Question

use the given data to complete parts (a) and (b).

(a) draw a scatter diagram of the data. choose the correct answer below.

compute the linear correlation coefficient. the linear correlation coefficient for the four pieces of data is
(round to three decimal places as needed.)

Explanation:

Step1: Calculate the means of \(x\) and \(y\)

The mean of \(x\) values: \(\bar{x}=\frac{2.1 + 3.8+3+4.7}{4}=\frac{13.6}{4}=3.4\)
The mean of \(y\) values: \(\bar{y}=\frac{3.8 + 1.5+3.6+4.9}{4}=\frac{13.8}{4}=3.45\)

Step2: Calculate the numerator and denominator of the correlation coefficient formula

The formula for the linear correlation coefficient \(r\) is \(r=\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})(y_{i}-\bar{y})}{\sqrt{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}\sum_{i = 1}^{n}(y_{i}-\bar{y})^{2}}}\)

For \(i = 1\): \((x_{1}-\bar{x})(y_{1}-\bar{y})=(2.1 - 3.4)(3.8-3.45)=(- 1.3)\times0.35=-0.455\)
\((x_{1}-\bar{x})^{2}=(2.1 - 3.4)^{2}=(-1.3)^{2}=1.69\)
\((y_{1}-\bar{y})^{2}=(3.8 - 3.45)^{2}=0.35^{2}=0.1225\)

For \(i = 2\): \((x_{2}-\bar{x})(y_{2}-\bar{y})=(3.8 - 3.4)(1.5 - 3.45)=0.4\times(-1.95)=-0.78\)
\((x_{2}-\bar{x})^{2}=(3.8 - 3.4)^{2}=0.4^{2}=0.16\)
\((y_{2}-\bar{y})^{2}=(1.5 - 3.45)^{2}=(-1.95)^{2}=3.8025\)

For \(i = 3\): \((x_{3}-\bar{x})(y_{3}-\bar{y})=(3 - 3.4)(3.6 - 3.45)=(-0.4)\times0.15=-0.06\)
\((x_{3}-\bar{x})^{2}=(3 - 3.4)^{2}=(-0.4)^{2}=0.16\)
\((y_{3}-\bar{y})^{2}=(3.6 - 3.45)^{2}=0.15^{2}=0.0225\)

For \(i = 4\): \((x_{4}-\bar{x})(y_{4}-\bar{y})=(4.7 - 3.4)(4.9 - 3.45)=1.3\times1.45 = 1.885\)
\((x_{4}-\bar{x})^{2}=(4.7 - 3.4)^{2}=1.3^{2}=1.69\)
\((y_{4}-\bar{y})^{2}=(4.9 - 3.45)^{2}=1.45^{2}=2.1025\)

\(\sum_{i = 1}^{4}(x_{i}-\bar{x})(y_{i}-\bar{y})=-0.455-0.78 - 0.06+1.885=0.6\)

\(\sum_{i = 1}^{4}(x_{i}-\bar{x})^{2}=1.69 + 0.16+0.16+1.69=3.7\)

\(\sum_{i = 1}^{4}(y_{i}-\bar{y})^{2}=0.1225+3.8025+0.0225+2.1025 = 6.05\)

Step3: Calculate the correlation coefficient

\(r=\frac{0.6}{\sqrt{3.7\times6.05}}=\frac{0.6}{\sqrt{22.385}}\approx\frac{0.6}{4.731}\approx0.127\)

Answer:

\(0.127\)