# Confidence Intervals and Sample Size

• Dec 8th 2008, 12:39 AM
ccdelia7
Confidence Intervals and Sample Size
What is the 2 sided confidence interval for the mean strength of a new structural column? The sample size is 25, the mean is 15,000 pounds, and the sample standard deviation is 200 lbs. Use a confidence level of 95 %.
• Dec 8th 2008, 12:49 AM
mr fantastic
Quote:

Originally Posted by ccdelia7
What is the 2 sided confidence interval for the mean strength of a new structural column? The sample size is 25, the mean is 15,000 pounds, and the sample standard deviation is 200 lbs. Use a confidence level of 95 %.

This is a simple substitution into a standard formula that should be in your textbook or class notes. The formula is here too: 1.3.5.2. Confidence Limits for the Mean.

Where exactly are you stuck?
• Dec 8th 2008, 01:10 AM
ccdelia7
Honestly, I just think my professor is not clear...I am trying to study for a final, and I cannot figure this problem out. This was an old homework problem, and I don't know how to start. Can you walk me through it?
• Dec 8th 2008, 01:11 AM
ccdelia7
I believe, from this data, beta is .05?
• Dec 8th 2008, 01:34 AM
mr fantastic
Quote:

Originally Posted by ccdelia7
Honestly, I just think my professor is not clear...I am trying to study for a final, and I cannot figure this problem out. This was an old homework problem, and I don't know how to start. Can you walk me through it?

The formula in the link I gave you is $\displaystyle \overline{Y} \pm t_{\alpha/2, \, N-1} \frac{s}{\sqrt{N}}$.

Substitute $\displaystyle \overline{Y} = 15,000$, $\displaystyle s = 200$ and $\displaystyle N = 25$.

$\displaystyle \alpha = 0.05$ and $\displaystyle N = 25$ mean that you require the value of $\displaystyle t_{\alpha/2, \, N-1} = t_{0.025, \, 24}$ (24 is the number of degrees of freedom). From a standard table of critical values I get $\displaystyle t_{0.025, \, 24} = 2.064$.

Substitute all this into the formula to get your confidence interval.
• Dec 8th 2008, 01:43 AM
ccdelia7
Please correct me if I'm wrong...

Lower limit = 14983.5
Upper limit = 15016.5

I substituted what you told me. That's my problem, I dont know where to start on a problem, because there are so many different tests...
• Dec 8th 2008, 01:53 AM
ccdelia7
I'm sorry, forgot the square root...

Lower Limit = 14917.6
Upper Limit = 15082.6
• Dec 8th 2008, 02:06 AM
mr fantastic
Quote:

Originally Posted by ccdelia7
I'm sorry, forgot the square root...

Lower Limit = 14917.6 Mr F says: I get 14917.44
Upper Limit = 15082.6

I'd be inclined to round to 14917 lbs and 15083 lbs.