If the correlation coefficient 'r' is 0.6 and the regression coefficient of x on y is 0.18, then the regression coefficient of y on x is:
2
The correlation coefficient \(r\) and the two regression coefficients are related by the identity:
\(r^2 = b_{xy} \cdot b_{yx}\)
Given: \(r = 0.6\) and \(b_{xy} = 0.18\) (regression coefficient of x on y).
Substituting into the identity:
\((0.6)^2 = 0.18 \times b_{yx}\)
\(0.36 = 0.18 \times b_{yx}\)
\(b_{yx} = \dfrac{0.36}{0.18} = 2\)
Therefore the regression coefficient of y on x is 2, which matches option 4.
Note: The formula \(b_{yx} = r \times b_{xy}\) is incorrect. The correct relation between r and the two regression coefficients is \(r = \pm\sqrt{b_{xy} \cdot b_{yx}}\).
If, in a series. there are m items whose ranks are common, then which factor is added for correction for each repeating value in both the series?
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The given table shows the ranking of ten students in two subjects mathematics and statistics.
Mathematics | Statistics |
3 | 6 |
5 | 4 |
8 | 9 |
4 | 8 |
7 | 1 |
10 | 2 |
2 | 3 |
1 | 10 |
6 | 5 |
9 | 7 |
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What is the line of regression of y on x?If, in a series. there are m items whose ranks are common, then which factor is added for correction for each repeating value in both the series?
If X 1, X 2, … X nis a simple random sample without replacement of size n from a finite population of N units with mean μ and variance σ 2, the covariance of (X i, Xj ) will be:
If x = X- \(\bar{X}\) and y = Y - \(\bar{Y}\) and the number of pairs (X, Y) is n, then the Karl Pearson's coefficient of correlation is: