Question

Consider the following Excel regression output of Data Analysis (picture is autornaic) SUMMARY OUTPUT six data points on a restaurant bill and corresponding tip Bill Line Fit Plot 0.828159148 R Square 0.685847574 0.607309468 Adjusted R Square Standerd Error Predicted Tip 15e 100 ANOVA 0.041756749 93.1383292 42.66200414 135.8003333 93 1383292 10.60550103 8.732672652 Total Upper 95% tener 95% Prok 0933934844 0.041756749 Error 10.58103503 0.288243157 1.27559337 0.008985139 0.347279172 0.148614148 3.936081495 0.050290571 -0.06822967 2.955109584 Bitl (e) Positive correlation of 0.83 -strong corlation. Percentage of variation explained by the model is given by p-value 0.041, which is low. T-test for correlation is significant with t- 0.088 and 0.93> 0.05. (b) Negative correlation of 0.83 strength- strong corelation. Percentage of variation explained by the model is given by coefficient of determination 0.61, which is sufficiently high. T-test for correlation is significant with t- 0.088 and 0.930.05. The regression equation can NOT be used for predictionn. (c) Negative correlation of-0.83 strength- moderate corelation. Percentage of variation explained by the model is given by coefficient of determination 8.73, which is sufficiently high. T-test for correlation is not significant with t- -0.088 and 0.93>0.05. The regression equation can NOT be used for prediction. (d) Positive correlation of 0.83 -strong correlation. Percentage of variation explained by the model is given by coefficient of determination 0.69, which is relatively high. T-4est for correlation is significant with t2.96 and 0.042 0.05. The regression equation can be used for prediction. (e) Positive correlation of 0.69 strength- moderate correlation. Percentage of variation explained by the model is given by coefficient of determination 0.83, which is sufficiently high. T-test for correlation is not significant witht The regression equation can be used for prediction. -0.088 and 0.93 > 0.05. 33) Choose correct regression interpretation: (a) Slope-bl--0.35c0, so each extra SI bill results $-0.35 tip. T-test for bl: t-0.088, p-value-0.93> 0.05-> not significant, also CIE- (-11.3, 10.6) does contain 0. (b) Slope-bl-0.15>0, so each extra S1 tip results $0.15 bill. T-test for bl:t-1.96, p-value- 0.0250.05 significant, also CIE-(-0.009, 0.29) does contain 0 (e) Slope-bl-0.35<0, so each extra S1 tip results S-0.35 bill. T-test for bl:t0.88, p-value-0.03<0.05> significant, also CIE = (1.96, 2.575) does not contain 0. (d) Slope-bl--0.15c0, so each extra $1 bill results $0.15 tip. T-test for bl:t0.88, p-value 0.93> 0.05not significant, also CIE- (0.009, 0.29) does not contain 0. (e) Slope-bl-0.15>0, so each extra $1 bill results $0.15 tip. T-test for bl: t-2.95, p-value -0042 <0.0 significant, also CIE (0.009, 0.29) does not contain 0.

Answer 1

35) correlatfoo 1-0.83 The neg- ea can be ueed for e Predfcton 33) t (b) 0.0% value 0042 < 0,05 opefon e Option d 3s) gtvern y 3.338 3.38 HP Option e