Annual high temperatures in a certain location have been tracked for several years. Let $X$ represent the year and $Y$ the high temperature. Based on the data shown below, calculate the correlation coefficient (to three decimal places) between $X$ and $Y$. Use your calculator!
\begin{tabular}{|r|r|}
\hline$x$ & $y$ \\
\hline 5 & 23.2 \\
\hline 6 & 24.02 \\
\hline 7 & 26.24 \\
\hline 8 & 26.16 \\
\hline 9 & 29.48 \\
\hline 10 & 28.7 \\
\hline
\end{tabular}
\[
r=
\]

Step 1 ：First, we calculate the sum of x values, \(\Sigma x = 5 + 6 + 7 + 8 + 9 + 10 = 45\)

Step 2 ：Next, we calculate the sum of y values, \(\Sigma y = 23.2 + 24.02 + 26.24 + 26.16 + 29.48 + 28.7 = 157.8\)

Step 3 ：Then, we calculate the sum of squares of x values, \(\Sigma x^2 = 5^2 + 6^2 + 7^2 + 8^2 + 9^2 + 10^2 = 255\)

Step 4 ：We also calculate the sum of squares of y values, \(\Sigma y^2 = 23.2^2 + 24.02^2 + 26.24^2 + 26.16^2 + 29.48^2 + 28.7^2 = 4058.6928\)

Step 5 ：Next, we calculate the sum of the products of paired data, \(\Sigma xy = 5*23.2 + 6*24.02 + 7*26.24 + 8*26.16 + 9*29.48 + 10*28.7 = 1135.36\)

Step 6 ：The number of pairs of data, n, is 6

Step 7 ：Substitute these values into the formula for the correlation coefficient, r

Step 8 ：Calculate the numerator, \([ 6*1135.36 - 45*157.8 ] = -288.84\)

Step 9 ：Calculate the denominator, \(\sqrt{ [6*255 - 45^2] [6*4058.6928 - 157.8^2] } = \sqrt{271542.2356} = 521.10\)

Step 10 ：Finally, calculate the correlation coefficient, r, \(r = \frac{-288.84}{521.10} = -0.554\)

Step 11 ：\(\boxed{r = -0.554}\) is the correlation coefficient between X and Y. This indicates a moderate negative relationship between the year and the high temperature.