Why does the question state standard deviation is unknown, when it is written in the question? See problem Below

The question "The dean randomly selected 25 students who took the calculus final exam and looked at their scores on it. In this sample the mean was 85.7 and the standard deviation was 4.2. The population of exam scores has a distribution that is approximately normal. Construct a 90% confidence interval for the mean score of all students who took this exam.

Round your answers to 2 places after the decimal point, if necessary. If it is not appropriate to construct a confidence interval in this situation, then enter "0" in both answer boxes below.

90% confidence interval: (84.26) (87.14) this answer is correct.

If you press the button "Show Detailed Solution" it says this

In this scenario we have the following given information:

n = 25 x with line above = 85.7 s = 4.2 (standard deviation symbol, sigma) is unknown The population is approximately normally distributed. 90% is the desired confidence level of the interval. it is much faster and easier to use the TInterval function on our calculators:

TInterval Inpt: Stats

"x" with a line above: 85.7 Sx: 4.2 n: 25 C-Level: 0.9

My question is why does the explanation say the (standard deviation symbol) is unknown when the standard deviation is clearly stated in the problem?

It could certainly have been written more clearly though the distinction you need to make here should already be clear in your mind by the time you're tackling this sort of question, or you're sure to get very muddled.

They mean that the population standard deviation ($$\sigma$$) is unknown.

It can of course be estimated from the data, e.g. by a sample standard deviation, ($$s$$) $$-$$ note the change of symbol, this change marks the distinction between the parameter and the estimate [1].

However if you're correctly reproducing the question and answer, it does muddle this important distinction itself at one point, which is bordering on the egregious in something that's supposed to be clarifying what's going on. Such slapdash writing in a detailed solution is not okay.

The noise in such an estimate has an effect on the behavior of the statistic you'd use if you had the population standard deviation so the distinction matters. This will be explained in detail in a reasonable textbook.

$$[$$Feel free to skip this parenthetic digression. I think a better answer would be along the lines of 'we don't actually know for certain whether a t confidence interval is appropriate, because there's insufficient information to judge whether the $$90\%$$ coverage could be attained sufficiently closely in these circumstances. If we assume that we did have such information, we would compute as follows [...]'. They clearly intend that the $$n=25$$ should be dispositive, but it is not, since the behavior of the interval could, for example, be impacted by substantial skewness in the population of scores -- imagine a test with mostly very easy questions but one or two very hard ones that were beyond almost every student; most students will score just below some very high score (like, oh, some value in the mid-80s, say, representing near to $$100\%$$ on the very easy questions), but there could be a pretty long tail to the left and a very short one on the right. In that case, $$n=25$$ (or indeed even substantially more) might not be quite sufficient for our specific purposes, on which we have been given no detail. It's an assumption we're making $$-$$likely reasonable, but an assumption nonetheless$$-$$ that something like this is not the case. If we wanted a one-sided interval such considerations become much more important, because the error in the other tail won't be there to help 'balance out' the error in the tail we use in a one-sided interval. It's less of an issue with a two-sided interval, but the potential problem remains. As a check, I was able to simulate plausible looking scores, with mean and standard deviation like those in the question, where in samples of size $$25$$ the coverage of "$$90\%$$" intervals was around $$80\%$$; lower coverage values would have been quite simple to obtain.$$]$$

[1]: conventionally, Greek letters are used for population parameters, Latin letters (normally lower case) for estimates.

TLDR: Glen_b is correct. What is known is the sample standard deviation 's' and sample mean 'x bar'. The population standard deviation sigma and population mean Mu are unknown.

That is the basic difference between mathematics and statistics. In math we are calculating known and exact quantities. In statistics, we estimate things we call parameters (e.g. mean and standard deviation) of an inaccessible population that are of course unknown, based on an accessible sample. This is because for most problems in science, social science, economics, and especially behavioral and health sciences, population parameters are generally unknown and the data we need to find them are impractical to obtain. (Think the average weight of all men in a given country or average size of all stars in the universe or how often the average person uses profanity)? It is not practical to get the data we need to calculate that.

What is practical is to take fairly selected representative sample to estimate a range for where the population's parameters we are estimating probably falls and how likely that parameter is outside that range. We consider a few things to do this:

• The sample mean
• sample standard deviation (a measure of how inconsistent the data is). Low standard deviation means we may be able to put the parameter in a narrower range and / or have higher confidence the parameter is in that range.
• sample size (More samples may mean we can have a narrower range and / or higher confidence the parameter is in that range.)
• overall distribution of the data
• Most ways to extrapolate a range and confidence for a population parameter that are taught in an introductory stats class will presume that the data are normally distributed, and if the data are not, it won't give you a correct range and confidence.
• This means that examples equally above and below the mean are equally common, and most examples are close to the mean while examples far from the mean are unusual in the data you have.
• For example, you may get incorrect extrapolations if you have a data set of the weights of citizens of a country where obesity is common. The data will be skewed to the right on on that bell curve. This means, in other words, examples above the mean will be more common than those below.
• @pjs Wow! Clearly I need to get caught up on sleep. Sorry for the screw - up. Apr 24, 2023 at 20:25