Question

It is known that, nationally, doctors working for health maintenance organizations (HMOs) average 13.5 years of experience in their specialties, with a standard deviation of 7.6 years. The executive director of an HMO in a western state is interested in determining whether or not its doctors have less experience than the national average. A random sample of 150 doctors from HMOs shows a mean of only 10.9 years of experience. What is the T

Answer 1

Answer:

The value of the test statistic is z = -4.19.

Step-by-step explanation:

The test statistic is:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

In which X is the sample mean, $\mu$ is the expected value for the population mean, $\sigma$ is the standard deviation and n is the size of the sample.

Average 13.5 years of experience in their specialties, with a standard deviation of 7.6 years.

This means that $\mu = 13.5, \sigma = 7.6$

A random sample of 150 doctors from HMOs shows a mean of only 10.9 years of experience.

This means that $n = 150, X = 10.9$

What is the test statistic?

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

$z = \frac{10.9 - 13.5}{\frac{7.6}{\sqrt{150}}}$

$z = -4.19$

The value of the test statistic is z = -4.19.