Regression and Correlation, round to at least 4 decimals. a= 0.05 Y._ ( Xi - X ) ( Y - Y ) (Xi - X) 2 (Yi - Y)2 ( Xi - X ) ( Y , - V ) Y ( Y - D) ( Y - V)2 X.= Y.= SSX= SStotal SSXY = SSreid n = (SS.Y.) Regression Linear Function: Hypothesis Testing. v=n-2= SSXY Sylx = SSresid = to = Itb/ > Itcl? Y/N Vn -2 SYlX * S = Conclusions: SSX b Y. = a+ b*X= +b* = Sb Coefficient of Determination: R2 = 1 - SSnesia SStotal = 1 - Correlation Hypothesis Testing v=n-2 = 1 -72 1 - 2 SSXY to = Itrl > Its/? Y/N n - 2 n - 2 VSSy * SSY Conclusions: t, = Syhcientists decided to estimate mass from measurements of plant volume with an allometric regression. To obtain the regression1 they sampled 40 plants from outside the experiment and measured both the volume (based on plant height and circumference) and mass of each plant. Data for 10 were randomly selected and are shown above. run a linear regression to predict plant mass due to volume for this data. 0. = 0.05. - State the null and alternative. - Conduct the test and report the equation for the regression, WEk, tb, and R2. Use at least three decimal places -What is the critical value based on u? - Conclude based on the results and state statistical relevance - What amount of variation (%) in mass does volume explain (R2)? - Determine the 95% condence interval of b The scientist used a linear regression in the above exercise because the plant volume that was recorded was accurate. often scientists measure two variables that contain high variability. If a predictor variable is measured with low accuracy or which variable is independent and dependent is uncertain; then a correlation test if often used. Assume this is now the scenario, thus linear regression should not be used. - Calculate correlation of plant volume and mass given 0. = 0.05. - State the appropriate test and Ho HA. Conduct the test. - Determine values of r and tr. - Determine the critical value based on a? Conclude using the results and relevance