Question

Question 9 [10 pts) Patients with chronic kidney failure may be treated by dialysis, in which a machine removes toxic wastes from the blood, a fundction normally performed by the kidneys. Kidney failure and dialysis can cause other changes, such as retention of phosphorus, that must be corrected by changes in diet. A study of the nutrition of dialysis patients measured the level of phosphorus in the blood of several patients on six occasions. Here are data for one patient (in milligrams of phosphorus per deciliter of blood): 5.1 5.2 4.5 4.9 5.7 6.3 The measurements are separated in time and can be considered a simple random sample of the patient's blood phosphorus level. Assume that this level varies normally with standard deviation = 0.9 mg/dl. a. Give a 95% confidence interval for the mean blood phosphorus level. b. The usual range of phosphorus in the blood is considered to be 2.6 to 4.8 mg/dl. Is there strong evidence that this patient has a mean phosphorus level that exceeds 4.8?

Answer 1

(a) The computation table: x x; - -0.18333 -0.08333 -0.78333 -0.38333 0.416667 1.016667 (x,- ) 0.033611 0.006944 0.613611 0.146944 0.173611 1.033612 { x = 31.7 $(x,- * = 2.008333 - - - - - - - - - - - - - --------------------------- The formula for calculating the point estimate is IM Mi- = (31.7) = 5.283333 X = 15.283333 Therefore, the sample mean is Ž = 5.283333 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

The sample statistics: The sample size is n = 5. The sample mean is X = 5.283333 mg/dl. The population standard deviation is o = 0.9 mg/dl. The formula for calculating the standard deviation is 0.9 To 0.9 2.44949 = 0.367423 SE, >10.367423 The determination of the critical value: The level of significance is a =0.05. The critical values of the normal distribution is 2/2 = 20.0512 20.025 Since, the normal distribution function of the Excel 2010 is ] 20.025 1.959964 =normsdist(2) Therefore, the critical value of the normal distribution is 2002 ~ 1.959964 - - - - - - - - - - - - - - - - The 95% confidence interval estimate of the population mean: The formula for calculating the 95% confidence interval for the population mean is =X+za/2 (SEZ) =X+20052 (SEZ) = +z0.ces (SE2) =5283333+1.959964(0.367423) 5.283333+0.720137 =(4.563197,6.00347) +2x12 = (4.5632,6.0035). Interpretation: We can be 95% confident that the true population mean is between [4.5632 and 16.0035 Xtz aves on = 14.5632) 5 4 5(60035 - - - - - - - - - - - - - - - - - - - - - - - - - Conclusion: Since, the usual range of phosphorous in the blood is considered to be 2.6 to 4.8 mg/dl. The population mean lies in the confidence interval calculated of the part(a). We do not reject the null hypothesis Ho. There is not strong evidence the patient has a mean phosphorous level that exceeds 4.8 at the 5% level of significance.