Question

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Answer 1

Answer:

c) .22

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.

In this question:

$\sigma = 2.5, n = 500$

Then

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

$M = 1.96*\frac{2.5}{\sqrt{500}}$

$M = 0.22$