Question

Suppose the heights of seasonal pine saplings are normally distributed. If the population standard deviation is 14 millimeters, what minimum sample size is needed to be 95% confident that the sample mean is within 4 millimeters of the true population mean?

Answer 1

Answer:

The minimum sample size needed to be 95% confident that the sample mean is within 4 millimeters of the true population mean is 47.

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.

If the population standard deviation is 14 millimeters, what minimum sample size is needed to be 95% confident that the sample mean is within 4 millimeters of the true population mean?

This is n when $\sigma = 14, M = 4$. So

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

$4 = 1.96*\frac{14}{\sqrt{n}}$

$4\sqrt{n} = 1.96*14$

$\sqrt{n} = \frac{1.96*14}{4}$

$(\sqrt{n})^{2} = (\frac{1.96*14}{4})^{2}$

$n = 47$

The minimum sample size needed to be 95% confident that the sample mean is within 4 millimeters of the true population mean is 47.