Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This problem has me scared to my wits, mostly because there are more like it!

The mean of a random sample of n observations drawn from a N $ ( \mu ,\sigma^{2} ) $ distribution is denoted by $ \overline{\chi} $. Given that P $ (| \overline{\chi} - \mu| > 0.5\sigma ) \lt 0.05 $

Find the smallest value of n.

So I tried:
Since $\overline{\chi}$ ~ N $ \left ( \mu ,\frac{\sigma^2}{n} \right ) $
P $ (| \overline{\chi} - \mu| > 0.5\sigma ) \lt 0.05 $ Should be:
$$ \frac{0.5\sigma - \mu}{\sqrt{\sigma^{2}/n}} \lt 0.05$$ $$1-\phi\left(\frac{0.5\sigma - \mu}{\sqrt{\sigma^{2}/n}}\right) \lt 0.05 $$
($\phi$ is the normal distribution function, I lookup a value from the normal dist. table.) $$ \frac{0.5\sigma - \mu}{\sqrt{\sigma^{2}/n}} \lt 1.645$$

Now where do I go from here!? This is totally confusing me, especially that darn modulus. How do I use my result so far with this?
P $ (| \overline{\chi} - \mu| > 0.5\sigma )$

Thanks Gideon

share|improve this question
    
This seems to be Chebyshev's Inequality. In particular, we have: $P(|x-\mu| \geq k) \leq \frac{\sigma^2}{k^2}$. Also $P(|x-\mu| \geq k \sigma) \leq \frac{1}{k^2}$. –  PEV Nov 6 '10 at 14:28
    
You should also have an equation of how well matched the standard deviation of your measurements and the standard deviation of the real distribution are. So you can plug the range of sigmas into your last equation and get and n from that. –  Ross Millikan Nov 6 '10 at 14:31
    
@Ross =( Not sure what you mean? How do I get rid of the mod? I wish I could do something like this: P $(\overline{\chi} > \mu + 0.5\sigma )$ –  gideon Nov 6 '10 at 14:48
    
Your measurements have a standard deviation as well as a mean. That standard deviation is an estimate of the real standard deviation, but has some error. There is some discussion in Wikipedia under Gaussian distribution/estimation of parameters and probably more in your book. So if plug the range of mus and sigmas that are consistent with your measurement in your last equation you can get n. –  Ross Millikan Nov 6 '10 at 15:01
    
Do you mean that it says n should be >30 to suitably estimate a distribution with the central limit theorem. –  gideon Nov 6 '10 at 15:17
add comment

1 Answer

up vote 2 down vote accepted

This is not a good question because it actually has nothing to do with sampling theory (nor with the Chebyshev inequality, for that matter): it's quite artificial. Nevertheless, you have grasped its intent well. A good way to work through it is to reason in rather basic terms, rather than relying solely on mathematical manipulation; to wit:

  • The standard deviation of $\bar{\chi}$ is $1/\sqrt{n}$ times the SD of the parent distribution, $\sigma$.

  • Therefore we should express the cutoff $0.5 \sigma$ in terms of the SD of $\bar{\chi}$ itself:

    $$0.5 \sigma = 0.5 \sqrt{n} \left(\sigma / \sqrt{n}\right) = 0.5 \sqrt{n} SD_{\bar{\chi}}.$$

  • We know (or can look up in a table) that $5$% of the time any normally distributed variate is further than $1.96$ times its standard deviation from its mean. (The value $1.645$ corresponds to a probability of $10$%: the absolute value requires you to look at both tails, not just the right hand tail.)

  • Thus the question is merely asking you to solve the equation

$$0.5 \sqrt{n} \ge 1.96.$$

share|improve this answer
    
yay! This came somewhat handy in my paper today, which was SO awesum!! =) –  gideon Nov 8 '10 at 17:22
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.