Test to compare 5 subgroups with population parameter I have 5 subgroups , A,B,C,D,E each have their own mean and standard deviation.
 Subgroup  Mean  SD   n
 A         25.8  9.11 77
 B         32.6  10.4 448
 C         31.3  10.3 92
 D         25.9  7.7  2347
 E         29.2  8.9  10121   

I also have the population mean and standard deviation ($\mu=30, \sigma=10 $). What statistical test should I use , if my goal to check if the subgroup mean differs from population ? Thanks.
 A: A suitable variant of ANOVA will work here.
This problem is special because in ANOVA (a) we usually do not assume the population mean and SD (under the null hypothesis) and (b) the null hypothesis is that all group means are equal.  But the concepts and technology of ANOVA directly apply.
Indeed, the null hypothesis here is that each group is an independent sample from the given population with mean $\mu$ and variance $\sigma^2.$  In a group of size $n_i$ ($i$ will index the groups) from a such a population, the expected group mean is $\mu$ and its variance is $\sigma^2/n.$  These facts are simple consequences of the basic properties of expectation and variance.
Furthermore, when the population is approximately Normally distributed (or when the group size is sufficiently large), the sampling distribution of the group mean $\bar X_i$ also is approximately Normal.  This implies that the random variable
$$\chi^2_i = \frac{\left(\bar X_i - \mu\right)^2}{\sigma^2/n_i}$$
approximately has a $\chi^2(1)$ distribution.  Consequently, the sum of the $\chi^2_i$ for all the group means (of which there are $m,$ say) approximately has a $\chi^2(m)$ distribution.
For the alternative hypothesis $H_A:$ one or more of the groups is a sample of a population with a mean other than $\mu$ and the same variance $\sigma^2,$ a large value of this sum favors the alternative.  This gives the usual $\chi^2$ test, albeit with a slightly different chi-squared statistic and more degrees of freedom than usual.
Summary

Let the group counts be $n_i,$ the group means be $\bar X_i,$ the parent population mean be $\mu,$ and the parent population variance be $\sigma^2.$  The chi-squared statistic for testing $H_0:$ all groups are iid samples of the stipulated parent population against the $H_A$ above is
$$\chi^2 = \sum_{i=1}^m \frac{\left(\bar X_i - \mu\right)^2}{\sigma^2/n_i} .$$
The p-value of this test is the upper tail probability of $\chi^2,$ computing using a chi-squared distribution of $m$ degrees of freedom.

Example
For the data in the question, the five chi-squared statistics (one for each group, in order) are
 A        B        C        D        E
13.5828  30.2848   1.5548 394.5307  64.7744

Under the null hypothesis, each one should have a $\chi^2(1)$ distribution.  Typical values are between $0$ and $10$ (there is only a one in a thousand chance of being greater than $10$).  Only the third (group C) has a typical value.  The others are large.  Collectively they sum to $505,$ associated with an astronomically small p-value.  (Typical values of a $\chi^2(5)$ distribution range from $0.2$ to $20,$ more or less.)  We conclude these data did not arise as iid samples from an approximately Normal distribution of mean $30$ and SD $10,$ because all groups except C have means that are too far from $30.$
Remark
This version of ANOVA is nearly the same as performing $m$ separate Z-tests of the groups and combining their p-values using Fisher's Method..  The latter approach would also be valid.
