# Why is my answer wrong? (statistics)

1. Dec 10, 2012

### Artusartos

Let $\bar{X}$ be the mean of a random sample of size n from a distribution that is $N(\mu,9)$. Find n such that $P(\bar{X}-1 < \mu < \bar{X}+1)=.90$, approximately.

$$-Z^* < \frac{\bar{X}-\mu}{9/\sqrt(n)} < Z^*$$ where $-Z^*$ and $Z^*$ are the critical values.

So...

For the confidence interval, we have $$\bar{x} \pm z^*(\frac{9}{\sqrt(n)})$$

When I looked up the normal table for $z^*$, I found that it was approximately equal to 1.29

Since the question tells us that the confidence interval is $$\bar{X} \pm 1$$, so I just solved $$1 = (1.29)(\frac{9}{\sqrt(n)})$$...but my answer was wrong...it was supposed to be 24 or 25. Can anybody please help?

Last edited by a moderator: Dec 10, 2012
2. Dec 10, 2012

### Staff: Mentor

A probability of 0.9 to be inside gives a probability of 0.05 for both sides. You have to look for 0.95 and not 0.90. In addition, this value should be at the other side of your =.

3. Dec 10, 2012

### Ray Vickson

Besides what 'mfb' has told you, there is always the issue of what the '9' means in N(μ,9). Most commonly, the notation is $N(\mu, \sigma^2),$ so that $\sigma = 3.$ I have seen the other notation $N(\mu,\sigma)$ used occasionally, but it is rarer. You need to check the convention used by your textbook and/or course notes. (I won't spoil your fun by telling you the answer.)

Last edited: Dec 11, 2012