The level of various substances in the blood of kidney dialysis patients is of c

ID: 3224245 • Letter: T

Question

The level of various substances in the blood of kidney dialysis patients is of concern. A researcher did blood tests on a dialysis patient on six consecutive clinic visits. One variable measured was the level of phosphate in the blood. The data on this one patient, in milligrams of phosphate per deciliter (mg/dl) blood, are: 5.6 5.1 4.6 4.8 5.7 6.4 a. The normal range of values for blood phosphate level is 2.6 to 4.8 mg/dl. Is there evidence that the patient's mean level in fact falls above 4.8? b. Given your conclusion, are you at risk of making a Type I error or a Type II error? Provide your answer in the context of the problem. c. What sample size would you recommend, if it is important to detect and treat the elevated average phosphate level of 5.5? d. What assumptions did you have to make? e. Report a one-sided 95% confidence interval for the mean blood phosphate level.

Explanation / Answer

Part-a:

Null hypothesis H0: µ=4.8

Alterative hypothesis Ha: µ>4.8

Sample mean xbar= 5.37 using excel function =AVERAGE(5.6,5.1,4.6,4.8,5.7,6.4)

Sample standard deviation s=0.67 using excel function =stdev(5.6,5.1,4.6,4.8,5.7,6.4)

Sample size n=6

Test statistic t=(xbar-4.8)/(s/sqrt(n))

=(5.37-4.8)/(0.67/sqrt(6))

=2.08

Degree of freedom =n-1=6-1=5

p-value=0.0190 using excel function =TDIST(2.8,5,1)

As p-value<0.05, we reject the null hypothesis and conclude that there is evidence that the patients’ mean level in fact falls above 4.8.

Part-b

As we reject the null hypothesis, so we may have committed Type-I error of rejecting the true null hypothesis.

Part-c

Mean is required within error E=5.5-5.37=0.13

So, sample size n>=(z*s/E)2=(1.96*0.67/0.13)2=102.04

So, sample required is at least 103.

Part-d

We assumed the population distribution is normal

Part-e

Right tailed critical t=2.015

So right tailed 95% confidence interval =xbar+t*s/sqrt(n), +

= 5.37+2.015*0.67/sqrt(6) , +

=(5.92, +)