Rules to test validity of estimates I know three rules to check for normality, and verify estimates for standard deviation for sample means and sample proportions.
I. $n > 30 $
II. $np > 10; n(1-p) > 10$
III. $10\%$ rule ($n > 10\% N$)
What are the uses of each one?
What if I picked 9 squirrels from a population of known standard deviation (not known shape)? Can I assess normality and standard deviation of the sampling distribution of the sample mean?
 A: Rule I (the imfamous 'rule of 30') is sometimes given as a way to determine whether the critical
value for a two-sided t test is near 1.96 (which would be the critical value
for the corresponding z test). 
For example, suppose you have data from a normal or nearly-normal population 
with sample size $n = 31,$ mean $\bar X = 11.2,$ and and SD $S = 3.25.$
The population mean $\mu$ and population SD $\sigma$ are both unknown. You want want to test $H_0: \mu = 10$ against $H_a: \mu \ne 10.$ Because $\sigma$ is
unknown and estimated by $S,$ this is properly a t test. The $T$ statistic
is $T = \frac{\bar X = 10}{S.\sqrt{n}} = 2.06.$ The critical value from
a t table for a test at the 5% level, based on DF = 31-1 = 30, is $c = 2.049.$
Because $|T| = 2.06 > 2.049$ you can (just barely) reject $H_0.$
Some elementary texts say that it is OK to treat this t test as a z test
because $n = 31 \ge 30.$ If it were a t test, then the critical value
would be $c = 1.96$ and you would reject. If just happens that the
critical values for $\mathsf{T}(30)$ and $\mathsf{Norm}(0,1)$ are about
the same for tests at the 5% level. So the approximation works. But
that does not mean that the test is really a z test. There are many
valid objections for ever using this so-called' rule of 30:


*

*The rule does not work at all for significance levels 1% [$n>100?$] and 10%
[$n>15?$]. or not really for any level other than 5%.

*If you are using software for a z test, you will be asked for $\sigma,$ which is 
unknown. And it is a lie to enter the sample SD $S$ instead.

*If you go on to find P-values or to find the power of the test, there are more fundamental differences between t tests and z tests.

*Students who go beyond the basic course have to unlearn this potentially
misleading rule.
The one correct way to tell a t test from a z test is quite simple: If the population SD $\sigma$ is known
it is a z test; if $\sigma$ is unknown it is a t test; and the sample size $n$ has nothing to do with the distinction.
Rule II is usually OK. It is ordinarily used to say whether it is OK
to approximate binomial probabilities by using the normal distribution.
The specific statement varies from text to text, depending on fussiness.
Some say it is OK to use a normal approximation if both $np > 5$ and $n(1-p) > 5.$ Your version with $10$ instead of $5$ is more cautions. Both rules
work better if $p \approx 1/2.$ No such rule-of-thumb works all of the time,
but this one is pretty good. In any case, you should not expect more
than two-place accuracy from a normal approximation to binomial. Two examples:


*

*If $n = 3,\,$ $p=1/2,$ and $X \sim \mathsf{Binom}(n,p),$ then the rule fails, but $P(0.5 < X < 2.5) = P(X = 1) + P(X = 2) = 0.75.$ The approximating
normal distribution is $\mathsf{Norm}(np = 1.5, \sqrt{np(1-p)}= 0.866) = 0.7518,$ which is a good approximation.


In R statistical software n = 3; p = .5;  sum(dbinom(1:2, n, p))
returns the exact binomial probability 0.75.
While the code diff(pnorm(c(.5,2.5), n*p, sqrt(n*p*(1-p))))
returns the very good normal approximation 0.7517869


*

*If $n = 100, p = 0.1,$ and $Y \sim \mathsf{Binom}(n, p),$ then $np = 11$
and $P(Y \le 3) =  0.003426046$, but the normal approximation gives
0.008264727, which has a very large relative error, even though the
absolute error is in the second or third decimal place. So even the stricter
version of the rule does not always give ideal results.


Rule III does not have a direct connection to normality. It
shows when it is OK to use the binomial distribution (assuming sampling
with replacement) when the hypergeometric distribution (modeling
sampling without replacement) is exactly correct. (The only connection
with normality is that, in some circumstances. both distributions might be
approximated by normal. But this rule does not directly bear on whether
the normal approximation is accurate.) 
For example, suppose you want to know the probability of drawing exactly three
Aces in five draws from a 52-card deck. This is sampling without replacement
and the exact hypergeometric probability is  0.9982455, but three is less
than ten percent of 48, so the approximate binomial probability 0.9998357 is
very close.
In summary, I would never recommend Rule 1 for any purpose, Rule 2 is usually
a reasonable way of guessing whether it is safe to use a normal approximation
for a binomial probability, and Rule 3 has to do with the practical distinction
between binomial and hypergeometric distributions, with no direct
connection to normality.
Checking for normality: There are several ways to check whether
a reasonably large sample might have come from a normal population.
One is a normal probability plot (normal 'quantile plot' or 'normal Q-Q plot')
If the points on such a plot are randomly sampled from a normal population
the points on such a plot will fall nearly in a straight line (with the
understanding that straggling extreme points may fall quite a bit off the line). 
Here are such plots for three samples of size $n = 100:$ neither sample $X$ nor $Y$
comes from anormal population, but sample $Z$ does.

Another method is to use one of several tests for goodness-of-fit to a
normal distribution. One of the best ones is the Shapiro-Wilk test.
For the same three samples this test gave P-values 0.0000, 0.0004, and 0.9826,
respectively (to four places). P-values below 0.1 are pretty good evidence
for a non-normal population. 
For a sample as large as $n = 100,$ a large Shapiro-Wilk
P-value is usually evidence that the population is very nearly normal.
With a sample size as small as your suggested $n = 9,$ it would be very
difficult to say for sure whether or not the population is normal.
