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Here is my question (homework obviously):

A sample from a normal population produced variance 4.0. Find the size of the sample if the sample mean deviates from the population mean by no more than 2.0 with a probability of at least 0.95.

So I'm trying to find $n$, the sample size, having only $\hat{\sigma}$, the sample variance, and a bound on the distance between $\bar{x}$ and $\mu$. My intuition was normally in this situation we need to use the t distribution since $\hat{\sigma}$ is an unbiased estimate for $\sigma$ (we did all the proofs in class).The problem is the t distribution changes depending on $n$, the sample size, so which distribution (how many degrees of freedom) should I consult when looking up the t-values containing 95% of the probability mass? I tried it for different values of $n$, and then squared the values to compare them to the d.f. of the t distribution - the closest I could get was 0.6 off. (I took the t-value at $\alpha = 0.025$ (right-tail) for 5 d.f., implying $n$ is 6, and squaring the t-value gave me 6.61, which is a discrepancy of 0.61 (isn't this large?). The reason I squared the t-values becomes apparent if you "normalize" the bound on the means into a t-statistic. Am I going about this correctly? This doesn't seem right...

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Are you familiar with confidence intervals for the mean? –  Daniel Littlewood Jan 21 '13 at 18:38
    
yes, that's what i did –  pootieman Jan 21 '13 at 18:40

1 Answer 1

A 95% confidence interval for the mean a small sample of a normal population of unknown variance has the form $$\left[\bar x \pm t\frac{s}{\sqrt{n-1}}\right]$$ Where $n$ is the sample size and, in your case, $s=\sqrt{4}=2$. The $n-1$ makes the estimator unbiased. The maximum deviation from the sample mean is, then, $\frac{ts}{\sqrt{n-1}}$. If $T \sim t_{n-1}$, then $t$ is the value such that $P(T\le t)=0.975$.

For $\frac{2t}{\sqrt{n-1}}<2$, $t<\sqrt{n-1}$. By comparing with tables, we have that when:
$n=6$, $t=2.571>\sqrt{5}$
$n=7$, $t=2.447<\sqrt{6}$

So $n$ must be at least $7$.

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