point estimates and endpoints of confidence interval

**chrissy72** · June 10th 2009, 05:45 PM

distribution of x is N(mu,4) n=10:
55.95
56.54
57.58
55.13
57.48
56.06
59.93
58.30
52.57
58.46

point estimate?
endpoints for 95% confidence interval for mu?
probability when one is selected at random that it is less than 52?

i'm lost and would loooove some help..... thankyou

**Random Variable** · June 10th 2009, 06:15 PM

$\bar{x} \pm z_{0.025} \ \Big( \frac {\sigma}{\sqrt{n}}\Big)$

where $\bar{x}$ is the sample mean (which you have to calculate), $\sigma$ is the standard deviation, and $n$ is the sample size

**chrissy72** · June 10th 2009, 06:30 PM

i got a mean of 56.83, a variance of 3.814 and a standard deviation of 1.95295 but don't know what to do with them. where do you get the z0.025 from? thankyou for helping me!

**Random Variable** · June 10th 2009, 07:07 PM

$\sigma$ is the standard deviation of the population, not the sample. So in this problem $\sigma = \sqrt{4} = 2$ .

You get $z_{0.025}$ from a standard normal distribution table. There is one in just about every statistics textbook. It's the value of z such that P(Z>z) = 0.025. The value is 1.96.

Have you gone over this stuff in class?

**matheagle** · June 10th 2009, 08:33 PM

Chrissy, plug in 1.96 into Free One Tailed Area Under the Standard Normal Curve Calculator
and see that the area to the right of 1.96 is .0249978.

**chrissy72** · June 13th 2009, 01:15 PM

All I have is a textbook. No class. Would the average of the 10 values be considered a point estimate of $\mu$ ? Also, if the area to the right of 1.96 is 0.249978, would the endpoints of the 95% confidence interval be 1.9849978 on the right and 1.96 or 1.9350022 on the left?

Thanks. I really don't know what I am doing and my text is no help. I need an actual example to learn from but it is very hard to find. I don't have enough calculus OR statistic background to solve these problems without a better book OR an actual teacher. Any tutors out there? =)

**Random Variable** · June 13th 2009, 01:45 PM

From the data you calculated the sample mean ( $\bar{x}$ ) to be 56.83.

So a 95% confidence inteveral for the population mean ( $\mu$ ) is

$\Big[56.83 - 1.96* \frac{2}{\sqrt{10}}, \ 56.83 + 1.96*\frac{2}{\sqrt{10}}\Big]$

[55.59, 58.07]

Now either this interval contains $\mu$ or it doesn't. But if you constructed many such intervals by taking different samples of the same size, 95% of them would contain $\mu$ .

**chrissy72** · June 13th 2009, 01:59 PM

you are wonderful! Thankyou. So would that in fact be considered a point estimate (56.83)? I'm not clear on exactly what that is. My book only mentions that it is a point rather than an interval, but I don't know which point. Also, would the probability of randomly choosing a variable less than 52 be simply 0? Thanks once again! you have made this so much easier to understand.

**Random Variable** · June 13th 2009, 02:32 PM

EDIT:

If $X$ is $N(\mu, \sigma^{2})$ , then $\frac {X-\mu}{\sigma}$ is $N(0,1)$ (or just $Z$ ) .

So $P(X<52) = P\Big(\frac{X -\mu}{\sigma} < \frac{52 - \mu}{2} \Big) = P(Z<?)$

But that can't be answered unless you know $\mu$

**chrissy72** · June 13th 2009, 02:40 PM

I feel so silly, I think I can just do the old (52-56.83)/2=-2.415, but 2.415 is not on my table so 2.42 becomes 0.0078. I'm not sure how to write that out. Is it $\Phi2.42=0.0078$ ?

**matheagle** · June 13th 2009, 09:19 PM

you can plug 2.415 into....
Free One Tailed Area Under the Standard Normal Curve Calculator
and get .0078676

**Random Variable** · June 13th 2009, 09:37 PM

But I'm fairly certain (which probably means I'm wrong) that $\frac {X-\bar{x}}{\sigma}$ is not N(0,1).

**matheagle** · June 13th 2009, 09:55 PM

It has mean zero and it is a normal.
I doubt the variance is 1.
Is this X part of the sample, where $\bar X$ was derived from or a different observaton?

If it's part of the sample, then ....

$V\biggl({X_1-\bar X\over \sigma}\biggr)={V\biggl(X_1-\bar X\biggr) \over \sigma^2}$

$={V\biggl(X_1\bigl(1-{1\over n}\bigr)-{\sum_{k=2}^n X_k\over n}\biggr) \over \sigma^2}$

$={\bigl(1-{1\over n}\bigr)^2\sigma^2 +{n-1\over n^2}\sigma^2 \over \sigma^2}$

$=1-{1\over n}$

Otherwise ...

$V\biggl({X_{n+1}-\bar X\over \sigma}\biggr)={\sigma^2+{\sigma^2\over n} \over \sigma^2}=1+{1\over n}$

which is exactly what you get in a prediction interval, inside the square root of course.

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