Question

Use 6.84 days as a planning value for the population standard deviation. a. Assuming 95% confidence, what sample size would be required to obtain a margin of error of days

Answer 1

Answer:

The sample size required is $n = (\frac{1.96*6.84}{M})^2$, in which M is the desired margin of error. If n is a decimal number, it is rounded up to the next integer.

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 Z-table as such z has a p-value of $1 - \alpha$.

That 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.

Use 6.84 days as a planning value for the population standard deviation.

This means that $\sigma = 6.84$.

What sample size would be required to obtain a margin of error of M days?

This is n for the given value of M. So

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

$M = 1.96\frac{6.84}{\sqrt{n}}$

$M\sqrt{n} = 1.96*6.84$

$\sqrt{n} = \frac{1.96*6.84}{M}$

$(\sqrt{n})^2 = (\frac{1.96*6.84}{M})^2$

$n = (\frac{1.96*6.84}{M})^2$

The sample size required is $n = (\frac{1.96*6.84}{M})^2$, in which M is the desired margin of error. If n is a decimal number, it is rounded up to the next integer.