Question

Suppose the speeds of vehicles traveling on a highway are normally distributed and have a known population standard deviation of 7 miles per hour and an unknown population mean. A random sample of 32 vehicles is taken and gives a sample mean of 64 miles per hour.
Find the margin of error for the confidence interval for the population mean with a 98% confidence level. You may use a calculator or the common z values above. Round the final answer to two decimal places.

Z0.10 Z0.05 Z0.025 Z0.01 Z0.005
1.282 1.645 1.960 2.326 2.576

Answer 1

Answer:

The margin of error for the confidence interval for the population mean with a 98% confidence level is 2.88 miles per hour.

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 $z = 2.326$

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.

Find the margin of error for the confidence interval for the population mean with a 98% confidence level.

$M = 2.326*\frac{7}{\sqrt{32}} = 2.88$

The margin of error for the confidence interval for the population mean with a 98% confidence level is 2.88 miles per hour.