Question

1. the data in the table to the left shows the olympic 500-meter gold medal speed skating times.

year | 1980 | 1984 | 1988 | 1992 | 1994 | 1998
time (s) | 422 | 432 | 404 | 420 | 395 | 382
a) find the line of best fit:
b) estimate the 500-meter time for the 2020 olympics.

Explanation:

Step1: Define Variables

Let \( x \) be the year (we can use a transformed variable for simplicity, e.g., \( x = \text{year} - 1980 \)) and \( y \) be the time in seconds.
So the data points are:
\( (0, 422) \), \( (4, 432) \), \( (8, 404) \), \( (12, 420) \), \( (14, 395) \), \( (18, 382) \)

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

\( \bar{x} = \frac{0 + 4 + 8 + 12 + 14 + 18}{6} = \frac{56}{6} \approx 9.333 \)
\( \bar{y} = \frac{422 + 432 + 404 + 420 + 395 + 382}{6} = \frac{2455}{6} \approx 409.167 \)

Step3: Calculate Slope (\( m \))

\( m = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{\sum (x_i - \bar{x})^2} \)
First, calculate \( (x_i - \bar{x})(y_i - \bar{y}) \) for each point:
- For \( (0, 422) \): \( (0 - 9.333)(422 - 409.167) \approx (-9.333)(12.833) \approx -119.8 \)
- For \( (4, 432) \): \( (4 - 9.333)(432 - 409.167) \approx (-5.333)(22.833) \approx -121.8 \)
- For \( (8, 404) \): \( (8 - 9.333)(404 - 409.167) \approx (-1.333)(-5.167) \approx 6.9 \)
- For \( (12, 420) \): \( (12 - 9.333)(420 - 409.167) \approx (2.667)(10.833) \approx 28.9 \)
- For \( (14, 395) \): \( (14 - 9.333)(395 - 409.167) \approx (4.667)(-14.167) \approx -66.1 \)
- For \( (18, 382) \): \( (18 - 9.333)(382 - 409.167) \approx (8.667)(-27.167) \approx -235.5 \)

Sum of these: \( -119.8 - 121.8 + 6.9 + 28.9 - 66.1 - 235.5 \approx -497.4 \)

Now, calculate \( (x_i - \bar{x})^2 \) for each point:
- \( (0 - 9.333)^2 \approx 87.11 \)
- \( (4 - 9.333)^2 \approx 28.44 \)
- \( (8 - 9.333)^2 \approx 1.78 \)
- \( (12 - 9.333)^2 \approx 7.11 \)
- \( (14 - 9.333)^2 \approx 21.78 \)
- \( (18 - 9.333)^2 \approx 75.11 \)

Sum of these: \( 87.11 + 28.44 + 1.78 + 7.11 + 21.78 + 75.11 \approx 221.33 \)

So, \( m = \frac{-497.4}{221.33} \approx -2.25 \)

Step4: Calculate Y-Intercept (\( b \))

Using \( y = mx + b \), and \( \bar{y} = m\bar{x} + b \)
\( b = \bar{y} - m\bar{x} \approx 409.167 - (-2.25)(9.333) \approx 409.167 + 21.0 \approx 430.167 \)

So the line of best fit is \( y = -2.25x + 430.167 \), where \( x = \text{year} - 1980 \) or \( y = -2.25(\text{year} - 1980) + 430.167 \), simplifying to \( y = -2.25\text{year} + 430.167 + 4450 = -2.25\text{year} + 4880.167 \)

Step5: Estimate for 2020

For 2020, \( x = 2020 - 1980 = 40 \)
Using \( y = -2.25(40) + 430.167 = -90 + 430.167 = 340.167 \) (Wait, this seems off. Maybe calculation errors. Let's recalculate the slope more accurately.)

Wait, maybe a better way is to use linear regression formula. Let's use the original years (1980, 1984, 1988, 1992, 1994, 1998) as \( x \):

\( x: 1980, 1984, 1988, 1992, 1994, 1998 \)
\( y: 422, 432, 404, 420, 395, 382 \)

\( \bar{x} = \frac{1980 + 1984 + 1988 + 1992 + 1994 + 1998}{6} = \frac{11936}{6} \approx 1989.333 \)
\( \bar{y} = \frac{422 + 432 + 404 + 420 + 395 + 382}{6} = \frac{2455}{6} \approx 409.167 \)

\( \sum (x_i - \bar{x})(y_i - \bar{y}) \):
- (1980 - 1989.333)(422 - 409.167) = (-9.333)(12.833) ≈ -119.8
- (1984 - 1989.333)(432 - 409.167) = (-5.333)(22.833) ≈ -121.8
- (1988 - 1989.333)(404 - 409.167) = (-1.333)(-5.167) ≈ 6.9
- (1992 - 1989.333)(420 - 409.167) = (2.667)(10.833) ≈ 28.9
- (1994 - 1989.333)(395 - 409.167) = (4.667)(-14.167) ≈ -66.1
- (1998 - 1989.333)(382 - 409.167) = (8.667)(-27.167) ≈ -235.5

Sum: -119.8 -121.8 +6.9 +28.9 -66.1 -235.5 = (-119.8 -121.8) + (6.9 +28.9) + (-66.1 -235.5) = (-241.6) + (35.8) + (-301.6) = -241.6 +35.8 -301.6 = -507.4

\( \sum (x_i - \bar{x})^2 \):
- (1980 - 1989.333)^2 = (-9.333)^2 ≈ 87.11
- (1984 - 1989.333)^2 = (-5.333)^2 ≈ 28.44
- (1988 - 1989.333)^2 = (-1.333)^2 ≈ 1.78
- (1992 - 1989.333)^2 = (2.667)^2 ≈ 7.11
- (1994 - 1989.333)^2 = (4.667)^2 ≈ 21.78
- (1998 - 1989.333)^2 = (8.667)^2 ≈ 75.11

Sum: 87.11 +28.44 +1.78 +7.11 +21.78 +75.11 = 221.33

So slope \( m = \frac{-507.4}{221.33} \approx -2.292 \)

Then \( b = \bar{y} - m\bar{x} \approx 409.167 - (-2.292)(1989.333) \approx 409.167 + 2.292*1989.333 \)

Calculate 2.292*1989.333 ≈ 2.292*2000 - 2.292*10.667 ≈ 4584 - 24.45 ≈ 4559.55

So \( b \approx 409.167 + 4559.55 ≈ 4968.717 \)

So the line of best fit is \( y = -2.292x + 4968.717 \)

Now, for 2020, \( x = 2020 \)
\( y = -2.292*2020 + 4968.717 ≈ -4629.84 + 4968.717 ≈ 338.877 \)

Wait, but the times are decreasing, which makes sense (faster times). But let's check with the data. From 1980 (422) to 1998 (382), the time decreased by 40 seconds in 18 years, so about 2.22 seconds per year. So the slope is about -2.22, which matches our calculation. So for 2020, which is 40 years after 1980, the time would be 422 - 2.22*40 = 422 - 88.8 = 333.2 seconds. Our linear regression gives about 338.8, which is close. Maybe due to the data points (1984 has a higher time, 432, which is an outlier? Let's check the data again: 1980:422, 1984:432 (higher), 1988:404 (lower), 1992:420 (higher), 1994:395 (lower), 1998:382 (lower). So there's some variation, but the overall trend is decreasing.

So for part a) the line of best fit is approximately \( y = -2.29x + 4969 \) (or in terms of years since 1980, \( y = -2.29t + 422 \), where \( t \) is years since 1980, since at t=0 (1980), y=422, and slope -2.29).

For part b), using the line of best fit, when x=2020, y ≈ 339 seconds (approximate).

But maybe the problem expects a simpler approach, like using two points to find the slope. Let's take (1980, 422) and (1998, 382). The slope is (382 - 422)/(1998 - 1980) = (-40)/18 ≈ -2.222. Then the equation is \( y - 422 = -2.222(x - 1980) \). For 2020, \( x = 2020 \), so \( y = 422 - 2.222(40) = 422 - 88.88 = 333.12 \) seconds.

Answer:

a) The line of best fit is approximately \( y = -2.29x + 4969 \) (or \( y = -2.22(x - 1980) + 422 \))
b) The estimated time for 2020 is approximately 333 - 340 seconds (more accurately, using linear regression, around 339 seconds)

(Note: Due to calculation approximations, the exact value may vary slightly, but the method is linear regression for the line of best fit and then extrapolation for 2020.)