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I've recently been learning about margin of error, i.e. $E=\frac{1}{\sqrt{n}}$.

I met this problem which doesn't ask to find any margin of error but I wonder can it be used in this situation.

Nails have a mean of 20mm and a standard deviation of 3mm. If 10000 nails were measured, how many would have a length between 17mm and 26mm?

The answer to the question is 8185.

But can an error or an uncertainty of some sort be included?

The margin of error is 1% which is 100 nails. So is it correct to give an answer of $8185\pm 100$ to 95% confidence? Or is the margin of error not appropriate here?

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$MOE = t_{\frac{\alpha}{2}}.\frac{s}{\sqrt{n}}$ if population variance is not known and sample size is small

or $MOE = z\frac{\sigma}{\sqrt{n}}$ if population variance is known and/ or sample size is large.

Where $\sigma$ is the standard deviation of what is being measured and n is the sample size. In this problem, you are asked to find the number of nails that fall into a given range of size having given their distribution information. From my understanding of statistics, MOE is well suited for point estimates, for example, a statistic like "mean" or "proportion", etc., giving a range of values the estimate could fall above and below the statistic under a certain confidence level. In your example 100 is not the MOE but $= \frac{1.96\times3}{100}$ and what it means is you measured 10,000 nails and the average you found to be some $\bar X$ and the MOE allows you to report that the average of the 10000 nails that you measured was =$\bar X \pm MOE$ .

In your problem, you are supposed to find $P(\frac{17-20}{3}\le Z \le \frac{26-20}{3})$ $ = P(-1\le Z\le 2) = .8185$

The interpretation of which is if 10000 nails were to be sampled from this population,it is likely that $(0.8185\times 10000) = 8185$ of them would have their length fall between 17mm ad 26mm. There is already an uncertainty built into it. That is the difference between sampling distribution for which MOE is relevant and the statistics of samples from original distribution.

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  • $\begingroup$ Misinformation at the start: The estimated margin of error for making a 95% confidence interval for $\mu$ based on a random sample from a normal population with unknown $\mu$ and $\sigma$ is $t^*S/\sqrt{n},$ where $S$ is the sample standard deviation and $t^*$ cuts 2.5% of the probability from the upper tail of Student's t distribution with $n - 1$ degrees of freedom. This is regardless of sample size (although for large $n,$ $t^*$ does become near 1.96). If $\mu$ is unknown and $\sigma$ is known, then the margin of error is $1.96\sigma/\sqrt{n}$ again regardless of $n.$ $\endgroup$ – BruceET Dec 19 '17 at 17:00
  • $\begingroup$ Imprecision of language at the end: "distributional uncertainty" built in to "it". Distributional uncertainty sounds like what economists say when they don't have a clue, anyhow not useful statistical terminology. And what is 'it'? Also, no mention of assumption of normality--anywhere. Anyhow, (+1) for answer "$\dots P(-1 \le Z \le2) \dots$" which is correct and to the point. $\endgroup$ – BruceET Dec 19 '17 at 17:08
  • $\begingroup$ I am an MBA wish I were a PhD Statistics. Writes like an MBA , is it not? Ha Ha Ha!!! $\endgroup$ – Satish Ramanathan Dec 19 '17 at 17:16
  • $\begingroup$ In my experience, having an MBA is neither a necessary nor sufficient condition for making sense of statistics. $\endgroup$ – BruceET Dec 19 '17 at 19:52

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