I have the following question:
The heights of a random sample of $50$ college students showed a mean of
$174.5$ centimeters and a standard deviation of $6.9$ centimeters.Construct a $98%$ confidence interval for the mean height of all
college students.
While learning about estimating the population mean using the sample mean I learned about $2$ cases:
$1.$ The population standard deviation $\sigma$ is known: we use the Z distribution (standard normal) to estimate it
$2.$ The population standard deviation $\sigma$ is unknown, we use the T-distribution, under the condition that around population is normally or approximately normally distributed
However, this question doesn't fall under either of these as $\sigma$ is unknown and we do have any info about the distribution of the population, the author solved it using the Z distribution and just took the population standard deviation to be $6.9$
My question: Is this a common practice to do (taking the sample sd to be the same as the population sd when we don't have it)? Are there any conditions to do this? Or did I just misunderstabdd something?
Best Answer
According to CLT $$ \sqrt{n}(\bar{X}_n - \mu)\xrightarrow{D} N(0,\sigma^2), $$ for $n\to \infty$. The conditions for this to hold are that $X_1, ..., X_n$ are iid with finite variance (no normality is required). For $n=50$, you can use it as approximation, i.e., $\bar{X}_{50}$ is approximately $N(0, \sigma^2/50)$. As you don't know $\sigma ^2$, you can replace it with its estimator. The $t$ distribution pops up naturally where $X_i$s are normal, then $\bar{X}_n$ is exactly normal for every $n$, and the estimator of $\sigma^2$ is exactly $\chi^2$ up to a constant, hence the ratio between $\bar{X}_n$ and its standard deviation is - by definition - distributed $t$ with $n-1$ degrees of freedom. However, once you use the normal approximation, the ratio between $\bar{X}_n$ and $\hat{\sigma}/\sqrt{n}$ is no longer $t$, but approximately normal for every finite $n$.