ISTER_REG
Junior Member
- Joined
- Oct 28, 2020
- Messages
- 59
Hello,
I am currently dealing with this task, to which I unfortunately lack the intuition to solve this. I quote here once the exact task:
How large a sample should one take in order to be 95% confident that a population standard deviation will not differ from a sample standard deviation by more than 2%
Typically, I would start with this approach, for example (is "z" here right (normal dist.) or better t-distribution?):
[MATH]\bar{X} \pm z_c \frac{\sigma}{\sqrt{n}}[/MATH]
because I am only interested in the error here I could also reduce this to:
[MATH]z_c \frac{\sigma}{\sqrt{n}}[/MATH]
Because I know that it is a 95% confidence level, then I can conclude that [MATH]z_c = 1.96[/MATH].
Now the next thing I would try to do is to consider the 2% in something like this.
[MATH]1.96 \frac{\sigma}{\sqrt{n}} = 0.02[/MATH]
However, I do not get further, so I am grateful for your tips!
I am currently dealing with this task, to which I unfortunately lack the intuition to solve this. I quote here once the exact task:
How large a sample should one take in order to be 95% confident that a population standard deviation will not differ from a sample standard deviation by more than 2%
Typically, I would start with this approach, for example (is "z" here right (normal dist.) or better t-distribution?):
[MATH]\bar{X} \pm z_c \frac{\sigma}{\sqrt{n}}[/MATH]
because I am only interested in the error here I could also reduce this to:
[MATH]z_c \frac{\sigma}{\sqrt{n}}[/MATH]
Because I know that it is a 95% confidence level, then I can conclude that [MATH]z_c = 1.96[/MATH].
Now the next thing I would try to do is to consider the 2% in something like this.
[MATH]1.96 \frac{\sigma}{\sqrt{n}} = 0.02[/MATH]
However, I do not get further, so I am grateful for your tips!