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Chapter 6: Q6E (page 262)

Urinary angiotensinogen (AGT) level is one quantitative indicator of kidney function. The article 鈥淯rinary Angiotensinogen as a Potential Biomarker of Chronic Kidney Diseases鈥� (J. of the Amer. Society of Hypertension, \({\rm{2008: 349 - 354}}\)) describes a study in which urinary AGT level \({\rm{(\mu g)}}\) was determined for a sample of adults with chronic kidney disease. Here is representative data (consistent with summary quantities and descriptions in the cited article):

An appropriate probability plot supports the use of the lognormal distribution (see Section \({\rm{4}}{\rm{.5}}\)) as a reasonable model for urinary AGT level (this is what the investigators did).

a. Estimate the parameters of the distribution. (Hint: Rem ember that \({\rm{X}}\) has a lognormal distribution with parameters \({\rm{\mu }}\) and \({{\rm{\sigma }}^{\rm{2}}}\) if \({\rm{ln(X)}}\) is normally distributed with mean \({\rm{\mu }}\) and variance \({{\rm{\sigma }}^{\rm{2}}}\).)

b. Use the estimates of part (a) to calculate an estimate of the expected value of AGT level. (Hint: What is \({\rm{E(X)}}\)?)

Short Answer

Expert verified

(a) The parameters of the distribution is obtained as\({\rm{\mu :\bar x = 4}}{\rm{.4297}}\)and\({{\rm{\sigma }}^{\rm{2}}}{\rm{:}}{{\rm{s}}^{\rm{2}}}{\rm{ = 2}}{\rm{.2949}}\).

(b) Estimate of the expected value of AGT level is \({\rm{E(X)}} \approx {\rm{264}}{\rm{.3172}}\).

Step by step solution

01

Concept Introduction

The average of the given numbers is computed by dividing the total number of numbers by the sum of the given numbers.

The median is the middle number in a list of numbers that has been sorted ascending or descending, and it might be more descriptive of the data set than the average. When there are outliers in the series that could affect the average of the numbers, the median is sometimes utilised instead of the mean.

The standard deviation is a statistic that measures the amount of variation or dispersion in a set of numbers.

02

Parameters of Distribution

(a)

The value of\({\rm{n}}\)is given as\({\rm{n = 40}}\).

The data provided is 鈥�

\(\begin{array}{l}{\rm{2}}{\rm{.6,6}}{\rm{.2,7}}{\rm{.4,9}}{\rm{.6,11}}{\rm{.5,13}}{\rm{.5,14}}{\rm{.5,17,20,28}}{\rm{.8,29}}{\rm{.5,29}}{\rm{.5,41}}{\rm{.7,45}}{\rm{.7,}}\\{\rm{56}}{\rm{.2,56}}{\rm{.2,66}}{\rm{.1,66}}{\rm{.1,67}}{\rm{.6,74}}{\rm{.1,97}}{\rm{.7,141}}{\rm{.3,147}}{\rm{.9,177}}{\rm{.8,186}}{\rm{.2,}}\\{\rm{186}}{\rm{.2,190}}{\rm{.6,208}}{\rm{.9,229}}{\rm{.1,229}}{\rm{.1,288}}{\rm{.4,288}}{\rm{.4,346}}{\rm{.7,407}}{\rm{.4,426}}{\rm{.6,}}\\{\rm{575}}{\rm{.4,616}}{\rm{.6,724}}{\rm{.4,812}}{\rm{.8,1122}}\end{array}\)

Take the natural logarithm of each data value (for example:\({\rm{ln2}}{\rm{.6}} \approx {\rm{0}}{\rm{.9555}}\)) 鈥�

\(\begin{array}{l}{\rm{0}}{\rm{.9555,1}}{\rm{.8245,2}}{\rm{.0015,2}}{\rm{.2618,2}}{\rm{.4423,2}}{\rm{.6027,2}}{\rm{.6741,2}}{\rm{.8332,}}\\{\rm{2}}{\rm{.9957,3}}{\rm{.3604,3}}{\rm{.3844,3}}{\rm{.3844,3}}{\rm{.7305,3}}{\rm{.8221,4}}{\rm{.0289,4}}{\rm{.0289,}}\\{\rm{4}}{\rm{.1912,4}}{\rm{.1912,4}}{\rm{.2136,4}}{\rm{.3054,4}}{\rm{.5819,4}}{\rm{.9509,4}}{\rm{.9965,5}}{\rm{.1807,}}\\{\rm{5}}{\rm{.2268,5}}{\rm{.2268,5}}{\rm{.2502,5}}{\rm{.3419,5}}{\rm{.4342,5}}{\rm{.4342,n\& 5}}{\rm{.6643,5}}{\rm{.6643,}}\\{\rm{5}}{\rm{.8485,6}}{\rm{.0098,6}}{\rm{.0558,6}}{\rm{.3551,6}}{\rm{.4242,6}}{\rm{.5853,6}}{\rm{.7005,7}}{\rm{.0229}}\\{\rm{ln2}}{\rm{.6}} \approx {\rm{0}}{\rm{.9555}}\end{array}\)

A point estimate of the population mean is the sample mean.

The sample mean is the sum of all values divided by the number of values 鈥�

\(\begin{array}{l}{\rm{\bar x = }}\frac{{{\rm{0}}{\rm{.9555 + 1}}{\rm{.8245 + 2}}{\rm{.0015 + \ldots + 6}}{\rm{.5853 + 6}}{\rm{.7005 + 7}}{\rm{.0229}}}}{{{\rm{40}}}}\\{\rm{ = }}\frac{{{\rm{177}}{\rm{.1871}}}}{{{\rm{40}}}} \approx {\rm{4}}{\rm{.4297}}\end{array}\)

Create the following table 鈥�

Find the sum of numbers in the last column to get 鈥�

\(\sum {{{{\rm{(}}{{\rm{x}}_{\rm{i}}}{\rm{ - \bar x)}}}^{\rm{2}}}{\rm{ = 89}}{\rm{.5016}}} \)

The variance is the sum of squared deviations from the mean divided by\({\rm{n - 1}}\).

\(\begin{array}{c}{{\rm{s}}^{\rm{2}}}{\rm{ = }}\frac{{{\rm{89}}{\rm{.5016}}}}{{{\rm{40 - 1}}}}\\{\rm{ = }}\frac{{{\rm{89}}{\rm{.5016}}}}{{{\rm{39}}}}\\ \approx {\rm{2}}{\rm{.2949}}\end{array}\)

Therefore, the values obtained are\({\rm{\mu :\bar x = 4}}{\rm{.4297}}\)and\({{\rm{\sigma }}^{\rm{2}}}{\rm{:}}{{\rm{s}}^{\rm{2}}}{\rm{ = 2}}{\rm{.2949}}\).

03

Estimate of value of AGT

(b)

The value of\({\rm{n}}\)is given as\({\rm{n = 40}}\).

The data provided is 鈥�

\(\begin{array}{l}{\rm{2}}{\rm{.6,6}}{\rm{.2,7}}{\rm{.4,9}}{\rm{.6,11}}{\rm{.5,13}}{\rm{.5,14}}{\rm{.5,17,20,28}}{\rm{.8,29}}{\rm{.5,29}}{\rm{.5,41}}{\rm{.7,45}}{\rm{.7,}}\\{\rm{56}}{\rm{.2,56}}{\rm{.2,66}}{\rm{.1,66}}{\rm{.1,67}}{\rm{.6,74}}{\rm{.1,97}}{\rm{.7,141}}{\rm{.3,147}}{\rm{.9,177}}{\rm{.8,186}}{\rm{.2,}}\\{\rm{186}}{\rm{.2,190}}{\rm{.6,208}}{\rm{.9,229}}{\rm{.1,229}}{\rm{.1,288}}{\rm{.4,288}}{\rm{.4,346}}{\rm{.7,407}}{\rm{.4,426}}{\rm{.6,}}\\{\rm{575}}{\rm{.4,616}}{\rm{.6,724}}{\rm{.4,812}}{\rm{.8,1122}}\end{array}\)

Take the natural logarithm of each data value (for example:\({\rm{ln2}}{\rm{.6}} \approx {\rm{0}}{\rm{.9555}}\)) 鈥�

\(\begin{array}{l}{\rm{0}}{\rm{.9555,1}}{\rm{.8245,2}}{\rm{.0015,2}}{\rm{.2618,2}}{\rm{.4423,2}}{\rm{.6027,2}}{\rm{.6741,2}}{\rm{.8332,}}\\{\rm{2}}{\rm{.9957,3}}{\rm{.3604,3}}{\rm{.3844,3}}{\rm{.3844,3}}{\rm{.7305,3}}{\rm{.8221,4}}{\rm{.0289,4}}{\rm{.0289,}}\\{\rm{4}}{\rm{.1912,4}}{\rm{.1912,4}}{\rm{.2136,4}}{\rm{.3054,4}}{\rm{.5819,4}}{\rm{.9509,4}}{\rm{.9965,5}}{\rm{.1807,}}\\{\rm{5}}{\rm{.2268,5}}{\rm{.2268,5}}{\rm{.2502,5}}{\rm{.3419,5}}{\rm{.4342,5}}{\rm{.4342,n\& 5}}{\rm{.6643,5}}{\rm{.6643,}}\\{\rm{5}}{\rm{.8485,6}}{\rm{.0098,6}}{\rm{.0558,6}}{\rm{.3551,6}}{\rm{.4242,6}}{\rm{.5853,6}}{\rm{.7005,7}}{\rm{.0229}}\\{\rm{ln2}}{\rm{.6}} \approx {\rm{0}}{\rm{.9555}}\end{array}\)

A point estimate of the population mean is the sample mean.

The sample mean is the sum of all values divided by the number of values 鈥�

\(\begin{array}{l}{\rm{\bar x = }}\frac{{{\rm{0}}{\rm{.9555 + 1}}{\rm{.8245 + 2}}{\rm{.0015 + \ldots + 6}}{\rm{.5853 + 6}}{\rm{.7005 + 7}}{\rm{.0229}}}}{{{\rm{40}}}}\\{\rm{ = }}\frac{{{\rm{177}}{\rm{.1871}}}}{{{\rm{40}}}} \approx {\rm{4}}{\rm{.4297}}\end{array}\)

Create the following table 鈥�

Find the sum of numbers in the last column to get 鈥�

\(\sum {{{{\rm{(}}{{\rm{x}}_{\rm{i}}}{\rm{ - \bar x)}}}^{\rm{2}}}{\rm{ = 89}}{\rm{.5016}}} \)

The variance is the sum of squared deviations from the mean divided by\({\rm{n - 1}}\).

\(\begin{array}{c}{{\rm{s}}^{\rm{2}}}{\rm{ = }}\frac{{{\rm{89}}{\rm{.5016}}}}{{{\rm{40 - 1}}}}\\{\rm{ = }}\frac{{{\rm{89}}{\rm{.5016}}}}{{{\rm{39}}}}\\ \approx {\rm{2}}{\rm{.2949}}\end{array}\)

The mean of a lognormal distribution is given by the formula 鈥�

\({\rm{E(X) = }}{{\rm{e}}^{{\rm{\mu + }}{{\rm{\sigma }}^{\rm{2}}}{\rm{/2}}}}\)

Substituting the values and solving 鈥�

\(\begin{array}{c}{\rm{E(X)}} \approx {{\rm{e}}^{{\rm{4}}{\rm{.4297 + 2}}{\rm{.2949/2}}}}\\{\rm{ = }}{{\rm{e}}^{{\rm{5}}{\rm{.57715}}}} \approx {\rm{264}}{\rm{.3172}}\end{array}\)

Therefore, the value is obtained as \({\rm{E(X)}} \approx {\rm{264}}{\rm{.3172}}\).

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Most popular questions from this chapter

The accompanying data on flexural strength (MPa) for concrete beams of a certain type was introduced in Example 1.2.

\(\begin{array}{*{20}{r}}{{\rm{5}}{\rm{.9}}}&{{\rm{7}}{\rm{.2}}}&{{\rm{7}}{\rm{.3}}}&{{\rm{6}}{\rm{.3}}}&{{\rm{8}}{\rm{.1}}}&{{\rm{6}}{\rm{.8}}}&{{\rm{7}}{\rm{.0}}}\\{{\rm{7}}{\rm{.6}}}&{{\rm{6}}{\rm{.8}}}&{{\rm{6}}{\rm{.5}}}&{{\rm{7}}{\rm{.0}}}&{{\rm{6}}{\rm{.3}}}&{{\rm{7}}{\rm{.9}}}&{{\rm{9}}{\rm{.0}}}\\{{\rm{3}}{\rm{.2}}}&{{\rm{8}}{\rm{.7}}}&{{\rm{7}}{\rm{.8}}}&{{\rm{9}}{\rm{.7}}}&{{\rm{7}}{\rm{.4}}}&{{\rm{7}}{\rm{.7}}}&{{\rm{9}}{\rm{.7}}}\\{{\rm{7}}{\rm{.3}}}&{{\rm{7}}{\rm{.7}}}&{{\rm{11}}{\rm{.6}}}&{{\rm{11}}{\rm{.3}}}&{{\rm{11}}{\rm{.8}}}&{{\rm{10}}{\rm{.7}}}&{}\end{array}\)

Calculate a point estimate of the mean value of strength for the conceptual population of all beams manufactured in this fashion, and state which estimator you used\({\rm{(Hint:\Sigma }}{{\rm{x}}_{\rm{i}}}{\rm{ = 219}}{\rm{.8}}{\rm{.)}}\)

b. Calculate a point estimate of the strength value that separates the weakest 50% of all such beams from the strongest 50 %, and state which estimator you used.

c. Calculate and interpret a point estimate of the population standard deviation\({\rm{\sigma }}\). Which estimator did you use?\({\rm{(Hint:}}\left. {{\rm{\Sigma x}}_{\rm{i}}^{\rm{2}}{\rm{ = 1860}}{\rm{.94}}{\rm{.}}} \right)\)

d. Calculate a point estimate of the proportion of all such beams whose flexural strength exceeds\({\rm{10MPa}}\). (Hint: Think of an observation as a "success" if it exceeds 10.)

e. Calculate a point estimate of the population coefficient of variation\({\rm{\sigma /\mu }}\), and state which estimator you used.

Let\({\rm{X}}\)have a Weibull distribution with parameters\({\rm{\alpha }}\)and\({\rm{\beta }}\), so

\(\begin{array}{l}{\rm{E(X) = \beta \times \Gamma (1 + 1/\alpha )V(X)}}\\{\rm{ = }}{{\rm{\beta }}^{\rm{2}}}\left\{ {{\rm{\Gamma (1 + 2/\alpha ) - (\Gamma (1 + 1/\alpha )}}{{\rm{)}}^{\rm{2}}}} \right\}\end{array}\)

a. Based on a random sample\({{\rm{X}}_{\rm{1}}}{\rm{, \ldots ,}}{{\rm{X}}_{\rm{n}}}\), write equations for the method of moments estimators of\({\rm{\beta }}\)and\({\rm{\alpha }}\). Show that, once the estimate of\({\rm{\alpha }}\)has been obtained, the estimate of\({\rm{\beta }}\)can be found from a table of the gamma function and that the estimate of\({\rm{\alpha }}\)is the solution to a complicated equation involving the gamma function.

b. If\({\rm{n = 20,\bar x = 28}}{\rm{.0}}\), and\({\rm{\Sigma x}}_{\rm{i}}^{\rm{2}}{\rm{ = 16,500}}\), compute the estimates. (Hint:\(\left. {{{{\rm{(\Gamma (1}}{\rm{.2))}}}^{\rm{2}}}{\rm{/\Gamma (1}}{\rm{.4) = }}{\rm{.95}}{\rm{.}}} \right)\)

Each of 150 newly manufactured items is examined and the number of scratches per item is recorded (the items are supposed to be free of scratches), yielding the following data:

Assume that X has a Poisson distribution with parameter \({\bf{\mu }}.\)and that X represents the number of scratches on a randomly picked item.

a. Calculate the estimate for the data using an unbiased \({\bf{\mu }}.\)estimator. (Hint: for X Poisson, \({\rm{E(X) = \mu }}\) ,therefore \({\rm{E(\bar X) = ?)}}\)

c. What is your estimator's standard deviation (standard error)? Calculate the standard error estimate. (Hint: \({\rm{\sigma }}_{\rm{X}}^{\rm{2}}{\rm{ = \mu }}\), \({\rm{X}}\))

Of \({{\rm{n}}_{\rm{1}}}\)randomly selected male smokers, \({{\rm{X}}_{\rm{1}}}\) smoked filter cigarettes, whereas of \({{\rm{n}}_{\rm{2}}}\) randomly selected female smokers, \({{\rm{X}}_{\rm{2}}}\) smoked filter cigarettes. Let \({{\rm{p}}_{\rm{1}}}\) and \({{\rm{p}}_{\rm{2}}}\) denote the probabilities that a randomly selected male and female, respectively, smoke filter cigarettes.

a. Show that \({\rm{(}}{{\rm{X}}_{\rm{1}}}{\rm{/}}{{\rm{n}}_{\rm{1}}}{\rm{) - (}}{{\rm{X}}_{\rm{2}}}{\rm{/}}{{\rm{n}}_{\rm{2}}}{\rm{)}}\) is an unbiased estimator for \({{\rm{p}}_{\rm{1}}}{\rm{ - }}{{\rm{p}}_{\rm{2}}}\). (Hint: \({\rm{E(}}{{\rm{X}}_{\rm{i}}}{\rm{) = }}{{\rm{n}}_{\rm{i}}}{{\rm{p}}_{\rm{i}}}\) for \({\rm{i = 1,2}}\).)

b. What is the standard error of the estimator in part (a)?

c. How would you use the observed values \({{\rm{x}}_{\rm{1}}}\) and \({{\rm{x}}_{\rm{2}}}\) to estimate the standard error of your estimator?

d. If \({{\rm{n}}_{\rm{1}}}{\rm{ = }}{{\rm{n}}_{\rm{2}}}{\rm{ = 200, }}{{\rm{x}}_{\rm{1}}}{\rm{ = 127}}\), and \({{\rm{x}}_{\rm{2}}}{\rm{ = 176}}\), use the estimator of part (a) to obtain an estimate of \({{\rm{p}}_{\rm{1}}}{\rm{ - }}{{\rm{p}}_{\rm{2}}}\).

e. Use the result of part (c) and the data of part (d) to estimate the standard error of the estimator.

Let\({{\rm{X}}_{\rm{1}}}{\rm{,}}{{\rm{X}}_{\rm{2}}}{\rm{, \ldots ,}}{{\rm{X}}_{\rm{n}}}\)represent a random sample from a Rayleigh distribution with pdf

\({\rm{f(x,\theta ) = }}\frac{{\rm{x}}}{{\rm{\theta }}}{{\rm{e}}^{{\rm{ - }}{{\rm{x}}^{\rm{2}}}{\rm{/(2\theta )}}}}\quad {\rm{x > 0}}\)a. It can be shown that\({\rm{E}}\left( {{{\rm{X}}^{\rm{2}}}} \right){\rm{ = 2\theta }}\). Use this fact to construct an unbiased estimator of\({\rm{\theta }}\)based on\({\rm{\Sigma X}}_{\rm{i}}^{\rm{2}}\)(and use rules of expected value to show that it is unbiased).

b. Estimate\({\rm{\theta }}\)from the following\({\rm{n = 10}}\)observations on vibratory stress of a turbine blade under specified conditions:

\(\begin{array}{*{20}{l}}{{\rm{16}}{\rm{.88}}}&{{\rm{10}}{\rm{.23}}}&{{\rm{4}}{\rm{.59}}}&{{\rm{6}}{\rm{.66}}}&{{\rm{13}}{\rm{.68}}}\\{{\rm{14}}{\rm{.23}}}&{{\rm{19}}{\rm{.87}}}&{{\rm{9}}{\rm{.40}}}&{{\rm{6}}{\rm{.51}}}&{{\rm{10}}{\rm{.95}}}\end{array}\)
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