Question

olympics the table shows the number of gold medals won by the united states at the winter olympics during various years.

year	1992	1994	1998	2002	2006	2010
gold medals	5	6	6	10	9	9

source: international olympic committee
a. write the equation in slope - intercept form for the best - fit line for the data. let x represent the number of years since 1990. round to the nearest hundredth, if necessary.
y = x+
b. find the correlation coefficient. round to the nearest thousandth, if necessary.
r≈

Explanation:

Step1: Prepare data points

Let \(x\) be the number of years since 1990. So the data points \((x,y)\) are: \((2,5),(4,6),(8,6),(12,10),(16,9),(20,9)\)

Step2: Calculate necessary sums

Let \(n = 6\).
\(\sum_{i = 1}^{n}x_{i}=2 + 4+8 + 12+16+20=62\)
\(\sum_{i = 1}^{n}y_{i}=5 + 6+6 + 10+9+9=45\)
\(\sum_{i = 1}^{n}x_{i}^{2}=2^{2}+4^{2}+8^{2}+12^{2}+16^{2}+20^{2}=4 + 16+64+144+256+400 = 884\)
\(\sum_{i = 1}^{n}x_{i}y_{i}=2\times5+4\times6 + 8\times6+12\times10+16\times9+20\times9=10 + 24+48+120+144+180 = 526\)

Step3: Calculate slope \(m\)

The formula for the slope \(m\) of the best - fit line is \(m=\frac{n\sum_{i = 1}^{n}x_{i}y_{i}-\sum_{i = 1}^{n}x_{i}\sum_{i = 1}^{n}y_{i}}{n\sum_{i = 1}^{n}x_{i}^{2}-(\sum_{i = 1}^{n}x_{i})^{2}}\)
\[

$$\begin{align*} m&=\frac{6\times526-62\times45}{6\times884 - 62^{2}}\\ &=\frac{3156-2790}{5304-3844}\\ &=\frac{366}{1460}\\ &= 0.25 \end{align*}$$

\]

Step4: Calculate y - intercept \(b\)

The formula for the y - intercept \(b\) is \(b=\overline{y}-m\overline{x}\), where \(\overline{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}=\frac{62}{6}\approx10.33\) and \(\overline{y}=\frac{\sum_{i = 1}^{n}y_{i}}{n}=\frac{45}{6} = 7.5\)
\[

$$\begin{align*} b&=7.5-0.25\times10.33\\ &=7.5 - 2.58\\ &=4.92 \end{align*}$$

\]
The equation of the best - fit line in slope - intercept form is \(y = 0.25x+4.92\)

Step5: Calculate correlation coefficient \(r\)

First, calculate \(\overline{x}=\frac{62}{6}\approx10.33\), \(\overline{y}=\frac{45}{6}=7.5\)
\(S_{xx}=\sum_{i = 1}^{n}x_{i}^{2}-n\overline{x}^{2}=884-6\times(10.33)^{2}=884 - 6\times106.77=884 - 640.62 = 243.38\)
\(S_{yy}=\sum_{i = 1}^{n}y_{i}^{2}-n\overline{y}^{2}\)
\(\sum_{i = 1}^{n}y_{i}^{2}=5^{2}+6^{2}+6^{2}+10^{2}+9^{2}+9^{2}=25+36+36+100+81+81 = 359\)
\(S_{yy}=359-6\times7.5^{2}=359 - 6\times56.25=359 - 337.5 = 21.5\)
\(S_{xy}=\sum_{i = 1}^{n}x_{i}y_{i}-n\overline{x}\overline{y}=526-6\times10.33\times7.5=526 - 464.85 = 61.15\)
The correlation coefficient \(r=\frac{S_{xy}}{\sqrt{S_{xx}S_{yy}}}=\frac{61.15}{\sqrt{243.38\times21.5}}=\frac{61.15}{\sqrt{5232.67}}=\frac{61.15}{72.34}\approx0.845\)

Answer:

a. \(y = 0.25x+4.92\)
b. \(r\approx0.845\)