Calculate the correlation coefficient for the following data. Round your answer to the nearest thousandth.

\begin{tabular}{|c|c|}
\hline [tex]$x$[/tex] & [tex]$y$[/tex] \\
\hline 3 & 37.8 \\
\hline 5 & 75.8 \\
\hline 7 & 62.3 \\
\hline 11 & 52.8 \\
\hline 13 & 85.5 \\
\hline 16 & 41.2 \\
\hline 19 & 75.1 \\
\hline
\end{tabular}



Answer :

To calculate the correlation coefficient for the given data and round it to the nearest thousandth, follow these steps:

### Step 1: Gather the Data

List the given [tex]\( x \)[/tex] and [tex]\( y \)[/tex] values:

[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 3 & 37.8 \\ \hline 5 & 75.8 \\ \hline 7 & 62.3 \\ \hline 11 & 52.8 \\ \hline 13 & 85.5 \\ \hline 16 & 41.2 \\ \hline 19 & 75.1 \\ \hline \end{array} \][/tex]

### Step 2: Calculate the Means

Compute the mean (average) of [tex]\( x \)[/tex] and [tex]\( y \)[/tex]:

[tex]\[ \bar{x} = \frac{\sum x}{n} = \frac{3 + 5 + 7 + 11 + 13 + 16 + 19}{7} = \frac{74}{7} \approx 10.571 \][/tex]

[tex]\[ \bar{y} = \frac{\sum y}{n} = \frac{37.8 + 75.8 + 62.3 + 52.8 + 85.5 + 41.2 + 75.1}{7} = \frac{430.5}{7} \approx 61.5 \][/tex]

### Step 3: Compute the Covariance

First, find the products of deviations from the mean for each pair [tex]\((x_i, y_i)\)[/tex]:

[tex]\[ (x_i - \bar{x})(y_i - \bar{y}) \][/tex]

Then sum these products:

[tex]\[ \sum (x_i - \bar{x})(y_i - \bar{y}) \approx \sum ([(3 - 10.571)(37.8 - 61.5), (5 - 10.571)(75.8 - 61.5), \dots]) \][/tex]

### Step 4: Compute the Standard Deviations

Calculate the standard deviations of [tex]\( x \)[/tex] and [tex]\( y \)[/tex]:

[tex]\[ s_x = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}} \][/tex]

[tex]\[ s_y = \sqrt{\frac{\sum (y_i - \bar{y})^2}{n-1}} \][/tex]

### Step 5: Compute the Correlation Coefficient

Using the formula for the Pearson correlation coefficient:

[tex]\[ r = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{(n-1)s_x s_y} \][/tex]

### Step 6: Round the Result

Finally, after calculating the correlation coefficient, round it to the nearest thousandth.

The correlation coefficient for the given data is approximately 0.241 when rounded to the nearest thousandth.