Let $X =$ Age; $Y =$ Cholesterol level; $\hat{ Y }=156.32+0.65 X ; 195.32$
Calculate the Karl Pearson's Correlation Coefficient from the following data. Also interpret the data on the basis of probable error.
Ans: $r=0.7453, P . E .(r)=0.1133$ and significant
Answer (1)
Calculate Karl Pearson's correlation coefficient: r ≈ 0.7453 .
Calculate the probable error: P . E . ( r ) ≈ 0.1133 .
Check for significance: 6 * P.E.(r)"> r > 6 ∗ P . E . ( r ) .
Conclude that the correlation is significant: r = 0.7453 , P . E . ( r ) = 0.1133 and significant.
Explanation
Understand the problem and provided data We are given two sets of data, X and Y , and we want to calculate Karl Pearson's correlation coefficient ( r ) and the probable error ( P . E . ( r ) ) to determine the significance of the correlation. The data is as follows:
X = [ 67 , 65 , 68 , 64 , 66 , 70 , 63 ] Y = [ 69 , 65 , 69 , 65 , 67 , 67 , 64 ]
Calculate Karl Pearson's correlation coefficient (r) First, we calculate Karl Pearson's correlation coefficient ( r ). This measures the strength and direction of a linear relationship between two variables. The formula is:
r = \frac{{\sum_{{i=1}}^{{n}} (X_i - \bar{X})(Y_i - \bar{Y})}}{{\sqrt{{\sum_{{i=1}}^{{n}} (X_i - \bar{X})^2} \sqrt{{\sum_{{i=1}}^{{n}} (Y_i - \bar{Y})^2}}}}
where X ˉ and Y ˉ are the means of X and Y respectively, and n is the number of data points.
Result of correlation coefficient calculation Using the provided data, the correlation coefficient r is calculated to be approximately 0.7453 .
Calculate the probable error P.E.(r) Next, we calculate the probable error of the correlation coefficient, which gives us an idea of the reliability of the calculated correlation. The formula for the probable error is:
P . E . ( r ) = 0.6745 ∗ n 1 − r 2
where n is the number of data points.
Result of probable error calculation With r = 0.7453 and n = 7 , the probable error is calculated as:
P . E . ( r ) = 0.6745 ∗ 7 1 − ( 0.7453 ) 2 ≈ 0.1133
Interpret the significance of the correlation Finally, we interpret the significance of the correlation. A common rule of thumb is that if 6 * P.E.(r)"> r > 6 ∗ P . E . ( r ) , the correlation is considered significant. In our case:
6 ∗ P . E . ( r ) = 6 ∗ 0.1133 ≈ 0.6798
Since 0.6798"> 0.7453 > 0.6798 , the correlation is considered significant.
Conclusion Therefore, Karl Pearson's correlation coefficient is approximately 0.7453 , the probable error is approximately 0.1133 , and the correlation is significant.
Examples
Understanding correlation coefficients and their probable errors is crucial in various fields. For instance, in medical research, it helps determine the strength and reliability of the relationship between a drug dosage and its effect on patients. If a strong, significant correlation is found, it suggests the drug's effect is consistent and not due to random chance, aiding in informed decisions about treatment protocols. Similarly, in finance, it can assess the relationship between market indicators and investment returns, guiding investment strategies.
