Statistic HELP Hope someone can

  • Thread starter Thread starter johnnyICON
  • Start date Start date
  • Tags Tags
    Statistic
AI Thread Summary
To determine the minimum sample size needed to achieve a margin of error of 0.9 at a 90% confidence level, the standard deviation (sigma) must be estimated. The formula for sample size is n = (z*σ)/m, where z* is 1.645 for the 90% confidence level. Although sigma is unknown, it can be derived from the existing margin of error and sample size. The relationship shows that sample size is inversely proportional to the square of the margin of error. Understanding this relationship simplifies the calculation without needing to know sigma directly.
johnnyICON
Messages
79
Reaction score
0
Statistic HELP! Hope someone can :(

I know this is only a physics and math help forum, but I'm pretty sure that most of you have taken a course in statistics... or so I hope. I'll take a long shot anyway...

I have a question where I need to find the sample size required in order to get a margin of error that is 0.9.


Here is the question:
A random sample size of 25 has been taken from a population in order to estimate the population mean. The standard deviation of the population is known. The sample mean is 10.2, and the associated margin of error, at 90% level of confidence, is equal to 1.8 If you want to reduce the margin of error to one-half of this value, using the same level of confidence, what would be the minimum sample size required?

The only formula I know in relation to margin of error is:
n = \frac{z^*\sigma}{m}

  • where z* is the critical value derived the the confidence level, that being 1.645 for this matter.
  • sigma is the standard deviation, which I DO NOT KNOW!
  • m is the margin of error, which will be 0.9

I've had two question's like this already. My real problem is figuring out how to find sigma, the standard deviation. If anyone can help... PLEASE DO! Thanks.
 
Physics news on Phys.org
johnnyICON said:
The only formula I know in relation to margin of error is:
n = \frac{z^*\sigma}{m}

  • where z* is the critical value derived the the confidence level, that being 1.645 for this matter.
  • sigma is the standard deviation, which I DO NOT KNOW!
  • m is the margin of error, which will be 0.9
First of all, I think you have to square the right side:

(1) n = |\frac{z^*\sigma}{m}|^2

This allows you to derive \sigma from the margin of error (which only makes sense, since it was used to calculate the margin of error):

(2) \sigma = \sqrt{n}\frac{m}{z^*}

Then plug it back into (1) to find n for m = .9 . But you really don't need to know \sigma. All you need to do is know that n \propto \frac{1}{m^2}

AM
 
Last edited:
AWESOME MAN, THANKS.

I should of saw that... I guess all this stress and anger is blinding me! :bugeye:
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top