Question

What should a size of a sample be in order to have a margin of error of 4% for a 90% confidence level from a population with a standard deviation of 35?

Answer 1

Answer:

We need a sample size of 2,071,800 or higher.

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:

$\alpha = \frac{1-0.9}{2} = 0.05$

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 $1-0.05 = 0.95$, so $z = 1.645$

Now, we 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.

In this problem:

We need a sample size of n or higher, when $M = 0.04, \sigma = 35$. So

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

$0.04 = 1.645*\frac{35}{\sqrt{n}}$

$0.04\sqrt{n} = 1.645*35$

$0.04\sqrt{n} = 57.575$

$\sqrt{n} = 1439.375$

$\sqrt{n}^{2} = (1439.375)^{2}$

$n = 2,071,800$

We need a sample size of 2,071,800 or higher.