In a study of end-stage renal disease, the creatinine values were obtained for 12 end-stage renal patients undergoing hemodialysis. These creatinine values have a mean of 11.9 mg/dl and a standard deviation of 3.6 mg/dl. A nephrology nurse believes that the population mean creatinine is 8.4 mg/dl. What is the correct conclusion that we can make based on the 95% confidence interval?

Test Mean HypothesizedValue 8.4 Actual Estimate 11.9333 DF 11 Std Dev 3.61973 tTest Test Statistic 3.3814 Prob :- |t| 0.0061* Prob at 0.0031* Prob st 0.9969 4 567 8 910111213 Condence Intervals Parameter Estimate LowerCl UpperCl 1-Alpha Mean 11.93333 9.633468 14.2332 0.950 Std Dev 3.619727 2.56419?r 6.145855 0.950 O This particular interval does not encompass the value 8.4, the mean in the general population. Therefore, we conclude that the mean creatinine in this group of patients is not the same as that in the general population of patients of this type 0 This particular interval does encompass the value 8.4, the mean in the general population. Therefore, we fail to conclude that the mean creatinine in this group of patients differs from the mean of the general population of patients of this type 0 This particular interval does not encompass the value zero. Therefore, we conclude that the mean creatinine in this group of patients is not the same as that in the general population of patients of this type 0 This particular interval does encompass the value zero. Therefore, we fail to conclude that the mean creatinine in this group of patients differs from the mean of the general population of patients of this type