Obtain the linear correlation coefficient for the data. Round your answer to three decimal places.
-Two separate tests, x and y, are designed to measure a student's ability to solve problems. Several students are randomly selected to take both tests and their results are shown below. $\begin{array}{c|c|c|c|c|c|c|c|c|c}
\mathrm{x} & 48 & 52 & 58 & 44 & 43 & 43 & 40 & 51 & 59 \
\hline \mathrm{y} & 73 & 67 & 73 & 59 & 58 & 56 & 58 & 64 & 74
\end{array}$
A)$0.867$
B) $0.714$
C) $0.109$
D) $0.548$

Question

Quizplus · Accepted Answer

The linear correlation coefficient, often denoted as $r$, measures the strength and direction of a linear relationship between two variables. To calculate $r$, we use the formula:$$r = \frac{n(\sum xy) - (\sum x)(\sum y)}{\sqrt{[n\sum x^2 - (\sum x)^2][n\sum y^2 - (\sum y)^2]}}$$Given the data for $x$ and $y$, we first calculate the necessary sums:- $\sum x = 48 + 52 + 58 + 44 + 43 + 43 + 40 + 51 + 59 = 438$- $\sum y = 73 + 67 + 73 + 59 + 58 + 56 + 58 + 64 + 74 = 582$- $\sum xy = (48 \times 73) + (52 \times 67) + (58 \times 73) + (44 \times 59) + (43 \times 58) + (43 \times 56) + (40 \times 58) + (51 \times 64) + (59 \times 74)$- $\sum x^2 = 48^2 + 52^2 + 58^2 + 44^2 + 43^2 + 43^2 + 40^2 + 51^2 + 59^2$- $\sum y^2 = 73^2 + 67^2 + 73^2 + 59^2 + 58^2 + 56^2 + 58^2 + 64^2 + 74^2$Plugging these sums into the formula and calculating, we find that $r$ is approximately $0.867$, rounded to three decimal places. This calculation involves detailed arithmetic steps, including summing the products of $x$ and $y$, squaring each $x$ and $y$ value before summing, and applying these sums to the formula. The result indicates a strong positive linear relationship between $x$ and $y$.

Obtain the Linear Correlation Coefficient for the Data A) 0.8670.8670.867 B) 0.7140.7140.714

Obtain the Linear Correlation Coefficient for the Data A) $0.867$ B) $0.714$