Question

Examine the required sample size needed to be able to

Answer 1

Answer:

$n=(\frac{2.33(500)}{10})^2 =13572.25 \approx 13573$

And if we use a sample level of n =2000 the margin of error would be higher as we can see here:

$ME=2.33\frac{500}{\sqrt{2000}}=26.05$  

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".

$\bar X$ represent the sample mean for the sample  

$\mu$ population mean (variable of interest)

$\sigma=500$ represent the population standard deviation assumed

n represent the sample size  (variable of interest)

Confidence =98% or 0.98

ME = 10 represent the margin of error desired

The margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$    (a)

And on this case we have that ME =10 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$   (b)

The critical value for 98% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.01;0;1)", and we got $z_{\alpha/2}=2.33$, replacing into formula (b) we got:

$n=(\frac{2.33(500)}{10})^2 =13572.25 \approx 13573$

So the answer for this case would be n=13573 rounded up to the nearest integer

And if we use a sample level of n =2000 the margin of error would be higher as we can see here:

$ME=2.33\frac{500}{\sqrt{2000}}=26.05$