# Estimating sample size en poussant

If we have sample mean of $$n$$ observations as $$\bar{x}$$. It is given that maximum value of the whole set is $$M$$ and minimum value is $$0$$. How many more observations we need so that the estimate for average we have differs from the true average by not more than 10 with 99% confidence?

• is this a home work? – Aksakal Jan 29 at 16:35
• no interview question. was asked to think in terms of Hoeffding's inequality but I had no idea. – Prayalankar Jan 29 at 16:36
• are you only given an upper bound? – Aksakal Jan 29 at 16:44
• also, wth is en poussant? I'd be looking incredulous at the interviewer is they threw that at me. is this an academic job? – Aksakal Jan 29 at 17:10
• @Aksakal that was my contribution. Literally means "in passing". Implied in the sense of sequential or cumulative. As stated, the question suggests some sampling is already done, and a sample size calculation is to be performed for the total sample to achieve desired precision. Sample size re-estimation is a hot topic in clinical trials. – AdamO Jan 29 at 18:33

The most important glaring correction is to the notion of having a desired precision of 10 within the actual mean. You cannot calculate that. You can set the desired precision of the statistic to within 10, but it may be more than 10 units from the actual mean more than 1% of the time (suppose the estimator or design is biased).

Knowing the maximum may be irrelevant. For many distributions the average is the sufficient statistic for the mean, so the maximum $$M$$ is an ancillary statistic and gives no additional information. There are counter examples to this, but since you have not stated a specific distribution, we assume $$M$$ is ancillary as a base case.

The other unknown(s) in this case are the standard error of the sample mean and the actual variance of the distribution. Assume they are $$s^2$$ and $$\sigma^2$$. (the former can be used as a plug in estimate for the latter with degrees of freedom corrections).

So with the current sample of $$n$$ the 99% confidence-interval-half-width (CIHW) of $$\bar{x}$$ is $$\mathcal{Z}_{0.995} \sqrt{s^2}$$ and by the addition of $$q$$ observations, that value is estimated to be $$\mathcal{Z}_{0.995} \sqrt{\frac{n s^2 + q \sigma^2}{n+q}}$$.

Set this value to 10 and solve for q.

• if actual variance of the distribution is not know than is it possible to calculate? – Prayalankar Jan 29 at 16:25
• @Prayalankar yes, but change $\mathcal{Z}_{0.995}$ to $\mathcal{T}_{df, 0.995}$ where $df$ is the appropriate number of degrees of freedom. – AdamO Jan 29 at 16:29

If you have upper and lower bounds M and 0, then you can apply Popovivicu's upper bound on variance: $$\sigma^2<\frac 1 4M^2$$

Once you have the variance, apply usual sample mean distribution logic, i.e. the variance of a sample mean $$\bar x$$ to be $$\sigma^2_{\bar x}\sim\sigma^2/n$$. The rest is trivial.

$$\frac {10}{(\sigma^2_{\bar x}/n)}=2.58$$ $$n=0.258\sigma^2_{\bar x}=0.067M^2$$

# "Derivation"

Ok, how would I proceed to derive this? I'd imagine the distribution with highest variance (entropy) possible, that is bounded between M and 0. That must be Bernoulli: $$x=0:p=1/2$$ $$x=M:p=1/2$$ Then the mean is $$\mu=M/2$$ and the variance is $$\sigma^2=M^2/2-(M/2)^2=M^2/4$$