OInstructor-created question Question Help ▼ In a random sample of n = 50 patients, the mean creatine level .x = 1.2 mg/dL with unreported sample standard deviation. The mean creatine level u in the...

Extracted text: OInstructor-created question Question Help ▼ In a random sample of n = 50 patients, the mean creatine level .x = 1.2 mg/dL with unreported sample standard deviation. The mean creatine level u in the population of all patients is unknown with standard deviation a = 0.5 mg/dL. With 95% confidence compare the mean creatine level of the patient population to the 1.O mg/dL mean creatine level of the general population using the p-value method: Test Ho: u= 1.0 mg/dL Patient creatine level is same as that of general population. Versus Ha: u#1.0 mg/dL Patient creatine level is different than that of general population. Use all digits showing on your calculator in your intermediate calculations but show only the requested digits to the right of the decimal point in your final answers. (a) test statistic showing 2 digits to the right of the decimal: (b) p-value showing 4 digits to the right of the decimal: (c)Conclusion: O A. Reject Ho with 95% confidence and conclude that the mean creatine level of patient population is different than that of the general population. O B. Fail to reject Ho. Sample does not provide enough evidence to say that the mean creatine level of patient population is different than that of the general population. Advar About Click to select your answer(s) and then click Check Answer. All parts showing Clear All Check Answer