1 Answer

Ask a Question

Sean Kenny · Answer 1 · 2021-07-03T06:40:32+0000

Answer:

$n=((1.960(10))/(2))^2 =96.04 \approx 97$

So the answer for this case would be n=97 rounded up to the nearest integer . And this value is reaonable and possible in the reality.

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)

$\sigma$ represent the population standard deviation

n represent the sample size

Solution to the problem

The margin of error is given by this formula:

$ME=z_(\alpha/2)(\sigma)/(√(n))$ (a)

And on this case we have that ME =2 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

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

The critical value for 95% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.025;0;1)", and we got
$z_(\alpha/2)=1.960$ , replacing into formula (b) we got:

$n=((1.960(10))/(2))^2 =96.04 \approx 97$

So the answer for this case would be n=97 rounded up to the nearest integer . And this value is reaonable and possible in the reality.

0 Comments

Please log in or register to add a comment.

Please log in or register to answer this question.

1 Answer

0 Comments

Please log in or register to add a comment.

Other Questions