Question

Suppose the population standard deviation is σ = 5 , an SRS of n = 100 is obtained, and the confidence level is chosen to be 98%. The margin of error for estimating a mean μ is given by: 1.165. 0.1228. 1.228. 0.1165.

Answer 1

Answer:

1.165.

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:

$\alpha = \frac{1-0.98}{2} = 0.01$

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 $1-0.01 = 0.99$, so $z = 2.33$

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 problem, we have that:

$\sigma = 5, n = 100$. So

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

$M = 2.33*\frac{5}{\sqrt{100}} = 1.165$

So the correct answer is:

1.165.