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Chapter 6: Q6E (page 262)

Urinary angiotensinogen (AGT) level is one quantitative indicator of kidney function. The article “Urinary Angiotensinogen as a Potential Biomarker of Chronic Kidney Diseases” (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 –

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

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

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

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 –

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

Create the following table –

Find the sum of numbers in the last column to get –

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

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 –

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

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

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

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 –

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

Create the following table –

Find the sum of numbers in the last column to get –

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

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 –

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

Substituting the values and solving –

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

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

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Most popular questions from this chapter

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:

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

An estimator \({\rm{\hat \theta }}\) is said to be consistent if for any \( \in {\rm{ > 0}}\), \({\rm{P(|\hat \theta - \theta |}} \ge \in {\rm{)}} \to {\rm{0}}\) as \({\rm{n}} \to \infty \). That is, \({\rm{\hat \theta }}\) is consistent if, as the sample size gets larger, it is less and less likely that \({\rm{\hat \theta }}\) will be further than \( \in \) from the true value of \({\rm{\theta }}\). Show that \({\rm{\bar X}}\) is a consistent estimator of \({\rm{\mu }}\) when \({{\rm{\sigma }}^{\rm{2}}}{\rm{ < }}\infty \) , by using Chebyshev’s inequality from Exercise \({\rm{44}}\) of Chapter \({\rm{3}}\). (Hint: The inequality can be rewritten in the form \({\rm{P}}\left( {\left| {{\rm{Y - }}{{\rm{\mu }}_{\rm{Y}}}} \right| \ge \in } \right) \le {\rm{\sigma }}_{\rm{Y}}^{\rm{2}}{\rm{/}} \in \). Now identify \({\rm{Y}}\) with \({\rm{\bar X}}\).)

An investigator wishes to estimate the proportion of students at a certain university who have violated the honor code. Having obtained a random sample of\({\rm{n}}\)students, she realizes that asking each, "Have you violated the honor code?" will probably result in some untruthful responses. Consider the following scheme, called a randomized response technique. The investigator makes up a deck of\({\rm{100}}\)cards, of which\({\rm{50}}\)are of type I and\({\rm{50}}\)are of type II.

Type I: Have you violated the honor code (yes or no)?

Type II: Is the last digit of your telephone number a\({\rm{0 , 1 , or 2}}\)(yes or no)?

Each student in the random sample is asked to mix the deck, draw a card, and answer the resulting question truthfully. Because of the irrelevant question on type II cards, a yes response no longer stigmatizes the respondent, so we assume that responses are truthful. Let\({\rm{p}}\)denote the proportion of honor-code violators (i.e., the probability of a randomly selected student being a violator), and let\({\rm{\lambda = P}}\)(yes response). Then\({\rm{\lambda }}\)and\({\rm{p}}\)are related by\({\rm{\lambda = }}{\rm{.5p + (}}{\rm{.5)(}}{\rm{.3)}}\).

a. Let\({\rm{Y}}\)denote the number of yes responses, so\({\rm{Y\sim}}\)Bin\({\rm{(n,\lambda )}}\). Thus Y / n is an unbiased estimator of\({\rm{\lambda }}\). Derive an estimator for\({\rm{p}}\)based on\({\rm{Y}}\). If\({\rm{n = 80}}\)and\({\rm{y = 20}}\), what is your estimate? (Hint: Solve\({\rm{\lambda = }}{\rm{.5p + }}{\rm{.15}}\)for\({\rm{p}}\)and then substitute\({\rm{Y/n}}\)for\({\rm{\lambda }}\).)

b. Use the fact that\({\rm{E(Y/n) = \lambda }}\)to show that your estimator\({\rm{\hat p}}\)is unbiased.

c. If there were\({\rm{70}}\)type I and\({\rm{30}}\)type II cards, what would be your estimator for\({\rm{p}}\)?

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 \({{\rm{\sigma }}^{\rm{2}}}\)

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?
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