Question

Desert Samaritan Hospital in Mesa, Arizona keeps record of emergency traffic. Those records reveal that the times between arriving patients have a special type of reverse-J-shaped distribution called an exponential distribution. They also indicate that the mean time between arriving patients in \(8.7\) minutes, as is the standard deviation. Suppose that you observe a sample of \(10\) interarrival times.

a. Theoretically what are the mean and standard deviation of all possible sample means?

b. Use the technology of your choice to simulate \(1000\) samples of \(10\) interarrival times each.

c. Determine the mean of each of the \(1000\) samples you obtained in part (b).

d. Roughly what would you expect the mean and standard deviation of the \(1000\) sample means you obtained in part (c) to be? Explain your answers.

e. Determine the mean and standard deviation of the \(1000\) sample means you obtained in part (c).

f. Compare your answers in parts (d) and (e). Why are they different?

Short answer

Part a. Mean \(=8.7\)

Standard Deviation \(=2.7512\)

Part b. The answers can vary depending on the first number selected.

Part c. The answers can vary depending on the first number selected.

Part d. Mean \(=8.7\)

Standard Deviation \(=42.7512\)

Part e. The answers can vary depending on the first number selected.

Part f. Sampling Error

Step-by-step solution

Part a. Step 1. Given information

\(\mu =\lambda =8.7\)

\(\sigma=8.7\)

\(n=10\)

Part a. Step 2. Calculation

It is known that the mean of the sample means of distribution of a sample is same as the population mean.

\(\mu_{\bar{x}}=\mu=8.7\)

It is known that the standard deviation of the sample means is same as the population standard deviation divided by under root of the size of the sample

\(\sigma _{x}=\frac{\sigma}{\sqrt{n}}=\frac{8.7}{\sqrt{10}}\approx2.7512\)

Part b. Step 1. Explanation

The data values can be simulated from a normal distribution of

Mean \(=8.7\)

Standard Deviation \(=2.7512\)

using excel with following formula:

\(=-LN(1-RAND())*8.7\)

Each of the \(10\) sample values can be simulated using the above command, then, this can be repeated until \(1000\) samples each with \(10\) values are finished being calculated.

A snapshot of the result obtained is as per follows:

Part c. Step 1. Calculation

The mean can be calculated as the summation of all values divided by the quantity of values.

For example, the following calculation is done for the one of the columns:

\(\bar{x}=\frac{7.582461582+...+0.0767191717}{10}=8.529210815\)

This can be repeated for all the columns..

Part d. Step 1. Explanation

The answer is expected to be

Mean \(=8.7\)

Standard Deviation \(=42.7512\)

The reason here is that the \(1000\) sample means will have the similar distribution as the sampling distribution of sample mean.

Part e. Step 1. Explanation

The mean can be calculated as the summation of all values divided by the quantity of values.

\(\bar{x}=\frac{8.529210815+...+8.95736067}{1000}=8.70346197\)

\(s=\sqrt{\frac{(8.529210815-8.70346197)^{2}+...+(8.95736067-8.70346197)^{2}}{1000-1}}\approx 2.749376\)

Part f. Step 1. Explanation

The main difference between the results is parts (d) and (e) is because of sampling error. To be more precise, different samples would have different means and different standard deviations even though the values would be very close to the expected values.