Question

If you want to be 95​% confident of estimating the population mean to within a sampling error of plus or minus3 and the standard deviation is assumed to be 20​, what sample size is​ required? The sample size required is

Answer 1

Answer:

The sample size required is at least 171

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.95}{2} = 0.025$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$.

So it is z with a pvalue of $1-0.025 = 0.975$, so $z = 1.96$

Now, find the margin of error M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

What sample size is​ required?

A samle size of at least n is required, in which n is found when $M = 3, \sigma = 20$

So

$M = z*\frac{\sigma}{\sqrt{n}}$

$3 = 1.96*\frac{20}{\sqrt{n}}$

$3\sqrt{n} = 20*1.96$

$\sqrt{n} = \frac{20*1.96}{3}$

$(\sqrt{n})^{2} = (\frac{20*1.96}{3})^{2}$

$n = 170.7$

Rouding up

The sample size required is at least 171