Respuesta :
Answer:
[tex] Width = \bar X +z_{\alpha/2} \frac{\sigma}{\sqrt{n}} -(\bar X -z_{\alpha/2} \frac{\sigma}{\sqrt{n}}) = 2z_{\alpha/2} \frac{\sigma}{\sqrt{n}}[/tex]
And we want to reduce this width by a factor of 1/2 so the new margin of error would be:
[tex] ME= z_{\alpha/2} \frac{\sigma}{\sqrt{n}}[/tex]
And in order to do this we need to increase the sample size by a factor of 4 since:
[tex]2z_{\alpha/2} \frac{\sigma}{\sqrt{4n}} =\frac{2}{2}z_{\alpha/2} \frac{\sigma}{\sqrt{n}}[/tex]
Halving the length requires n to be increased by a factor of 4
If the sample zise is increased by a factor of 25 we got:
[tex] 2z_{\alpha/2} \frac{\sigma}{\sqrt{25n}} =\frac{2}{5}2z_{\alpha/2} \frac{\sigma}{\sqrt{n}}[/tex] the width is reduced by a factor of 5
Step-by-step explanation:
Previous concepts
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
Solution to the problem
For this case we know the the confidence interval for the mean is btween these two limits:
[tex] (\bar X -z_{\alpha/2} \frac{\sigma}{\sqrt{n}} ,\bar X +z_{\alpha/2} \frac{\sigma}{\sqrt{n}}) [/tex]
And for this case the width is given by:
[tex] Width = \bar X +z_{\alpha/2} \frac{\sigma}{\sqrt{n}} -(\bar X -z_{\alpha/2} \frac{\sigma}{\sqrt{n}}) = 2z_{\alpha/2} \frac{\sigma}{\sqrt{n}}[/tex]
And we want to reduce this width by a factor of 1/2 so the new margin of error would be:
[tex] ME= z_{\alpha/2} \frac{\sigma}{\sqrt{n}}[/tex]
And in order to do this we need to increase the sample size by a factor of 4 since:
[tex]2z_{\alpha/2} \frac{\sigma}{\sqrt{4n}} =\frac{2}{2}z_{\alpha/2} \frac{\sigma}{\sqrt{n}}[/tex]
Halving the length requires n to be increased by a factor of 4
If the sample zise is increased by a factor of 25 we got:
[tex] 2z_{\alpha/2} \frac{\sigma}{\sqrt{25n}} =\frac{2}{5}2z_{\alpha/2} \frac{\sigma}{\sqrt{n}}[/tex] the width is reduced by a factor of 5
Using confidence interval concepts, it is found that:
- For the width to be halved, the sample size has to be increased by a factor of 4, that is, it has to be multiplied by 4.
- If the sample size is increased by a factor of 25, the width is 1/5 of the original width.
The width of a confidence interval is twice the margin of error, thus:
[tex]W = 2z\frac{\sigma}{\sqrt{n}}[/tex]
In which:
- z is critical value.
- [tex]\sigma[/tex] is the standard deviation.
- n is the sample size.
First, we want the width to be halved.
- It is inversely proportional to the square root of the sample size.
- Thus, for the width to be halved, the sample size has to be increased by a factor of 4, that is, it has to be multiplied by 4.
Then, if the sample size is increased by a factor of 25, [tex]\sqrt{25} = 5[/tex], thus, the width will be one-fifth of the original width.
A similar problem is given at https://brainly.com/question/22618167
