The residuals on the Versus Fits plot are all randomly (response is Delivery Time Minutes = Y) distributed above and below 0. w Homoscedasticity is .. . not violated on this 0 graph. Residual -1 It can be assumed this -2 graph has homoscedasticity. -5 30 40 50 60 70 Fitted Value . The residuals on the Residuals Versus Delivery Time Minutes = Y plot are all randomly (response is Delivery Time Minutes = Y) distributed above and below 0. w .. . Homoscedasticity is not violated on this 0 graph. Residual -1- It can be assumed this -2 graph has -3 homoscedasticity. 30 40 50 60 70 Delivery Time Minutes = YOrder sizes and shipping costs for last twelve months Customer Number of Cases Delivery Time Minutes =X =Y 52 32.1 53 33.0 64 34.8 O VOUAWN- 73 36.2 85 37.8 95 37.8 97 38.5 8 103 39.7 9 116 38.5 10 121 41.9 11 143 44.2 12 157 47.1 13 161 43.0 14 184 49.4 15 202 57.2 16 218 56.8 17 243 60.6 18 254 61.2 19 267 58.2 20 275 63.1 21 287 65.6 22 298 67.3 23 300 69.50 24 305 69.75 25 307 69.75Daniel Clevett Professor Jeffery Culver DSCI 346 Matrix Plot of Number of Cases = X, Delivery Time Minutes = Y 99% CI for Pearson Correlation What goes in the box 60 Delivery Time Minutes = \\ r = 0.990 CI = (0.970, 0.997) 100 200 300 Number of Cases = X Correlation Problem Definition: Hypothesis: Ho: p=0 H1: p#0 Decision Rule: If my P-value is less than alpha = 0.1 the null hypothesis will be rejected Test Dainuies Dosesan CorralationsPairwise Pearson Correlations Sample 1 Samplez N Cnnelalinn 99% a My PViue DEliVHy Tune Minutes = Number of C355 = 25 0.990 (0.970, 0.997] [UNI] X Y The pvalue of 0.000 is 0 If my Pvalue is less than alpha = 0.1 the null hypothesis will be rejected. Analysis of Variance Source OF Seq SS Contribution Adj SS Adj MS F-Value P-Value Regression 1 405153 98.02% 4051.53 4051.53 1135.80 0.000 Number of Cases = X 1 4051.53 98.02% 4051.53 4051.53 11315.80 0.000 Error 23 82.04 1.98% 32.04 3.57 Total 24 4133.57 100.00% Model Summary 5 R-g R-sggadi! PRESS R-sggpred) Alc: BIC 1.888158 98.02% 97.93% 93.9832 97.73% 107.80 110.31 The p-value of 0.000 is