Question

Determine the minimum sample size required when you want to be 75% confident that the sample mean is within 30 units of the population mean assume the population standard deviation of 327.8 in a normally distributed population

Answer 1

Answer:

The minimum sample size is 158.

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:

$\alpha = \frac{1-0.75}{2} = 0.125$

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 $1-0.125 = 0.875$, so $z = 1.15$

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.

In this question:

The minimum sample size is n.

n is found when M = 30.

We have that $\sigma = 327.8$

So

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

$30 = 1.15*\frac{327.8}{\sqrt{n}}$

$30\sqrt{n} = 1.15*327.8$

$\sqrt{n} = \frac{1.15*327.8}{30}$

$(\sqrt{n})^{2} = (\frac{1.15*327.8}{30})^{2}$

$n = 157.9$

Rounding up

The minimum sample size is 158.