Confidence intervals

• January 26th 2010, 09:07 AM
PensFan10
Confidence intervals
im not asking to do the problems because i really want to learn this. i just need a little help understanding the material. i haven't done this stuff in quite some time and my book was on back order.

Thanks a lot.

1) A random sample size of 4 has a sample mean of 75 and a sample standard deviation of 8. Find the 99% confidence interval of for the population mean.

2) A population is known to have a SD of 30. You wish to estimate the population mean with a margin of error of 10 at the 95% confidence level. What is the minimum sample size required?

3) A random sample size of 144 has a sample mean of 300 and a sample SD of 24. Find the 90% conf. interval of the population mean.

4) You take a random sample of 36 from a population of mean 48 and SD of 12. find the probability that the sample mean is between 49 and 50.

My thoughts:

For number 1, i know its a small sample size so that means I need to use degrees of freedom ( i think). degrees of freedom is n-1 which is just 3. i am not really sure what to do with this information though. the formulate is: ybar plus/minus t(alpha/2)S(ybar)

ybar = sample mean = 75
t(alpha/2) = not really sure since i think i have to use degrees of freedom here
S(ybar) = sample sd i think.

For number 2, I have no idea how to deal with margin of error.

For number 3, its close to number one, buts its a large sample size so I dont have to use degrees of freedom. I looked up z value of .90 and got 1.645 divide that by 2. for alpha/2 = .8225 y bar = 300 and Sd of ybar = 24 so would it be...

300 plus/minus .8225(24) => 280.26 to 319.74

For number 4, I found the P(49-48/(12/sqrt(36)) less than equal to z less than equal to (50-48/(12/sqrt(36)). that gave P(.5 less than equal to z less than equal to 1) and then used the normal dist. chart to look up z values. i think this is right because it is a large sample size.
• January 26th 2010, 05:42 PM
statmajor
$\overline{X} +/- 1.96 \frac{s}{\sqrt{n}}$ is a CI at the 95% confidence level.

The corresponding marginal error is $1.96 \frac{s}{\sqrt{n}}$ (just equate the sample variance to the margin of error and solve for n)

Everything else seems right (haven`t checked your numerical calculations, but you have the right idea)

By the way, in part 1, is the population normal distirbution?