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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?

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  • $\begingroup$ If the contest of the question is to estimate the normal population mean $\mu$ using the mean of a random sample, then, regrettably, authors often get sloppy about saying that each new population is normal. This is especially true with human heights and with exam scores, which are often assumed to be normal. // So I have to say it is 'common' to assume normality without explicitly saying so. Shouldn't be true, but seems to be. $\endgroup$
    – BruceET
    May 19 at 13:34
  • $\begingroup$ Thank you for your input! $\endgroup$
    – Sergio
    May 19 at 14:09
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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 $n$.

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