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What does the symbol mean? n ∑ Xi i =1 Please provide an example. Suppose X is variable. The following sample observations were randomly selected. X: 4, 5, 3, 6, 10 Y: 4, 6, 5, 7, 7 Obsevation Number 1 2 3 4 5 Total X Y 4 5 3 6 10 4 6 5 7 7 𝑋2 28 29 Number of observations 𝑛 = 5 5 ∑ 𝑋𝑖 = 28 𝑖=1 5 ∑ 𝑌𝑖 = 29 𝑖=1 5 ∑ 𝑋𝑖 2 = 186 𝑖=1 5 ∑ 𝑌𝑖 2 = 175 𝑖=1 5 ∑ 𝑋𝑖 𝑌𝑖 = 173 𝑖=1 Coefficient of correlation 𝑌2 16 25 9 36 100 186 XY 16 36 25 49 49 175 16 30 15 42 70 173 𝑟= ∑ 𝑋𝑌 − (∑ 𝑋)(∑ 𝑌) √(∑ 𝑋 2 − (∑ 𝑋)2 /𝑛)(∑ 𝑌 2 − (∑ 𝑌)2 /𝑁) = 173 − (28)(29)/5 √(186 − 282 /5)(175 − 292 /5) 173 − 812/5 = √(186 − 784/5)(175 − 841/5) 173 − 162.4 = √(186 − 156.8)(175 − 168.2) = 10.6 √(29.2)(6.8) = 10.6 √198.56 = 10.6 = 0.7522 14.0911 The correlation coefficient is 0.7522. The value is positive. So the variables are positively or directly correlated. An increase in one variable is accompanied an increase in the other variable and vice versa. The coefficient of determination is 𝑟 2 = 0.75222 = 0.5659 56.59% of the variation in Y is explained by the variation in X. Determine the coefficient of correlation and the coefficient of determination. Interpret the results. The following regression equation was computed from a sample of 20 observations: Y’ = 15 - 5X SSE was found to be 100 and SST total 400. Determine the standard error of estimate. Determine the coefficient of determination. Determine the coefficient of correlation. (Caution: Watch the sign!) This question is due on day 5. 𝑦 = 𝛽0 + 𝛽1 𝑥 𝑦̂ = 𝛽̂0 + 𝛽̂𝑥 = 15 − 5𝑥 𝛽̂0 = 15 𝛽̂1 = −5 Standard error of estimate 𝑆𝑆𝐸 100 100 𝑆𝑦𝑥 = √ =√ =√ = √5.5556 = 2.357 𝑛−2 20 − 2 18 Coefficient of determination = 𝑆𝑆𝑇−𝑆𝑆𝐸 𝑆𝑆𝑇 = 400−100 400 300 = 400 = 0.75 Coefficient of correlation is negative, since 𝛽̂1 is negative. The magnitude is √𝑟 2 = √0.75 = 0.866 The coefficient of correlation is 𝑟 = −0.866