Thread: point estimates and endpoints of confidence interval

1. point estimates and endpoints of confidence interval

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

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

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

3. 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!

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

You get $\displaystyle z_{0.025}$ 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?

5. 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.

6. All I have is a textbook. No class. Would the average of the 10 values be considered a point estimate of $\displaystyle \mu$? 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? =)

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

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

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

[55.59, 58.07]

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

8. 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.

9. EDIT:

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

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

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

10. 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 $\displaystyle \Phi2.42=0.0078$?

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

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

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

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

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

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

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

$\displaystyle =1-{1\over n}$

Otherwise ...

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

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