and asked the amount of time (in hours) they spend for his course per week. The data are given below.
10, 8, 9, 7, 11, 13
Estimate the standard error of the estimated mean time spent in a week for this course by the students who are taking this course.
Ok my work:
I've found $\hat{\mu}=\bar{x}=\cfrac{29}{3}$
$\hat{\sigma}=s=2.16 $ approximately
so $\hat{\sigma_\hat{\mu}}=s_\hat{\mu}=\cfrac{\hat{\sigma}}{\sqrt{n-1}}=\cfrac{2.16}{\sqrt{5}} = .966$ approximately
so final answer should be $.966$ hours
But the book's answer says that the answer is $0.882$ hours
You get this answer via $\cfrac{2.16}{\sqrt{6}} = .882$ approximately
But this makes no sense to me. This would be population standard deviation for $\bar{x}$ NOT sample standard error/deviation for $\bar{x}$. Am I mistaken? Shouldn't it be $\sqrt{n-1}$ in the denominator not just $\sqrt{n}$ because it's a sample? If not why? Can someone explain this to me so that I understand the concept?
The reason why there is difference between the variance of the sample data whether the mean is known or not could be explained with the degree of freedom. Assume we have $N$ independent data from a dataset, if the mean is unknown, it should be estimated as $$ \overline x = \frac{1}{N}\sum_i^N x_i $$ the expectation of discrepancy between sample variance and population variance is $$ E \{ \sigma^2 - s^2 \} = E\left \{ \frac{1}{N}\sum_{i=1}^N (x_i-\mu)^2 - \frac{1}{N}\sum_{i=1}^N(x_i-\overline x)^2 \right \} \\ = E\left \{ \frac{1}{N}\sum_{i=1}^N \left ( ( x_i^2 - 2x_i\mu+\mu^2) - (x_i^2 - 2x_i\overline x + \overline x^2) \right) \right \}\\ = E\left \{ \frac{1}{N}\sum_{i=1}^N \left ( \mu^2 - \overline x^2 + 2x_i(\overline x - \mu) \right) \right \} = E\left \{ \mu^2 - \overline x^2 + \frac{1}{N}\sum_{i=1}^N \left (2x_i(\overline x - \mu) \right) \right \} = E\left \{ (\overline x - \mu)^2 \right \} = var(\overline x) = \frac{\sigma^2}{N} $$ So the expected value of sample variance have the following relation with the actual variance, $$ E \{ s^2 \} = \sigma^2 - \frac{\sigma^2}{N} = \frac {N-1}{N} \sigma^2 $$ However, if our population only consists of the given numbers, then actual mean and estimated one would be identical and thus, there will be no difference between $\sigma$ and $s$.