Question

The ages of registered voters in Smith County are normally distributed with a population standard deviation of 3 years and an unknown population mean. A random sample of 18 voters is taken and results in a sample mean of 55 years. Find the margin of error for a 95% confidence interval for the population mean. 20.01 20.10 1.282 20.05 1.645 20.025 1.960 20.005 2.576 2.326 You may use a calculator or the common z values above. • Round the final answer to two decimal places.

Answer 1

Answer:

The margin of of error for a 95% confidence interval for the population mean is of 1.39 years.

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.

The margin of error is:

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

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

Population standard deviation of 3 years

This means that $\sigma = 3$

Sample of 18 voters

This means that $n = 18$

Margin of error:

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

$M = 1.96\frac{3}{\sqrt{18}}$

$M = 1.39$

The margin of of error for a 95% confidence interval for the population mean is of 1.39 years.