Question

A researcher surveyed 220 residents of a city about the number of hours they

Answer 1

Answer:

Margin of error of 0.0485 hours.

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 $1 - \alpha$.

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.

In this question:

$\sigma = 0.35, n = 220$

The margin of error is of:

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

$M = 1.96\frac{0.35}{\sqrt{200}}$

$M = 0.0485$

Margin of error of 0.0485 hours.