Question

Suppose the germination periods, in days, for grass seed are normally distributed. If the population standard deviation is 3 days, what minimum sample size is needed to be 90% confident that the sample mean is within 1 day of the true population mean? z0.101.282z0.051.645z0.0251.960z0.012.326z0.0052.576 Use the table above for the z-score, and be sure to round up to the nearest integer.

Answer 1

Answer:

The minimum sample size is needed to be 90% confident that the sample mean is within 1 day of the true population mean is 25.

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, 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 3 days, what minimum sample size is needed to be 90% confident that the sample mean is within 1 day of the true population mean?

This is n when $\sigma = 3, M = 1$. So

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

$1 = 1.645*\frac{3}{\sqrt{n}}$

$\sqrt{n} = 3*1.645$

$(\sqrt{n})^{2} = (3*1.645)^{2}$

$n = 24.3$

Rouding up to the nearest integer, 25.

The minimum sample size is needed to be 90% confident that the sample mean is within 1 day of the true population mean is 25.