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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

Suppose a certain type of fertilizer has an expected yield per acre of \({{\rm{\mu }}_{\rm{2}}}\)with variance \({{\rm{\sigma }}^{\rm{2}}}\)whereas the expected yield for a second type of fertilizer is with the same variance \({{\rm{\sigma }}^{\rm{2}}}\).Let \({\rm{S}}_{\rm{1}}^{\rm{2}}\) and \({\rm{S}}_{\rm{2}}^{\rm{2}}\)denote the sample variances of yields based on sample sizes \({{\rm{n}}_{\rm{1}}}\)and \({{\rm{n}}_{\rm{2}}}\),respectively, of the two fertilizers. Show that the pooled (combined) estimator

\({{\rm{\hat \sigma }}^{\rm{2}}}{\rm{ = }}\frac{{\left( {{{\rm{n}}_{\rm{1}}}{\rm{ - 1}}} \right){\rm{S}}_{\rm{1}}^{\rm{2}}{\rm{ + }}\left( {{{\rm{n}}_{\rm{2}}}{\rm{ - 1}}} \right){\rm{S}}_{\rm{2}}^{\rm{2}}}}{{{{\rm{n}}_{\rm{1}}}{\rm{ + }}{{\rm{n}}_{\rm{2}}}{\rm{ - 2}}}}\)

is an unbiased estimator of \({{\rm{\sigma }}^{\rm{2}}}\)

When the sample standard deviation S is based on a random sample from a normal population distribution, it can be shown that \({\rm{E(S) = }}\sqrt {{\rm{2/(n - 1)}}} {\rm{\Gamma (n/2)\sigma /\Gamma ((n - 1)/2)}}\)

Use this to obtain an unbiased estimator for \({\rm{\sigma }}\) of the form \({\rm{cS}}\). What is \({\rm{c}}\) when \({\rm{n = 20}}\)?

At time \({\rm{t = 0, 20}}\) identical components are tested. The lifetime distribution of each is exponential with parameter \({\rm{\lambda }}\). The experimenter then leaves the test facility unmonitored. On his return \({\rm{24}}\) hours later, the experimenter immediately terminates the test after noticing that \({\rm{y = 15}}\) of the \({\rm{20}}\) components are still in operation (so \({\rm{5}}\) have failed). Derive the mle of \({\rm{\lambda }}\). (Hint: Let \({\rm{Y = }}\) the number that survive \({\rm{24}}\) hours. Then \({\rm{Y}} \sim {\rm{Bin(n,p)}}\). What is the mle of \({\rm{p}}\)? Now notice that \({\rm{p = P(}}{{\rm{X}}_{\rm{i}}} \ge {\rm{24)}}\), where \({{\rm{X}}_{\rm{i}}}\) is exponentially distributed. This relates \({\rm{\lambda }}\) to \({\rm{p}}\), so the former can be estimated once the latter has been.)

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 \({\rm{cX}}\).)

The shear strength of each of ten test spot welds is determined, yielding the following data (psi):

\(\begin{array}{*{20}{l}}{{\rm{392}}}&{{\rm{376}}}&{{\rm{401}}}&{{\rm{367}}}&{{\rm{389}}}&{{\rm{362}}}&{{\rm{409}}}&{{\rm{415}}}&{{\rm{358}}}&{{\rm{375}}}\end{array}\)

a. Assuming that shear strength is normally distributed, estimate the true average shear strength and standard deviation of shear strength using the method of maximum likelihood.

b. Again assuming a normal distribution, estimate the strength value below which\({\rm{95\% }}\)of all welds will have their strengths. (Hint: What is the\({\rm{95 th}}\)percentile in terms of\({\rm{\mu }}\)and\({\rm{\sigma }}\)? Now use the invariance principle.)

c. Suppose we decide to examine another test spot weld. Let\({\rm{X = }}\)shear strength of the weld. Use the given data to obtain the mle of\({\rm{P(X拢400)}}{\rm{.(Hint:P(X拢400) = \Phi ((400 - \mu )/\sigma )}}{\rm{.)}}\)
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