# point estimates and endpoints of confidence interval

#### chrissy72

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

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

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

chrissy72

#### chrissy72

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

$$\displaystyle \sigma$$ 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?

chrissy72

#### chrissy72

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? =)

#### Random Variable

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 $$\displaystyle \mu$$.

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chrissy72

#### chrissy72

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

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 $$\displaystyle \mu$$

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#### chrissy72

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