I am sorry for a possibly confusing title. I do not get the following (using an example from mathisfun.com):
- In a sample of 40 men, we get a mean height of 175 cm. The standard deviation is 20 cm. If I get it right, then (assuming normal distribution), 68% of men will have a height between 155 and 195 cm (i.e. from - 1 SD to + 1 SD). Is that true?
- Now calculating the 95% confidence interval, the result will be 175 ± 6.2 cm. As 95% CI uses the Z value of 1.96, it is pretty much 2 SDs. I just do not get it - if 68% of men will have their height between 155 and 195 (± 1 SD), I would expect that with 2 SDs, it should be between 135 and 205! What I am missing please?
A page where I am reffering to, with the explanation and formulas: https://www.mathsisfun.com/data/confidence-interval.html
EDIT: In the following example, the interval is calculated without dividing by the number of subjects:
140 children had a mean urinary lead concentration of 2.18 µmol24hr, with standard deviation 0.87. The points that include 95% of the observations are 2.18 ± (1.96 × 0.87), giving a range of 0.48 to 3.89
EDIT:
You are mistaking the distribution of the data and the distribution of the mean.
Assuming the data of the men is indeed normally distributed, then 68% of the men in your sample would have a height between $\mu \pm 1 \sigma = 175 \pm 20$.
But if we want to find the range of the true mean that have "generated" the data, we need to create a confidence interval for the mean.
For the mean you need to use the formular
$[CI_{-}, CI_{+}] = [\mu- Z\cdot \frac{\sigma}{\sqrt{n}}, \mu + Z\cdot \frac{\sigma}{\sqrt{n}}]$.
See the full example below for both cases:
$\mu = 175$
$\sigma =20$
$n = 40$
and the fomular
and your Z-value for a 68% CI is almost 1, (
qnorm(1-(1-0.68)/2)in R for the correct value) thus the 68% CI interval is$175 \pm \frac{20}{\sqrt{40}} = [171.84,178.16]$
where the 95% is
$175 \pm 1.96 \cdot \frac{20}{\sqrt{40}} = [168.8,181.2] = 1.75\pm 6.2$
As we can see, the 95% is greater than the 68% as expected