Question

The travel-to-work time for residents in Ventura County is unknown. Assume the population variance is 39. How large should a sample be if the margin of error is 1 minute for a 93% confidence interval

Answer 1

Answer:

A sample size of 128 is needed.

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:

$\alpha = \frac{1-0.93}{2} = 0.035$

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 $1-0.035 = 0.965$, so $z = 1.81$

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(square root of the variance) and n is the size of the sample.

How large should a sample be if the margin of error is 1 minute for a 93% confidence interval

We need a sample size of n, which is found when $M = 1$. We have that $\sigma = \sqrt{39}$. So

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

$1 = 1.81*\frac{\sqrt{39}}{\sqrt{n}}$

$\sqrt{n} = 1.81\sqrt{39}$

$(\sqrt{n})^2 = (1.81\sqrt{39})^2$

$n = 127.8$

Rounding up

A sample size of 128 is needed.