] + midwid          # upper bound
+     CI[i,3] <- CI[i,1] <= 5 & CI[i,2] >= 5  # does it cover the true mean?
+ }
> sum(CI[,3])/5000                            # % of CI's covering the true mean
[1] 0.949

But when you draw a particular sample, and calculate the sample mean and its confidence interval, this interval either covers the parameter value or it does not. You can't say that $P(\text{my interval covers the true mean})=95\%$, you can only say that your interval is a set of plausible values for the true mean, with a 95% "confidence level" (not probability, see here).

However, if you say that your 23 numbers represent the 100% of the population (in your comment to spdrnl'answer), then confidence intervals are meaningless, because you know the true mean (and the true standard deviation).

If you wish to know which values lie between the 2.5th percentile and the 97.5th percentile you can just use the R quantile function:

> (x <- c(4,2,4,3,1,6,2,9,2,12,6,5,7,8,5,9,6,4,6,7,9,10,5))
 [1]  4  2  4  3  1  6  2  9  2 12  6  5  7  8  5  9  6  4  6  7  9 10  5
> (q <- quantile(x, probs=c(0.025,0.975)))
 2.5% 97.5% 
 1.55 10.90 
> x[x >= q[1] & x <= q[2]]
 [1]  4  2  4  3  6  2  9  2  6  5  7  8  5  9  6  4  6  7  9 10  5

EDIT

If you want to calculate a 95% CI on your data, here is how I would do in R:

> x <- c(4,2,4,3,1,6,2,9,2,12,6,5,7,8,5,9,6,4,6,7,9,10,5)
> (x_mean <- mean(x))
[1] 5.73913
> (x_sd <- sd(x))
[1] 2.86384
> (halfwidth <- qt(0.975,22)*x_sd/sqrt(23))
[1] 1.238417
> (ci <- c(x_mean - halfwidth, x_mean + halfwidth))
[1] 4.500713 6.977548

Your CI is different:

> c(x_mean-1.96*x_sd/sqrt(23), x_mean+1.96*x_sd/sqrt(23))
[1] 4.568713 6.909548

but you can use $1.96$ instead of the 97.5th percentile of a Student $t$ variable with $n-1=22$ degrees of freedom:

> qt(0.975,22)
[1] 2.073873

only if you know that $2.86$ is the population (i.e. the "true") standard deviation.[1]

As to the interpretation, a CI is not a (sub)range of your data. The people value spanning in between that CI do not matter. A CI is just a range of plausible means: you can say that the population ("true") mean is plausibly between 4.5 and 7.

So by checking several averages of 22 person at a random, that mean has always to be in the CI level. Is that the case?

Yes, almost... By checking several averages (and standard deviations) of 23 person at random, the sample mean will tend to be in a 95% CI (which will be different for each sample) in 95% of the samples (see my first code above). This is why you believe that your CI is plausible.


[1] Try using the CONFIDENCE.T function instead of CONFIDENCE or CONFIDENCE.NORM (see here, then here). In Python, numpy.std returns the population standard deviation, if you are looking for the sample standard deviation, you can supply an optional ddof parameter (see https://stackoverflow.com/questions/34050491/standard-deviation-in-numpy):

>>> import numpy as np
>>> x = np.array([4,2,4,3,1,6,2,9,2,12,6,5,7,8,5,9,6,4,6,7,9,10,5])
>>> np.std(x)
2.800891027548941
>>> np.std(x, ddof=1)
2.863840258755363
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  • $\begingroup$ Thanks @Sergio. My knowledge is not like yours, hence I got 1/4 of what you said, but at least I have some indications. If I have to distill something basic, in this case the mean, I guess (and please correct me if I'm wrong) that I should look at the results in this way. Get the mean, calculate the CI, and then eventually the 95% confidence. The higher the results of the confidence level is, the more it indicates that my mean could represent the whole population. Is that the case? If so, should I look at a lower value or an higher value to understand how far I am? $\endgroup$ – Andrea Moro Jul 25 '20 at 13:23
  • $\begingroup$ @AndreaMoro You choose the confidence level, then calculate CI. The narrower the CI is, the more it indicates that the sample mean is close to the population mean. If you don't know the standard deviation, you need an estimate, $s_n$, and then your CI is $\overline{x}\pm t_{n-1;1-\alpha/2}s_n/\sqrt{n}$. If $n$ (the sample size) is small, then your CI is wide. If $n$ is large, your CI is narrow. See <stat.yale.edu/Courses/1997-98/101/confint.htm> or <stat.purdue.edu/~mlevins/docs/stat511/Lec10.pdf>. $\endgroup$ – Sergio Jul 25 '20 at 13:44
  • $\begingroup$ I'm getting into something, but numbers don't add up if you look at my example. Whether I calculate the CI adding/subtracting the confidence value (which I understand it's the correct value to be used) or the value of the standard deviation (incorrect), the people value spanning in between that CI is either 7 or 14. And those are not next to the 95% of the confidence level I have considered. That's the part that I cannot understand. If this number has to give me confidence that the 95% of the people in the subset has to be in a range. What am I doing wrong? See update $\endgroup$ – Andrea Moro Jul 25 '20 at 16:08
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    $\begingroup$ @AndreaMoro I've edited my answer. HTH. $\endgroup$ – Sergio Jul 25 '20 at 16:59
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    $\begingroup$ @AndreaMoro. Yes, I've added Python code to my answer. $\endgroup$ – Sergio Jul 26 '20 at 8:59
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You are confounding two different notions, lets split these apart.

First there is the normal distribution, which can be used for example to model the heights of people. The distribution is characterized by the mean and the standard deviation. Most people will be close to the mean.

Second there are confidence intervals. Confidence intervals serve to indicate the confidence ("precision") of a statistic or parameter. An example of a statistic or parameter is for example the mean. A confidence interval for the mean would actually indicate how much confidence you have in the calculation of the mean. Note that this mean is calculated using a sample, and the confidence interval then can give information on the possible values of the "real" population mean. The field of statistics in general is about how to generalize from samples to populations.

What you are interested in is the range of lengths which contains 95% of the people, which is the first case. For this one would calculate the interval around the mean of the observations/values by adding and subtracting 1.96 * standard deviation. The Z value for 95% confidence is Z=1.96. The Z value is a property of the normal distribution.

HTH.

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  • $\begingroup$ Thanks for returning @spdrnl. Let's start by saying this example represents the 100% of the population. That said, I assume I have to take the 1.96 from the standard distribution table, calculating the percentage I want and finding the relative value, right? With this interval I can compare each value with the individual z-score to understand which one falls in the range. Should I need to find the value of the 1 standard dev, would that be safe doing 1.96 * stdev / 2? Confidence intervals, u said it measures the level of confidence of the mean: How is this happening? What that 1.12 tells? $\endgroup$ – Andrea Moro Jul 24 '20 at 14:57
  • $\begingroup$ For finding the 1 std deviation interval one would add or substract 1 * std deviation from the mean, this would contain 68% of the observations. I am not familiar with Excel, so I cannot comment on the individual cells unfortunately. The notion that the sample contains 100% of the population is not standard. $\endgroup$ – spdrnl Jul 24 '20 at 15:24
  • $\begingroup$ Ok, and what about the Confidence intervals, u said it measures the level of confidence of the mean: How is this happening? What that 1.12 tells? I'm actually doing a Data Science course, so if you prefer enriching your answer with something that is Python related (preferably) or R, showing some output result and table to help me understand, that would be massively appreciated. $\endgroup$ – Andrea Moro Jul 24 '20 at 15:35
  • $\begingroup$ I am not familiar with Excel, so I cannot comment on the individual cells unfortunately. $\endgroup$ – spdrnl Jul 24 '20 at 15:40
  • $\begingroup$ Are you able to exapand on your answer by providing a Python or R example? $\endgroup$ – Andrea Moro Jul 24 '20 at 16:14

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