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Question: For \[n=25\] \[\sum{x}=125,\sum{{{x}^{2}}}=650,\sum{y}=100,\sum{{{y}^{2}}}=460,\sum{xy}=508\], then ...

For n=25n=25 ∑x=125,∑x2=650,∑y=100,∑y2=460,∑xy=508\sum{x}=125,\sum{{{x}^{2}}}=650,\sum{y}=100,\sum{{{y}^{2}}}=460,\sum{xy}=508, then the correlation coefficient is
(a) 0.99
(b) 0.207
(c) 0.66
(d) 0.89

Explanation

Solution

We solve this problem simply by using the correlation formula of data. We know that the correlation factor is calculated to measure how strong the relationship between two parameters. The correlation formula of a data having ′x′'x' and ′y′'y' as two parameters is given as
r=n∑xy−(∑x×∑y)(n∑x2−(∑x)2)(n∑y2−(∑y)2)r=\dfrac{n\sum{xy}-\left( \sum{x}\times \sum{y} \right)}{\sqrt{\left( n\sum{{{x}^{2}}}-{{\left( \sum{x} \right)}^{2}} \right)\left( n\sum{{{y}^{2}}}-{{\left( \sum{y} \right)}^{2}} \right)}}
By using this formula we substitute the required values given and find the correlation coefficient of the data.

Complete step-by-step solution:
We are given that for a data of n=25n=25, ′x′'x' and ′y′'y' are two parameters of a certain data.
We are given with the value of data as
∑x=125,∑x2=650,∑y=100,∑y2=460,∑xy=508\sum{x}=125,\sum{{{x}^{2}}}=650,\sum{y}=100,\sum{{{y}^{2}}}=460,\sum{xy}=508
Let us assume that ′r′'r' as the correlation coefficient.
We know that the formula of the correlation coefficient is given as
r=n∑xy−(∑x×∑y)(n∑x2−(∑x)2)(n∑y2−(∑y)2)r=\dfrac{n\sum{xy}-\left( \sum{x}\times \sum{y} \right)}{\sqrt{\left( n\sum{{{x}^{2}}}-{{\left( \sum{x} \right)}^{2}} \right)\left( n\sum{{{y}^{2}}}-{{\left( \sum{y} \right)}^{2}} \right)}}
Now, by substituting the required values that are given in the question in the above formula we get
⇒r=(25×508)−(125×100)((25×650)−(125)2)((25×460)−(100)2)\Rightarrow r=\dfrac{\left( 25\times 508 \right)-\left( 125\times 100 \right)}{\sqrt{\left( \left( 25\times 650 \right)-{{\left( 125 \right)}^{2}} \right)\left( \left( 25\times 460 \right)-{{\left( 100 \right)}^{2}} \right)}}
By multiplying the respective numbers and squaring the respected numbers we get

& \Rightarrow r=\dfrac{\left( 12700 \right)-\left( 12500 \right)}{\sqrt{\left( \left( 16250 \right)-\left( 15625 \right) \right)\left( \left( 11500 \right)-\left( 10000 \right) \right)}} \\\ & \Rightarrow r=\dfrac{200}{\sqrt{625\times 1500}} \\\ \end{aligned}$$ Now, by evaluating the result in the square root we get $$\begin{aligned} & \Rightarrow r=\dfrac{200}{983.615} \\\ & \Rightarrow r=0.20633 \\\ \end{aligned}$$ By rounding off the above number to third decimal number we get $$\Rightarrow r\simeq 0.207$$ Therefore, we can say that the correlation coefficient of the given data is 0.207. From this correlation coefficient, we can say that for every unit increase in the value of $$'x'$$ there is an increase of 0.207 in the value of $$'y'$$. **So, option (b) is the correct answer.** **Note:** In this question the main mistake a student can do is in the calculation part. Here, we can see that there is a lot of calculation required for solving the problem that is $$\Rightarrow r=\dfrac{\left( 25\times 508 \right)-\left( 125\times 100 \right)}{\sqrt{\left( \left( 25\times 650 \right)-{{\left( 125 \right)}^{2}} \right)\left( \left( 25\times 460 \right)-{{\left( 100 \right)}^{2}} \right)}}$$ Students may take the values incorrectly like not following the BODMAS rule. First we have to evaluate the values inside the brackets then we get for outside brackets. This part will be mistaken by many students that results in the wrong answer. Solving the problem with taking care of calculation is the only main part to get the answer correctly.