MLB Player Salaries & Dark Chocolate's Vascular Health Impact

Understanding the distribution of sample means is crucial in statistics. Let's analyze two distinct scenarios.

MLB Player Salaries in 2012

In 2012, there were 855 major league baseball players. The mean salary was \(\mu = 3.44\) million dollars, with a standard deviation of \(\sigma = 4.70\) million dollars. We will examine random samples of size \(n = 50\) players to understand the distribution of their mean salaries.

Distribution of Sample Means

To describe the shape, center, and spread of the distribution of sample means, we apply the Central Limit Theorem (CLT). The CLT states that for sufficiently large sample sizes (typically \(n \ge 30\)), the sampling distribution of the sample mean will be approximately normal, irrespective of the population distribution's shape.

1. Shape

Given our sample size \(n = 50\), which is greater than 30, the shape of the distribution of sample means will be:

Approximately normal (bell-shaped).

2. Center

The mean of the sampling distribution of the sample mean is equal to the population mean:

\(\mu_{\bar{x}} = \mu = 3.44\) million dollars.

Therefore, the center of the distribution is:

3.44 million dollars.

3. Spread

The standard deviation of the sampling distribution of the sample mean, known as the standard error (SE), is calculated as:

\(\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{4.70}{\sqrt{50}} \approx \frac{4.70}{7.071} \approx 0.6649\) million dollars.

Thus, the spread (standard deviation of sample means) is:

Approximately 0.665 million dollars.

Summary for MLB Salaries:

• Shape: Approximately normal
• Center: 3.44 million dollars
• Spread: Approximately 0.665 million dollars

Dark Chocolate and Vascular Health

A study investigated the impact of daily dark chocolate consumption on vascular health. Eleven individuals consumed 46 grams of dark chocolate daily for two weeks. Their vascular health was measured before and after the period. Higher numbers indicate greater vascular health. The mean increase observed was 1.3, with a standard deviation of 2.32. The data are assumed to be reasonably symmetric with no extreme values.

a). 90% Confidence Interval for Mean Increase

We will calculate and interpret a 90% confidence interval for the mean increase in vascular health.

Given:

• Sample size \(n = 11\)
• Sample mean \(\bar{x} = 1.3\)
• Sample standard deviation \(s = 2.32\)
• Degrees of freedom \(df = n - 1 = 10\)
• The \(t^*\) value for 90% confidence and 10 degrees of freedom is approximately 1.812.

Calculation:

• Standard Error (SE):

\(SE = \frac{s}{\sqrt{n}} = \frac{2.32}{\sqrt{11}} \approx \frac{2.32}{3.317} \approx 0.699\)

• Margin of Error (ME):

\(ME = t^* \cdot SE = 1.812 \cdot 0.699 \approx 1.267\)

• Confidence Interval:

\(\bar{x} \pm ME = 1.3 \pm 1.267 = (0.033, 2.567)\)

Interpretation: We are 90% confident that the true mean increase in vascular health after two weeks of eating dark chocolate is between 0.033 and 2.567 units.

b). Confidence in Positive Mean Change

Yes, we can be 90% confident that the mean change for everyone would be positive. Since the entire confidence interval (0.033, 2.567) is above 0, it suggests that the true mean increase in vascular health is likely positive.

c). Margin of Error

The margin of error for the confidence interval is approximately 1.267.