The IQs from 181 boys aged between 6-7 years old were measured. Calculate its mean's confidence interval for $\alpha = 5\%$ The IQ from 181 boys aged between 6-7 years old were measured. The mean IQ
is 108.08, and the standard deviation is 14.38. 
(a) Determine the confidence interval with confidence coefficient $95\%$ for the mean IQ of the boys' sample.
(b) Interpret the confidence interval with words.
(c) Was it necessary to assume that the IQs were normally distributed in this case? Why?
MY ATTEMPT
(a) I am not sure if the problem's information is related to the actual population variance or to the sample variance. In the first case, we should use the pivotal quantity $\sqrt{n}(\overline{X} - \mu)/\sigma$ whilst in the second case we should use the pivotal quantity $\sqrt{n}(\overline{X}-\mu)/S$. However in both cases we should use the normal distribution, since the sample is large enough. Can someone help me on the subject? Am I making any conceptual mistake?
(b) Any help is appreciated.
(c) As previously mentioned, I don't think it is necessary to assume the IQs are normally distributed, because of central limit theorem. However, I am not sure about it. Can someone give me a hint?
Thanks in advance!
 A: From the wording of the problem, I surmise that the sample size is $n = 181,$ the sample mean is $\bar X= 108.08$ and the sample standard deviation is $S = 14.38.$ 
If the population standard deviation $\sigma$ is unknown and estimated by $S,$ then you should use a t interval, regardless of the sample size. 
(It is true that you will get a confidence interval that is 'almost correct' by pretending that $\sigma = S$ and using the normal distribution. However, if you are using
software, there is a clear distinction---between z-procedures and t-procedures---in how the information is entered.)
A 95% confidence interval for $\mu$ is of the form
$$\bar X \pm t^*S/\sqrt{n},$$ where $t^*$ cuts probability $0.025$ from the upper tail of Student's t distribution with
$n - 1$ degrees of freedom. In this problem $t^* = 1.973.$ (This is very close to the number $1.96$ that cuts probability $0.025$
from the upper tail of the standard normal distribution.)
Computation in R:
qt(.975, 180)
[1] 1.973231

Here are results from Minitab's t interval procedure, for your data:
One-Sample T 

  N    Mean  StDev  SE Mean       95% CI
181  108.08  14.38     1.07  (105.97, 110.19)

I will leave it to you to match the formula above for the 95% confidence interval and the numerical result from Minitab.
Note: IQ scores from most populations are very nearly normally distributed. This is partly because an IQ score can be viewed as the sum of scores on a moderately large number of questions that are of approximately equal value and approximately independent; partly because the attributes measured by IQ scores may be approximately normally distributed in the human population; and also, partly because the people who make up IQ tests fudge the question selection and scoring a little to make overall scores even more nearly normal than they might naturally be.
However, you are correct that even if IQ scores were not
approximately normal, the mean $\bar X$ of scores of 181 subjects would be close to normal and both z and t tests would give reliable results.
