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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

Consider a random sample \({{\rm{X}}_{\rm{1}}}{\rm{,}}{{\rm{X}}_{\rm{2}}}.....{\rm{,}}{{\rm{X}}_{\rm{n}}}\) from the shifted exponential pdf

\({\rm{f(x;\lambda ,\theta ) = }}\left\{ {\begin{array}{*{20}{c}}{{\rm{\lambda }}{{\rm{e}}^{{\rm{ - \lambda (x - \theta )}}}}}&{{\rm{x}} \ge {\rm{\theta }}}\\{\rm{0}}&{{\rm{ otherwise }}}\end{array}} \right.\). Taking \({\rm{\theta = 0}}\) gives the pdf of the exponential distribution considered previously (with positive density to the right of zero). An example of the shifted exponential distribution appeared in Example \({\rm{4}}{\rm{.5}}\), in which the variable of interest was time headway in traffic flow and \({\rm{\theta = }}{\rm{.5}}\) was the minimum possible time headway. a. Obtain the maximum likelihood estimators of \({\rm{\theta }}\) and \({\rm{\lambda }}\). b. If \({\rm{n = 10}}\) time headway observations are made, resulting in the values \({\rm{3}}{\rm{.11,}}{\rm{.64,2}}{\rm{.55,2}}{\rm{.20,5}}{\rm{.44,3}}{\rm{.42,10}}{\rm{.39,8}}{\rm{.93,17}}{\rm{.82}}\), and \({\rm{1}}{\rm{.30}}\), calculate the estimates of \({\rm{\theta }}\) and \({\rm{\lambda }}\).

Consider randomly selecting \({\rm{n}}\) segments of pipe and determining the corrosion loss (mm) in the wall thickness for each one. Denote these corrosion losses by \({{\rm{Y}}_{\rm{1}}}{\rm{,}}.....{\rm{,}}{{\rm{Y}}_{\rm{n}}}\). The article 鈥淎 Probabilistic Model for a Gas Explosion Due to Leakages in the Grey Cast Iron Gas Mains鈥� (Reliability Engr. and System Safety (\({\rm{(2013:270 - 279)}}\)) proposes a linear corrosion model: \({{\rm{Y}}_{\rm{i}}}{\rm{ = }}{{\rm{t}}_{\rm{i}}}{\rm{R}}\), where \({{\rm{t}}_{\rm{i}}}\) is the age of the pipe and \({\rm{R}}\), the corrosion rate, is exponentially distributed with parameter \({\rm{\lambda }}\). Obtain the maximum likelihood estimator of the exponential parameter (the resulting mle appears in the cited article). (Hint: If \({\rm{c > 0}}\) and \({\rm{X}}\) has an exponential distribution, so does \({\rm{cX}}\).)

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: \({\rm{\sigma }}_{\rm{X}}^{\rm{2}}{\rm{ = \mu }}\), \({\rm{X}}\))

Let\({{\rm{X}}_{\rm{1}}}{\rm{,}}{{\rm{X}}_{\rm{2}}}{\rm{, \ldots ,}}{{\rm{X}}_{\rm{n}}}\)be a random sample from a pdf\({\rm{f(x)}}\)that is symmetric about\({\rm{\mu }}\), so that\({\rm{\backslash widetildeX}}\)is an unbiased estimator of\({\rm{\mu }}\). If\({\rm{n}}\)is large, it can be shown that\({\rm{V (\tilde X)\gg 1/}}\left( {{\rm{4n(f(\mu )}}{{\rm{)}}^{\rm{2}}}} \right)\).

a. Compare\({\rm{V(\backslash widetildeX)}}\)to\({\rm{V(\bar X)}}\)when the underlying distribution is normal.

b. When the underlying pdf is Cauchy (see Example 6.7),\({\rm{V(\bar X) = \yen}}\), so\({\rm{\bar X}}\)is a terrible estimator. What is\({\rm{V(\tilde X)}}\)in this case when\({\rm{n}}\)is large?

Let\({\rm{X}}\)represent the error in making a measurement of a physical characteristic or property (e.g., the boiling point of a particular liquid). It is often reasonable to assume that\({\rm{E(X) = 0}}\)and that\({\rm{X}}\)has a normal distribution. Thus, the pdf of any particular measurement error is

\({\rm{f(x;\theta ) = }}\frac{{\rm{1}}}{{\sqrt {{\rm{2\pi \theta }}} }}{{\rm{e}}^{{\rm{ - }}{{\rm{x}}^{\rm{2}}}{\rm{/2\theta }}}}\quad {\rm{ - \yen < x < \yen}}\)

(Where we have used\({\rm{\theta }}\)in place of\({{\rm{\sigma }}^{\rm{2}}}\)). Now suppose that\({\rm{n}}\)independent measurements are made, resulting in measurement errors\({{\rm{X}}_{\rm{1}}}{\rm{ = }}{{\rm{x}}_{\rm{1}}}{\rm{,}}{{\rm{X}}_{\rm{2}}}{\rm{ = }}{{\rm{x}}_{\rm{2}}}{\rm{, \ldots ,}}{{\rm{X}}_{\rm{n}}}{\rm{ = }}{{\rm{x}}_{\rm{n}}}{\rm{.}}\)Obtain the mle of\({\rm{\theta }}\).
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