Answer
Verified
439.8k+ views
Hint: We need to find the correlation coefficient between the length and weight and interpret the result. We will be using Karl Pearson’s coefficient of correlation which is given by \[{{r}_{xy}}=\dfrac{n\sum{{{x}_{i}}{{y}_{i}}-\sum{{{x}_{i}}\sum{{{y}_{i}}}}}}{\sqrt{n{{\sum{{{x}_{i}}^{2}-\left( \sum{{{x}_{i}}} \right)}}^{2}}}\sqrt{n{{\sum{{{y}_{i}}^{2}-\left( \sum{{{y}_{i}}} \right)}}^{2}}}}\] . We will denote $x$ as length and $y$ as weight. Since there are five items, $n=5$ . A table is drawn to find the required terms to be substituted in the Karl Pearson’s coefficient of correlation equation. Finally, we will interpret the degree of correlation as perfect, strong, moderate, low or no correlation.
Complete step by step answer:
We need to find the correlation coefficient between the length and weight.
Karl Pearson’s coefficient of correlation is given as
\[{{r}_{xy}}=\dfrac{n\sum{{{x}_{i}}{{y}_{i}}-\sum{{{x}_{i}}\sum{{{y}_{i}}}}}}{\sqrt{n{{\sum{{{x}_{i}}^{2}-\left( \sum{{{x}_{i}}} \right)}}^{2}}}\sqrt{n{{\sum{{{y}_{i}}^{2}-\left( \sum{{{y}_{i}}} \right)}}^{2}}}}\]
Where \[n\] is the number of items
\[{{x}_{i}}\] is the total length
\[{{y}_{i}}\] is the total weight.
Now let us find these.
Therefore, \[\sum{{{x}_{i}}}=20,\sum{{{y}_{i}}}=25,\sum{{{x}_{i}}^{2}}=110,\sum{{{y}_{i}}^{2}}=159,\sum{{{x}_{i}}}{{y}_{i}}=130,n=5\]
Now let us substitute these in \[{{r}_{xy}}=\dfrac{n\sum{{{x}_{i}}{{y}_{i}}-\sum{{{x}_{i}}\sum{{{y}_{i}}}}}}{\sqrt{n{{\sum{{{x}_{i}}^{2}-\left( \sum{{{x}_{i}}} \right)}}^{2}}}\sqrt{n{{\sum{{{y}_{i}}^{2}-\left( \sum{{{y}_{i}}} \right)}}^{2}}}}\]
We will get \[{{r}_{xy}}=\dfrac{5\times 130-20\times 25}{\sqrt{5\times 110-{{20}^{2}}}\sqrt{5\times 159-{{25}^{2}}}}\]
Multiplying the terms in numerator and denominator, we will get
\[{{r}_{xy}}=\dfrac{650-500}{\sqrt{550-400}\sqrt{795-625}}\]
Now let us subtract the values in the numerator and denominator. We will get
\[{{r}_{xy}}=\dfrac{150}{\sqrt{150}\sqrt{170}}\]
Now let us split the numerator. We know that $\sqrt{a}\times \sqrt{a}=a$ . So the above equation becomes
\[{{r}_{xy}}=\dfrac{\sqrt{150}\sqrt{150}}{\sqrt{150}\sqrt{170}}\]
Cancelling common terms from numerator and denominator, we will get
\[{{r}_{xy}}=\dfrac{\sqrt{150}}{\sqrt{170}}\]
We know that $\dfrac{\sqrt{a}}{\sqrt{b}}=\sqrt{\dfrac{a}{b}}$ . So the above equation becomes
\[{{r}_{xy}}=\sqrt{\dfrac{150}{170}}\]
Cancelling zeroes from numerator and denominator, we will get
\[{{r}_{xy}}=\sqrt{\dfrac{15}{17}}\]
Now let us further simplify by dividing. The above equation becomes
\[{{r}_{xy}}=\sqrt{0.8823}\]
Now take the square root of the term, we will get
\[{{r}_{xy}}=0.9393\]
The correlation coefficient value is in between $\pm 0.5$and $\pm 1$ . Hence there exists a high degree of correlation or strong correlation between \[x\] and $y$ .
Note: The degree of correlation can be determined as follows:
There will be perfect correlation between two variables if the value of correlation coefficient is near $\pm 1$, that is, as one variable increases, the other variable tends to also increase (if positive) or decrease (if negative).
There will be a high degree of correlation or strong correlation between two variables if the value of correlation coefficient lies between $\pm 0.5$and $\pm 1$ .
There will be a moderate degree of correlation or medium correlation between two variables if the value of correlation coefficient lies between $\pm 0.30$and $\pm 0.49$ .
There will be a low degree of correlation or small correlation between two variables if the value of correlation coefficient lies below $\pm 0.29$ .
There will be no correlation between two variables if the value of correlation coefficient is zero.
Complete step by step answer:
We need to find the correlation coefficient between the length and weight.
Karl Pearson’s coefficient of correlation is given as
\[{{r}_{xy}}=\dfrac{n\sum{{{x}_{i}}{{y}_{i}}-\sum{{{x}_{i}}\sum{{{y}_{i}}}}}}{\sqrt{n{{\sum{{{x}_{i}}^{2}-\left( \sum{{{x}_{i}}} \right)}}^{2}}}\sqrt{n{{\sum{{{y}_{i}}^{2}-\left( \sum{{{y}_{i}}} \right)}}^{2}}}}\]
Where \[n\] is the number of items
\[{{x}_{i}}\] is the total length
\[{{y}_{i}}\] is the total weight.
Now let us find these.
$x$ | $y$ | ${x}^{2}$ | ${y}^{2}$ | $xy$ | |
1 | 1 | 1 | 1 | 1 | |
2 | 3 | 4 | 9 | 6 | |
4 | 6 | 16 | 36 | 24 | |
5 | 7 | 25 | 49 | 35 | |
8 | 8 | 64 | 64 | 64 | |
Total | 20 | 25 | 110 | 159 | 130 |
Therefore, \[\sum{{{x}_{i}}}=20,\sum{{{y}_{i}}}=25,\sum{{{x}_{i}}^{2}}=110,\sum{{{y}_{i}}^{2}}=159,\sum{{{x}_{i}}}{{y}_{i}}=130,n=5\]
Now let us substitute these in \[{{r}_{xy}}=\dfrac{n\sum{{{x}_{i}}{{y}_{i}}-\sum{{{x}_{i}}\sum{{{y}_{i}}}}}}{\sqrt{n{{\sum{{{x}_{i}}^{2}-\left( \sum{{{x}_{i}}} \right)}}^{2}}}\sqrt{n{{\sum{{{y}_{i}}^{2}-\left( \sum{{{y}_{i}}} \right)}}^{2}}}}\]
We will get \[{{r}_{xy}}=\dfrac{5\times 130-20\times 25}{\sqrt{5\times 110-{{20}^{2}}}\sqrt{5\times 159-{{25}^{2}}}}\]
Multiplying the terms in numerator and denominator, we will get
\[{{r}_{xy}}=\dfrac{650-500}{\sqrt{550-400}\sqrt{795-625}}\]
Now let us subtract the values in the numerator and denominator. We will get
\[{{r}_{xy}}=\dfrac{150}{\sqrt{150}\sqrt{170}}\]
Now let us split the numerator. We know that $\sqrt{a}\times \sqrt{a}=a$ . So the above equation becomes
\[{{r}_{xy}}=\dfrac{\sqrt{150}\sqrt{150}}{\sqrt{150}\sqrt{170}}\]
Cancelling common terms from numerator and denominator, we will get
\[{{r}_{xy}}=\dfrac{\sqrt{150}}{\sqrt{170}}\]
We know that $\dfrac{\sqrt{a}}{\sqrt{b}}=\sqrt{\dfrac{a}{b}}$ . So the above equation becomes
\[{{r}_{xy}}=\sqrt{\dfrac{150}{170}}\]
Cancelling zeroes from numerator and denominator, we will get
\[{{r}_{xy}}=\sqrt{\dfrac{15}{17}}\]
Now let us further simplify by dividing. The above equation becomes
\[{{r}_{xy}}=\sqrt{0.8823}\]
Now take the square root of the term, we will get
\[{{r}_{xy}}=0.9393\]
The correlation coefficient value is in between $\pm 0.5$and $\pm 1$ . Hence there exists a high degree of correlation or strong correlation between \[x\] and $y$ .
Note: The degree of correlation can be determined as follows:
There will be perfect correlation between two variables if the value of correlation coefficient is near $\pm 1$, that is, as one variable increases, the other variable tends to also increase (if positive) or decrease (if negative).
There will be a high degree of correlation or strong correlation between two variables if the value of correlation coefficient lies between $\pm 0.5$and $\pm 1$ .
There will be a moderate degree of correlation or medium correlation between two variables if the value of correlation coefficient lies between $\pm 0.30$and $\pm 0.49$ .
There will be a low degree of correlation or small correlation between two variables if the value of correlation coefficient lies below $\pm 0.29$ .
There will be no correlation between two variables if the value of correlation coefficient is zero.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Which are the Top 10 Largest Countries of the World?
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Difference Between Plant Cell and Animal Cell
Give 10 examples for herbs , shrubs , climbers , creepers
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
How do you graph the function fx 4x class 9 maths CBSE
Write a letter to the principal requesting him to grant class 10 english CBSE