Question

The Athletic Coping Skills Inventory (ACSI) is a test to measure psychological skills believed to influence athletic performance. Researchers E. Estanol er al. studied the relationship between ACSI scores and eating disorders in dancers in the article "Mental Skills as Protective Attributes Against Eating Disorder Risk in Dancers. The study found that dancers ACSI scores are normally distributed with mean \(49.34\) points and standard deviation \(10.14\) points.

a. Use the technology of your choice to simulate \(1000\) samples of four dancers ACSI scores each.

b. Find the sample mean of each of the \(1000\) samples obtained in part (a).

c. Obtain the mean, the standard deviation and a histogram of the \(1000\) sample means.

d. Theoretically what are the mean, standard deviation and distribution of all possible sample means for samples of size \(4\)?

e. Compare your answers from parts (c) and (d).

Short answer

Part a. The samples will be like

Part b. The mean will be \(\mu=[49.3615, 49.1953, 48.7853, 49.1773]\)

Part c. The standard deviation and mean will be

\(\mu_{bar{x}}=49.6020\)

\(\sigma_{bar{x}}=\frac{10.14}{\sqrt{4}}=5.07\)

Part d. \(\mu_{\bar{x}}=\mu=49.34

\(\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}=\frac{10.14}{\sqrt{4}}=5.07\)

Part e. We can see that the mean part (c) is slightly lower than the theoretical mean of part (d), whereas the standard deviation of part (c) is slightly higher than part (d).

However, we do note that the mean and standard deviation of part (c) and (d) are close in value.

Step-by-step solution

Part a. Step 1. Given information

The sample size, mean and standard deviation is given.

\(\mu=mean=49.34\)

\(\sigma=standard deviation=10.14\)

\(n=sample size=4\)

Part a. Step 2. Calculation

So, according to the question I have used MATLAB to generate the samples.

We use the function \(“norminv (rand (1000,4), 49.34, 10.14)”\).

After using the above function in MATLAB, we will get the \(1000\) sample for \(4\) dancers.

Program:

Query:

• First, we have defined the number of samples.
• Then using function “norminv” to find out the samples for dancers.

Part b. Step 1. Calculation

So, according to the question I have used MATLAB to generate the samples.

We use the function \(“norminv (rand (1000,4), 49.34, 10.14)”\).

Then calculate the mean

\(\mu=\frac{\sum_{i-1}^{1000}x_{i}}{n}\)

After solving the equation, we will get

\(\mu=[49.3615, 49.1953, 48.7853, 49.1773]\)

Program:

Query:

• First, we have defined the number of samples.
• Then using function “norminv” to find out the samples for dancers.
• Then calculate the mean using function “mean”.

Part c. Step 1. Calculation

So, according to the question I have used MATLAB to generate the samples.

We use the function \(“norminv (rand (1000,4), 49.34, 10.14)”\).

Then calculate the mean

\(\mu=\frac{\sum_{i-1}^{1000}x_{i}}{n}\)

After solving the equation, we will get

\(\mu=[49.3615, 49.1953, 48.7853, 49.1773]\)

Calculate the mean of means

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

\(\mu_{bar{x}}=49.6020\)

and standard deviation

\(\sigma_{\bar{x}}=\frac{10.14}{\sqrt{4}}=5.07\)

Program:

Query:

• First, we have defined the number of samples.
• Then using function “norminv” to find out the samples for dancers.
• Then calculate the mean using function “mean”.
• Calculate the mean of means and standard deviation.
• Sketch a histogram.

Part d. Step 1. Explanation

The mean of the sampling distribution is equal to the population mean.

\(\mu_{\bar{x}}=\mu=49.34\)

The standard deviation of the sampling distribution is equal to the population standard deviation divide by the square root of the sample size.

\(\sigma_{\bar{x}}=\frac{10.14}{\sqrt{4}}=5.07\)

Part e. Step 1. Explanation

The mean of the sampling distribution is equal to the population mean.

\(\mu_{\bar{x}}=\mu=49.6020\)

\(\mu_{\bar{x}}=\mu=49.34\)

The standard deviation of the sampling distribution is equal to the population standard deviation divide by the square root of the sample size.

\(\sigma_{\bar{x}}=\frac{10.14}{\sqrt{4}}=5.07\)