Questions: The following table contains data from a sample of ten students regarding the numbers of parking tickets they received during one semester and their monthly incomes (including allowances from parents as well as paychecks from employment as income). Construct a 95% confidence interval for the y-intercept of the regression line. Round your answers to two decimal places, if necessary. Parking Tickets and Monthly Income Number of Tickets, x Monthly Income (in Dollars), y ------ 1 1600 8 3300 5 2800 4 2100 4 2300 3 1900 1 1000 0 1100 5 2500 1 1000

Solution Steps

Calculate the means of the independent variable \( x \) and the dependent variable \( y \):

\[ \bar{x} = \frac{1}{n} \sum_{i=1}^{n} x_i = 3.2 \]

\[ \bar{y} = \frac{1}{n} \sum_{i=1}^{n} y_i = 1960.0 \]

Calculate the correlation coefficient \( r \):

\[ r = 0.9642 \]

Calculate the numerator and denominator for the slope \( \beta \):

Numerator:

\[ \sum_{i=1}^{n} x_i y_i - n \bar{x} \bar{y} = 79800 - 10 \cdot 3.2 \cdot 1960.0 = 17080.0 \]

Denominator:

\[ \sum_{i=1}^{n} x_i^2 - n \bar{x}^2 = 158 - 10 \cdot (3.2)^2 = 55.6 \]

Thus, the slope \( \beta \) is calculated as:

\[ \beta = \frac{17080.0}{55.6} = 307.1942 \]

Calculate the intercept \( \alpha \):

\[ \alpha = \bar{y} - \beta \bar{x} = 1960.0 - 307.1942 \cdot 3.2 = 976.9784 \]

The line of best fit is given by the equation:

\[ y = 976.9784 + 307.1942x \]

Calculate the standard error of the estimate \( s_{yx} \):

\[ s_{yx} = 222.80100486719084 \]

Calculate the standard error of the intercept \( se_{\alpha} \):

\[ se_{\alpha} = 118.7704201877319 \]

Determine the critical value for \( t \) at a 95% confidence level with \( n-2 \) degrees of freedom. The margin of error is calculated as:

\[ \text{Margin of Error} = t_{\text{critical}} \cdot se_{\alpha} \]

Thus, the 95% confidence interval for the intercept is:

\[ (\alpha - \text{Margin of Error}, \alpha + \text{Margin of Error}) = (703.09, 1250.86) \]

Final Answer