Question

accidents and drivers ages for a random selection of drivers, their ages and the number of automobile accidents that they had over a 3 - year period are shown. is there a linear relationship between the variables? use age for the independent variable and accidents for the dependent variable. age accidents 21 5 22 6 40 1 35 2 41 3 50 0 28 4 25 4 send data to excel the correlation coefficient for the data is r = - 0.923 and α = 0.01. should regression analysis be done? the regression analysis should not be done. the regression analysis should be done. find the equation of the regression line. round the coefficients to at least three decimal places. y = a + bx a = b = find the predicted number of accidents for a person age 30. round to the nearest whole accident. a 30 - year - old driver would be predicted to have accidents.

Explanation:

Step1: Identify Variables and Data

Let \( x \) be Age (independent variable) and \( y \) be Accidents (dependent variable). The data points are:
\((21, 5)\), \((22, 6)\), \((40, 1)\), \((35, 2)\), \((41, 3)\), \((50, 0)\), \((28, 4)\), \((25, 4)\).

Step2: Calculate Mean of \( x \) and \( y \)

First, find \( \bar{x} \) (mean of \( x \)) and \( \bar{y} \) (mean of \( y \)):
\( \bar{x} = \frac{21 + 22 + 40 + 35 + 41 + 50 + 28 + 25}{8} = \frac{282}{8} = 35.25 \)
\( \bar{y} = \frac{5 + 6 + 1 + 2 + 3 + 0 + 4 + 4}{8} = \frac{25}{8} = 3.125 \)

Step3: Calculate Slope \( b \)

The formula for slope \( b \) in linear regression is:
\( b = r \cdot \frac{s_y}{s_x} \), but we can also use the formula:
\( b = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{\sum (x_i - \bar{x})^2} \)

First, compute \( (x_i - \bar{x})(y_i - \bar{y}) \) and \( (x_i - \bar{x})^2 \) for each data point:

• For \( (21, 5) \): \( (21 - 35.25)(5 - 3.125) = (-14.25)(1.875) = -26.71875 \); \( (-14.25)^2 = 203.0625 \)
• For \( (22, 6) \): \( (22 - 35.25)(6 - 3.125) = (-13.25)(2.875) = -38.1875 \); \( (-13.25)^2 = 175.5625 \)
• For \( (40, 1) \): \( (40 - 35.25)(1 - 3.125) = (4.75)(-2.125) = -10.09375 \); \( (4.75)^2 = 22.5625 \)
• For \( (35, 2) \): \( (35 - 35.25)(2 - 3.125) = (-0.25)(-1.125) = 0.28125 \); \( (-0.25)^2 = 0.0625 \)
• For \( (41, 3) \): \( (41 - 35.25)(3 - 3.125) = (5.75)(-0.125) = -0.71875 \); \( (5.75)^2 = 33.0625 \)
• For \( (50, 0) \): \( (50 - 35.25)(0 - 3.125) = (14.75)(-3.125) = -46.09375 \); \( (14.75)^2 = 217.5625 \)
• For \( (28, 4) \): \( (28 - 35.25)(4 - 3.125) = (-7.25)(0.875) = -6.34375 \); \( (-7.25)^2 = 52.5625 \)
• For \( (25, 4) \): \( (25 - 35.25)(4 - 3.125) = (-10.25)(0.875) = -8.96875 \); \( (-10.25)^2 = 105.0625 \)

Now, sum the numerator (\( \sum (x_i - \bar{x})(y_i - \bar{y}) \)):
\( -26.71875 - 38.1875 - 10.09375 + 0.28125 - 0.71875 - 46.09375 - 6.34375 - 8.96875 = -136.875 \)

Sum the denominator (\( \sum (x_i - \bar{x})^2 \)):
\( 203.0625 + 175.5625 + 22.5625 + 0.0625 + 33.0625 + 217.5625 + 52.5625 + 105.0625 = 809.5 \)

Thus, \( b = \frac{-136.875}{809.5} \approx -0.1691 \) (rounded to four decimals for accuracy).

Step4: Calculate Intercept \( a \)

Using \( a = \bar{y} - b\bar{x} \):
\( a = 3.125 - (-0.1691)(35.25) \approx 3.125 + 5.961 \approx 9.086 \)

Step5: Regression Equation

The regression equation is \( \hat{y} = a + bx \approx 9.086 - 0.169x \) (rounded to three decimal places: \( a \approx 9.086 \), \( b \approx -0.169 \)).

Step6: Predict for Age 30

Substitute \( x = 30 \) into the regression equation:
\( \hat{y} = 9.086 - 0.169(30) = 9.086 - 5.07 = 4.016 \approx 4 \) (rounded to nearest whole number).

Answer:

• Regression equation: \( \hat{y} \approx 9.086 - 0.169x \) (so \( a \approx 9.086 \), \( b \approx -0.169 \))
• Predicted accidents for 30-year-old: \( 4 \)