Consider the output below, in which number of hours worked by Oregonians is used to predict their income from wages.

Regression of variable WAGP: Goodness of fit statistics (WAGP): 0.110 DW 2.102 Cp 2.000 AIC 1013.923 SBC 1017.666 PC 0.967 Analysis of variance (WAGP): Source DF Sum of squares Mean squares F Pr > F Model 1 8159872827.348 8159872827.348 5.697 0.021 Error 46 65890863839.318 1432410083.463 Corrected 47 74050736666.667 Total Computed against model Y=Mean(Y) Model parameters (WAGP): Lower Upper Pr > Source Value Standard error t bound bound (95%) (95%) Intercept 5196.581 17048.476 0.305 0.762 -29120.222 39513.383 WKHP 1008.814 422.671 2.387 0.021 158.020 1859.607Cansidering the output below, in which the number of hours worked by Oregonians is used to predict their income from wages, [ Select ] 2. How manyr people's ages and incomes were considered? I [Select] 1. What is the null hypothesis for the ANOVA test below? [ Select] l 3. What "percent of the story\" does this model tell? That is. what percent of the variability is accounted for? 4.Are the slope and l or interview "signicant"? :VI 5. Interpret the meaning of the slope. i 53'9\"] 6. What income does this model predict for a worker who puts in 40 hours a week? ii. 7. One person works 20 hrs/wk earns $25,000. What is the residual for this data point? $ l' 8. Because the p-value is small,; [53'9\"] ; but because the r squared value is fairiylr small. ' [salad]