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

Respuesta :

Answer:

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

Step-by-step explanation:

The test statistic is:

[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

In which X is the sample mean, [tex]\mu[/tex] is the expected value for the population mean, [tex]\sigma[/tex] 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 [tex]\mu = 13.5, \sigma = 7.6[/tex]

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

This means that [tex]n = 150, X = 10.9[/tex]

What is the test statistic?

[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

[tex]z = \frac{10.9 - 13.5}{\frac{7.6}{\sqrt{150}}}[/tex]

[tex]z = -4.19[/tex]

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