Submission 2 (1/13 points) Monday, September 20, 2021 11:45 PM EDT An article reported on an experiment in which 120 patients with similar clinical features were randomly divided into a control group and a treatment group, each consisting of 60 patients. The sample mean ICU stay (days) and sample standard deviation for the treatment group were 19.3 and 39.4, respectively, whereas these values for the control group were 13.7 and 15.4. (a) Calculate a point estimate for the difference, in days, between true average ICU stay for the treatment and control groups. (Use \"treatment a x Does this estimate suggest that there is a signicant difference between true average stays under the two conditions? I This answer has not been graded yet. _ \"control') (b) Answer the question posed in part (a) by carrying out a formal test of hypotheses. (Use a significance level of 0.05.) State the relevant hypotheses. HO: \"treatment _ \"control = 0 H a: \"treatment \"control 2 0 H0: \"treatment _ \"control = 0 Ha: \"treatment _ \"control 0 HO: \"treatment _ \"control = 0 Ha: \"treatment _ \"control $ 0 H0: \"treatment _ \"control = 0 H a: \"treatment _ \"control 5 0 I Calculate the test statistic and P-value. ( Round your test statistic to two decimal places and your P-value to four decimal places. ) Z = 1.98 X P-value = 0.0000 X State the conclusion in the problem context. O Reject Ho. The data does not suggest that there is a significant difference between true average stays under the two conditions. O Fail to reject H. The data does not suggest that there is a significant difference between true average stays under the two conditions. O Fail to reject Ho. The data suggests that there is a significant difference between true average stays under the two conditions. O Reject Ho. The data suggests that there is a significant difference between true average stays under the two conditions. X Is the result different from what you conjectured in part (a)? (No Response) This answer has not been graded yet. (c) Does it appear that ICU stay for patients given the ventilation treatment is normally distributed? Explain your reasoning. The sample mean length of stay minus two standard deviations for patients receiving the treatment is (No Response) days. This value is (No Response) | 0 which is (No Response) with the assumption that the lengths of stay are normally distributed. So, it (No Response) that the length of stay for patients receiving the treatment is normally distributed. (d) Estimate true average length of stay for patients given the ventilation treatment in a way that conveys information about precision and reliability. (Use a 95% confidence level. Round your answers to two decimal places.) -25.34 X , -14.87 x days You may need to use the appropriate table in the Appendix of Tables to answer this
