Sure, let's analyze the given data step-by-step to understand how we arrived at the final result.
Given data:
[tex]\[
\begin{array}{l|rrrrrrr}
x & 57 & 53 & 59 & 61 & 53 & 56 & 60 \\
\hline
y & 156 & 164 & 163 & 177 & 159 & 175 & 151
\end{array}
\][/tex]
Step 1: Calculate the means of [tex]\( x \)[/tex] and [tex]\( y \)[/tex]
The mean of [tex]\( x \)[/tex]:
[tex]\[
\bar{x} = \frac{\sum_{i=1}^n x_i}{n} = \frac{57 + 53 + 59 + 61 + 53 + 56 + 60}{7} = 57.0
\][/tex]
The mean of [tex]\( y \)[/tex]:
[tex]\[
\bar{y} = \frac{\sum_{i=1}^n y_i}{n} = \frac{156 + 164 + 163 + 177 + 159 + 175 + 151}{7} \approx 163.57
\][/tex]
Step 2: Calculate the covariance of [tex]\( x \)[/tex] and [tex]\( y \)[/tex]
The covariance formula is:
[tex]\[
\text{cov}(x, y) = \frac{\sum_{i=1}^n (x_i - \bar{x})(y_i - \bar{y})}{n-1}
\][/tex]
After calculating, we find:
[tex]\[
\text{cov}(x, y) \approx 3.33
\][/tex]
Step 3: Calculate the standard deviations of [tex]\( x \)[/tex] and [tex]\( y \)[/tex]
The standard deviation of [tex]\( x \)[/tex]:
[tex]\[
\sigma_x = \sqrt{\frac{\sum_{i=1}^n (x_i - \bar{x})^2}{n-1}} \approx 3.21
\][/tex]
The standard deviation of [tex]\( y \)[/tex]:
[tex]\[
\sigma_y = \sqrt{\frac{\sum_{i=1}^n (y_i - \bar{y})^2}{n-1}} \approx 9.55
\][/tex]
Step 4: Calculate the Pearson correlation coefficient
The Pearson correlation coefficient [tex]\( r \)[/tex] is given by:
[tex]\[
r = \frac{\text{cov}(x, y)}{\sigma_x \sigma_y}
\][/tex]
Using the covariance and standard deviations calculated:
[tex]\[
r \approx \frac{3.33}{3.21 \times 9.55} \approx 0.109
\][/tex]
Thus, the Pearson correlation coefficient between [tex]\( x \)[/tex] and [tex]\( y \)[/tex] is approximately [tex]\( 0.109 \)[/tex].
Conclusion
The Pearson correlation coefficient of 0.109 indicates a very weak positive linear relationship between the variables [tex]\( x \)[/tex] and [tex]\( y \)[/tex].
