Question

A survey recently reported that the mean national annual expenditure for inpatient and outpatient services of all persons over 64 years of age was $5,423 with a standard deviation of $979. A random sample of 352 persons over age 64 living in Sudbury had an average expense of $5,516. We want to test whether the mean inpatient and outpatient expense of all Sudbury residents over age 64 is higher than the national average of $5,423. Calculate the P-value for this test.

Answer 1

Using the z-distribution, it is found that the p-value for this test is of 0.0375.

At the null hypothesis, it is tested if the expenses are the same as the national average, that is, of $5,423. Hence:

$H_0: \mu = 5423$

At the alternative hypothesis, it is tested if it is higher, hence:

$H_1: \mu > 5423$

We have the standard deviation for the population, thus, the z-distribution is used. The test statistic is given by:

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

The parameters are:

• $\overline{x}$ is the sample mean.
• $\mu$ is the value tested at the null hypothesis.
• $\sigma$ is the standard deviation of the sample.
• n is the sample size.

For this problem, the values of the parameters are: $\overline{x} = 5516, \mu = 5423, \sigma = 979, n = 352$

The value of the test statistic is:

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

$z = \frac{5516 - 5423}{\frac{979}{\sqrt{352}}}$

$z = 1.78$

The p-value is the probability of finding a sample mean above $5,516, which is 1 subtracted by the p-value of z = 1.78.

• Looking at the z-table, z = 1.78 has a p-value of 0.9625.

1 - 0.9625 = 0.0375

Hence, the p-value for this test is of 0.0375.

A similar problem is given at brainly.com/question/25549474