Question

In: Statistics and Probability

The level of various substances in the blood of kidney dialysis patients is of concern because...

The level of various substances in the blood of kidney dialysis patients is of concern because kidney failure and dialysis can lead to nutritional problems. A researcher performed blood tests on several dialysis patients on 6 consecutive clinic visits. One variable measured was the level of phosphate in the blood. Phosphate levels for an individual tend to vary normally over time. The data on one patient, in milligrams of phosphate per deciliter (mg/dl) of blood, are given below.

5.2 4.6 6.3 4.3 5.6 5.1

(a) Calculate the sample mean x and its standard error.
x =  

Sx =

(b) Use the t procedures to find the margin of error for a 90% confidence interval for this patient's mean phosphate level.
margin of error =  

(c) Use the t procedures to give a 90% confidence interval for this patient's mean phosphate level.
90% CI = (______ , _______ )

Solutions

Expert Solution

Solution :

a) The sample mean is given as follows :

Where, xi's are sample values and n is sample size.

We have, n = 6

The sample mean is 5.1833 mg/dl.

The standard error of the sample mean is given as follows:

Where, s is sample standard deviation and n is sample size.

We have, n = 6 and x̄ = 5.1833

The standard error is 0.29145.

b) The margin of error for a 90% confidence interval for the population mean is given as follows :

Where, E is margin of error, s is sample standard deviation, n is sample size and t(0.10/2, n-1) is critical t value to construct 90% confidence interval.

We have, s = 0.71391, n = 6

Using t-table we get, t(0.10/2, 6-1) = 2.0150

Hence, margin of error is,

The margin of error for a 90% confidence interval for this patient's mean phosphate level is 0.5873.

c) The 90% confidence interval for population mean is given as follows :

Where, x̄ is sample mean and E is margin of error for the 90% confidence interval of population mean.

From part (a) and (c) we have,

x̄ = 5.1833 and E = 0.5873

Hence, 90% confidence interval for this patient's mean phosphate level is,

The 90% confidence interval for this patient's mean phosphate level is (4.596 mg/dl, 5.7706 mg/dl).

Please rate the answer. Thank you.


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