Question

you want to create a 99 confidence interval with a margin of error of 25 assuming that the population standard deviation is equal to 1.5 what's the minimum size of the Rams Tampa

Answer 1

Answer:

The minimum sample size is 239.

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:

$\alpha = \frac{1 - 0.99}{2} = 0.005$

Now, we have to find z in the Z-table as such z has a p-value of $1 - \alpha$.

That is z with a pvalue of $1 - 0.005 = 0.995$, so Z = 2.575.

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.

Population standard deviation is equal to 1.5

This means that $\sigma = 1.5$

Margin of error of 0.25

This means that $M = 0.25$

What's the minimum size of the sample?

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

$0.25 = 2.575\frac{1.5}{\sqrt{n}}$

$0.25\sqrt{n} = 2.575*1.5$

$\sqrt{n} = \frac{2.575*1.5}{0.25}$

$(\sqrt{n})^2 = (\frac{2.575*1.5}{0.25})^2$

$n = 238.7$

Rounding up:

The minimum sample size is 239.