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Use a calculator to find the [tex]r[/tex]-value of these data. Round the value to three decimal places.

\begin{tabular}{|c|c|}
\hline
[tex]$x$[/tex] & [tex]$y$[/tex] \\
\hline
5 & 19 \\
\hline
7 & 17 \\
\hline
10 & 16 \\
\hline
15 & 12 \\
\hline
19 & 7 \\
\hline
\end{tabular}

A. 0.985
B. -0.971
C. 0.971
D. -0.985



Answer :

GinnyAnswer
To find the [tex]\( r \)[/tex]-value, or Pearson correlation coefficient, for the given data set, follow these steps:

1. List the given data points:
[tex]\[ \begin{array}{c|c} x & y \\ \hline 5 & 19 \\ 7 & 17 \\ 10 & 16 \\ 15 & 12 \\ 19 & 7 \\ \end{array} \][/tex]

2. Understand the Pearson correlation coefficient formula:
[tex]\[ r = \frac{n(\sum xy) - (\sum x)(\sum y)}{\sqrt{ [n \sum x^2 - (\sum x)^2][n \sum y^2 - (\sum y)^2] }} \][/tex]
where [tex]\( n \)[/tex] is the number of data points.

3. Plug in the given data into the formula to find [tex]\( r \)[/tex].
- Calculate the sums:
[tex]\[ \sum x = 5 + 7 + 10 + 15 + 19 = 56 \][/tex]
[tex]\[ \sum y = 19 + 17 + 16 + 12 + 7 = 71 \][/tex]
[tex]\[ \sum xy = (5 \times 19) + (7 \times 17) + (10 \times 16) + (15 \times 12) + (19 \times 7) = 95 + 119 + 160 + 180 + 133 = 687 \][/tex]
[tex]\[ \sum x^2 = 5^2 + 7^2 + 10^2 + 15^2 + 19^2 = 25 + 49 + 100 + 225 + 361 = 760 \][/tex]
[tex]\[ \sum y^2 = 19^2 + 17^2 + 16^2 + 12^2 + 7^2 = 361 + 289 + 256 + 144 + 49 = 1099 \][/tex]

4. Substitute into the Pearson formula to get [tex]\( r \)[/tex]:
[tex]\[ r = \frac{5(687) - (56)(71)}{\sqrt{[5(760) - 56^2][5(1099) - 71^2]}} \][/tex]
Simplify the numerator and denominator:
[tex]\[ \text{Numerator} = 3435 - 3976 = -541 \][/tex]
[tex]\[ \text{Denominator} = \sqrt{ [5 \times 760 - 56^2][5 \times 1099 - 71^2] } = \sqrt{ [3800 - 3136][5495 - 5041] } = \sqrt{664 \times 454} \approx \sqrt{301456} \approx 549.05 \][/tex]
[tex]\[ r = \frac{-541}{549.05} \approx -0.985338 \][/tex]

5. Round the computed [tex]\( r \)[/tex]-value to three decimal places:
[tex]\[ r \approx -0.985 \][/tex]

The correct answer is hence option D: [tex]\(-0.985\)[/tex].

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