Using your calculator, run a regression analysis on thefollowing bivariate set of data with y as the responsevariable.
x y
71.2 23.1
90.8 122.9
88.8 82.9
57.7 18.3
71.4 8.6
60.4 25.9
60.2 -43.1
88.5 77.4
68.2 -5.9
41.1 -83
61.4 -2.5
87.1 39.4
Find the correlation coefficient and report it accurate to threedecimal places. r =
What proportion of the variation in y can be explained by thevariation in the values of x? Report answer as a percentageaccurate to one decimal place. (If the answer is 0.84471, then itwould be 84.5%...you would enter 84.5 without the percent symbol.)r² = %
Based on the data, calculate the regression line (each value tothree decimal places) y = x +
Predict what value (on average) for the response variable willbe obtained from a value of 82.2 as the explanatory variable. Use asignificance level of ? = 0.05 to assess the strength of the linearcorrelation.
What is the predicted response value? (Report answer accurate toone decimal place.) y =
