Question

use the given data set to complete parts (a) through (c) below. (use α = 0.05.)

x	y
8	6.77
13	12.74
9	7.11
11	7.81
14	8.83
6	6.08
4	5.38
12	8.15
7	6.43
5	5.72

click here to view a table of critical values for the correlation coefficient.

b. find the linear correlation coefficient, r, then determine whether there is sufficient evidence to support the claim of a linear correlation between the two variables.
the linear correlation coefficient is r = \boxed{}. (round to three decimal places as needed.)

Explanation:

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

Let \(x = [10,8,13,9,11,14,6,4,12,7,5]\) and \(y=[7.46,6.77,12.74,7.11,7.81,8.83,6.08,5.38,8.15,6.43,5.72]\)
\(\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}=\frac{10 + 8+13+9+11+14+6+4+12+7+5}{11}=\frac{99}{11} = 9\)
\(\bar{y}=\frac{\sum_{i = 1}^{n}y_{i}}{n}=\frac{7.46+6.77+12.74+7.11+7.81+8.83+6.08+5.38+8.15+6.43+5.72}{11}=\frac{82.48}{11}\approx7.498\)

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

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

Calculate \((x_{i}-\bar{x})(y_{i}-\bar{y})\):
For \(x = 10,y = 7.46\): \((10 - 9)(7.46-7.498)=1\times(- 0.038)=-0.038\)
For \(x = 8,y = 6.77\): \((8 - 9)(6.77 - 7.498)=(-1)\times(-0.728) = 0.728\)
For \(x = 13,y = 12.74\): \((13 - 9)(12.74 - 7.498)=4\times5.242 = 20.968\)
For \(x = 9,y = 7.11\): \((9 - 9)(7.11 - 7.498)=0\times(-0.388)=0\)
For \(x = 11,y = 7.81\): \((11 - 9)(7.81 - 7.498)=2\times0.312 = 0.624\)
For \(x = 14,y = 8.83\): \((14 - 9)(8.83 - 7.498)=5\times1.332 = 6.66\)
For \(x = 6,y = 6.08\): \((6 - 9)(6.08 - 7.498)=(-3)\times(-1.418)=4.254\)
For \(x = 4,y = 5.38\): \((4 - 9)(5.38 - 7.498)=(-5)\times(-2.118)=10.59\)
For \(x = 12,y = 8.15\): \((12 - 9)(8.15 - 7.498)=3\times0.652 = 1.956\)
For \(x = 7,y = 6.43\): \((7 - 9)(6.43 - 7.498)=(-2)\times(-1.068)=2.136\)
For \(x = 5,y = 5.72\): \((5 - 9)(5.72 - 7.498)=(-4)\times(-1.778)=7.112\)

\(\sum_{i = 1}^{11}(x_{i}-\bar{x})(y_{i}-\bar{y})=-0.038 + 0.728+20.968+0+0.624+6.66+4.254+10.59+1.956+2.136+7.112=55.0\)

Calculate \((x_{i}-\bar{x})^{2}\):
\((10 - 9)^{2}=1\), \((8 - 9)^{2}=1\), \((13 - 9)^{2}=16\), \((9 - 9)^{2}=0\), \((11 - 9)^{2}=4\), \((14 - 9)^{2}=25\), \((6 - 9)^{2}=9\), \((4 - 9)^{2}=25\), \((12 - 9)^{2}=9\), \((7 - 9)^{2}=4\), \((5 - 9)^{2}=16\)
\(\sum_{i = 1}^{11}(x_{i}-\bar{x})^{2}=1+1+16+0+4+25+9+25+9+4+16=110\)

Calculate \((y_{i}-\bar{y})^{2}\):
For \(y = 7.46\): \((7.46 - 7.498)^{2}=(-0.038)^{2}=0.001444\)
For \(y = 6.77\): \((6.77 - 7.498)^{2}=(-0.728)^{2}=0.529984\)
For \(y = 12.74\): \((12.74 - 7.498)^{2}=(5.242)^{2}=27.478564\)
For \(y = 7.11\): \((7.11 - 7.498)^{2}=(-0.388)^{2}=0.150544\)
For \(y = 7.81\): \((7.81 - 7.498)^{2}=(0.312)^{2}=0.097344\)
For \(y = 8.83\): \((8.83 - 7.498)^{2}=(1.332)^{2}=1.774224\)
For \(y = 6.08\): \((6.08 - 7.498)^{2}=(-1.418)^{2}=2.010724\)
For \(y = 5.38\): \((5.38 - 7.498)^{2}=(-2.118)^{2}=4.485924\)
For \(y = 8.15\): \((8.15 - 7.498)^{2}=(0.652)^{2}=0.425104\)
For \(y = 6.43\): \((6.43 - 7.498)^{2}=(-1.068)^{2}=1.140624\)
For \(y = 5.72\): \((5.72 - 7.498)^{2}=(-1.778)^{2}=3.161284\)

\(\sum_{i = 1}^{11}(y_{i}-\bar{y})^{2}=0.001444+0.529984+27.478564+0.150544+0.097344+1.774224+2.010724+4.485924+0.425104+1.140624+3.161284=41.25\)

Step3: Calculate \(r\)

\(r=\frac{55.0}{\sqrt{110\times41.25}}=\frac{55.0}{\sqrt{4537.5}}=\frac{55.0}{67.376}\approx0.816\)

Answer:

\(r\approx0.816\)