Log-rank test performed by our linear programming homework help experts

Our linear programming homework help experts used the log-rank test to determine if the survival functions are significantly different. Linear regression is not suitable because it cannot handle censored data, which is the main feature of the survival function, which thus necessitates the use of the specialized statistical techniques.

  1. The best model to select is model 2 because all variables in the model are significant. model 1 is understated as it omits a significant variable in form of log (WBC) while model 3is overstated because it includes the interaction of log WBC and standard treatment which is not statistically significant.
  2. The result from model 2 shows that standard treatment has significantly higher hazards than patients who were given a new treatment. From these results, our linear programming homework helpers argue that patients who receive standard treatment have 3.65 times hazard, of leukemia, of patients who receive new treatment and from the population, we expect the hazard of leukemia for Patients who receive standard treatment to be between 1.59 and 8.34 times of Patients who receive a new treatment. Therefore, we conclude that the new treatment is very effective in treating leukemia.

Exploratory data analysis explained by our online linear programming experts

The side-by-side boxplot for SBP for type A is shown above and based on this data, our linear programming experts deduce that the average SBP of type A is slightly greater than that of non-type A.

The graph above presents the scatter plot of SBP against age with a lowess curve embedded. According to our online linear programming tutors, it would be right to say that there is no relationship between age and SBP as the curve is nearly horizontal.

The graph above presents the scatter plot of SBP against weight with a lowess curve embedded. There seems to be a positive relationship between age and SBP as the curve is upward sloping.

Model selection

The final model using the backward selection method is:


The diagnostic plot is shown by the graph above, the residuals versus fitted plot show that the residuals are not random around the 0 lines which cast doubt on the appropriateness of the linearity assumption, moreover, the plot shows that variances of error terms are not equal and that there are some outliers. The residuals versus independent variables also appear to be non-random which suggests that the linear model is not appropriate. Finally, the Q-Q plot suggests that the residuals depart from normality and this is confirmed by the Anderson-Darling test which p-value is 2.2e-16<0.05 which means we reject the null hypothesis of normality. The failure of normality is something of concern because the inferences made are valid only when the assumption holds, however, since the data set is sufficiently large, the central limit theory suggests that the residuals will approach normality, which means we do not need to be concerned about this.


The result above shows that the adjusted mean difference in SBP for type A and non-type A is 8.808 and the 95% CI is between 0.654 and 16.964. The mean difference is large and statistically significant which means the mean difference in SBP between type A and non-type A is of practical significance.



The result above shows that the SBP of  a typical person with type A personality is 129.41 with a 95% CI is between 128.70 and 130.12 while the SBP of  a typical person with non-type A personality is 127.88 with a 95% CI between 121.17 and 128.6