Respuesta :
Answer:
[tex]n = 2662.56\sigma^2[/tex]
Step-by-step explanation:
We are given the following in the question:
Significance level = 0.01
Width of interval = 0.1
Population variance = [tex]\sigma^2[/tex]
We have to find the sample size so that the width of the confidence interval is no larger than 0.1
[tex]z_{critical}\text{ at}~\alpha_{0.01} = \pm 2.58[/tex]
Formula for sample size:
[tex]n = \displaystyle\frac{z^2\sigma^2}{E^2}[/tex]
where E is the margin of error. Since the confidence interval width is 0.1,
[tex]E = 0.05[/tex]
Putting these values in the equation:
[tex]n = \displaystyle\frac{(2.58)^2\sigma^2}{(0.05)^2} = 2662.56\sigma^2[/tex]
So, the above expression helps us to calculate the sample size so that the width of the confidence interval is no larger than 0.1 for different sample variances.
