alfonso surveyed several students in his school. he asked them to give the grade that they are in and the…

alfonso surveyed several students in his school. he asked them to give the grade that they are in and the number of pets they have at home. the results are shown in the table below.\n| grade | number of pets |\n| ---- | ---- |\n| 5 | 0 |\n| 6 | 4 |\n| 2 | 2 |\n| 1 | 3 |\n| 4 | 5 |\n| 2 | 1 |\n| 4 | 2 |\n| 1 | 2 |\n| 4 | 0 |\n| 3 | 5 |\n| 5 | 2 |\nfind the correlation coefficient for the data. round to the nearest hundredth.

alfonso surveyed several students in his school. he asked them to give the grade that they are in and the number of pets they have at home. the results are shown in the table below.\n| grade | number of pets |\n| ---- | ---- |\n| 5 | 0 |\n| 6 | 4 |\n| 2 | 2 |\n| 1 | 3 |\n| 4 | 5 |\n| 2 | 1 |\n| 4 | 2 |\n| 1 | 2 |\n| 4 | 0 |\n| 3 | 5 |\n| 5 | 2 |\nfind the correlation coefficient for the data. round to the nearest hundredth.

Answer

Answer:

Let the grade be $x$ - variable and the number of pets be $y$ - variable. First, we calculate the following sums for $n = 11$ data - points:

Let (x={5,6,2,1,4,2,4,1,4,3,5}) and (y = {0,4,2,3,5,1,2,2,0,5,2})

  1. Calculate (\sum_{i = 1}^{n}x_{i}), (\sum_{i = 1}^{n}y_{i}), (\sum_{i = 1}^{n}x_{i}^{2}), (\sum_{i = 1}^{n}y_{i}^{2}), and (\sum_{i = 1}^{n}x_{i}y_{i})

    • (\sum_{i=1}^{11}x_{i}=5 + 6+2 + 1+4+2+4+1+4+3+5=37)
    • (\sum_{i=1}^{11}y_{i}=0 + 4+2+3+5+1+2+2+0+5+2=26)
    • (\sum_{i = 1}^{11}x_{i}^{2}=5^{2}+6^{2}+2^{2}+1^{2}+4^{2}+2^{2}+4^{2}+1^{2}+4^{2}+3^{2}+5^{2}=25 + 36+4 + 1+16+4+16+1+16+9+25 = 159)
    • (\sum_{i = 1}^{11}y_{i}^{2}=0^{2}+4^{2}+2^{2}+3^{2}+5^{2}+1^{2}+2^{2}+2^{2}+0^{2}+5^{2}+2^{2}=0 + 16+4+9+25+1+4+4+0+25+4 = 92)
    • (\sum_{i = 1}^{11}x_{i}y_{i}=5\times0+6\times4+2\times2+1\times3+4\times5+2\times1+4\times2+1\times2+4\times0+3\times5+5\times2=0 + 24+4+3+20+2+8+2+0+15+10 = 88)

  2. Calculate the means of (x) and (y):

    • (\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}=\frac{37}{11}\approx3.36)
    • (\bar{y}=\frac{\sum_{i = 1}^{n}y_{i}}{n}=\frac{26}{11}\approx2.36)

  3. Use the formula for the correlation coefficient (r): [r=\frac{n\sum_{i = 1}^{n}x_{i}y_{i}-\sum_{i = 1}^{n}x_{i}\sum_{i = 1}^{n}y_{i}}{\sqrt{n\sum_{i = 1}^{n}x_{i}^{2}-(\sum_{i = 1}^{n}x_{i})^{2}}\sqrt{n\sum_{i = 1}^{n}y_{i}^{2}-(\sum_{i = 1}^{n}y_{i})^{2}}}]

Substitute the values: [n = 11,\sum_{i = 1}^{n}x_{i}y_{i}=88,\sum_{i = 1}^{n}x_{i}=37,\sum_{i = 1}^{n}y_{i}=26,\sum_{i = 1}^{n}x_{i}^{2}=159,\sum_{i = 1}^{n}y_{i}^{2}=92]

[n\sum_{i = 1}^{n}x_{i}y_{i}-\sum_{i = 1}^{n}x_{i}\sum_{i = 1}^{n}y_{i}=11\times88-37\times26=968 - 962=6]

[n\sum_{i = 1}^{n}x_{i}^{2}-(\sum_{i = 1}^{n}x_{i})^{2}=11\times159 - 37^{2}=1749-1369 = 380]

[n\sum_{i = 1}^{n}y_{i}^{2}-(\sum_{i = 1}^{n}y_{i})^{2}=11\times92-26^{2}=1012 - 676=336]

[r=\frac{6}{\sqrt{380}\sqrt{336}}]

(\sqrt{380}\approx19.50), (\sqrt{336}\approx18.33)

[r=\frac{6}{19.50\times18.33}=\frac{6}{357.44}\approx0.02]

Explanation:

Step1: Calculate sums

Calculate (\sum x), (\sum y), (\sum x^{2}), (\sum y^{2}), (\sum xy)

Step2: Calculate means

Find (\bar{x}=\frac{\sum x}{n}), (\bar{y}=\frac{\sum y}{n})

Step3: Apply correlation - coefficient formula

Use (r=\frac{n\sum xy-\sum x\sum y}{\sqrt{n\sum x^{2}-(\sum x)^{2}}\sqrt{n\sum y^{2}-(\sum y)^{2}}})

Step4: Substitute and calculate

Substitute values and find (r\approx0.02)