Question

can you do 32 and 33 for me plz ? just 2 multiple choices thanks

Consider the following Excel regression output Date Analysis (picture is automatic) SUMMARY OUTPUT output of six data points on a restaurant bill and corresponding tip. Bill Line Fit Plot R Square 0.828159148 0.685847574 0.607309468 3.265807868 R Square Stendard Error Total 10.66550103 Coefficients Standard Evor 0,347279172 .936081493 D.08872967 0.9 9551584 32) Choose correct correlation interpretation: (a) Positive correlation of 0.83- strong corelation. 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 0.932 0.05 (b) Negative correlation of 0.83 strength - strong correlation. 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 0.93 >0.05. The regression equation can NOT be used for prediction (c) Negative correlation of -0.83 strength -moderate correlation. 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-test for correlation is significant with t-2.96 0.05. The regression equation can be used for prediction. (e) Positive correlation of 0.69 given by coefficient of determination 0.83, which is sufficiently high. T-test for correlation i -0.088 and 0.93> 0.05. The regression equation can be used for prediction. and 0.042 Percentage of variation explained by the model is strength - moderate correlation. 33) Choose correct regression interpretation: (a) Slope-bi 0.35e0, so each extra $1 bill results $-035 tip. T-test for bl: t--0088, p-value-093> 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 0.025<0.0s significant, also CIE-(-0.009, 0.29) does contain 0. (c) Slope-bl--0.35c0, so each extra S1 tip results S-0.35 bill. T-test for bl: t--0.88, 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 S0.15 tip. T-test for bl:t0.88, p-value -0.93> 0.05- not significant, also CIE - (0.009, 0.29) does not contain 0. (e) Slope-bl-0.15 0, so each extra SI bill results $0.15 tip. T-test for bl: t 2.95, p-value 0.042 0.05> significant, also CIE- (0.009, 0.29) does not contain 0

Answer 1

Solution32:

R=0.83

R sq=0.69

there exits a strong positive correlation .

R sq=0.69 that is 69% variation in tip  is explained by bill

MARK OPTION D

Solution32:

slope=0.15 t=2.96 p=0.042

p<0.05

There exists a linear relationship between tip and bill.

MARK OPTION E