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

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

Explanation:

We have that to find our
`\alpha` level, that is the subtraction of 1 by the confidence interval divided by 2. So:

`\alpha = (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*(\sigma)/(√(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*(7)/(√(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.