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A sample of five items is taken from the production firm. Length and weight of the five items are given below (Given \[\sqrt{0.8823}=0.9393\] ). Calculate the correlation coefficient between length and weight and interpret the result.
Length(cm)12458
Weight(gm)13678

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Last updated date: 20th Jun 2024
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Answer
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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.
$x$$y$${x}^{2}$${y}^{2}$$xy$
11111
23496
46163624
57254935
88646464
Total2025110159130

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.