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The standard deviation (SD) is a measure of the variability, or dispersion, between individual data values and the mean, whereas the standard error of the mean (SEM) is a measure of how far the sample mean (average) of the data is expected to differ from the genuine population mean. Always, the SEM is smaller than the SD.

The total audited values are denoted by\({\theta _T}\).

First, using simply\(N\)and\(\bar X\)the following statistic can be employed.

\({{\rm{\hat \theta }}_{{{\rm{T}}_{\rm{1}}}}}{\rm{ = N \times \bar X}}\)

Which is simply the \(N\) invoice multiplied by the average.

Second, using\({\rm{N \times \bar X}}\), and\(\bar D\), the following statistic can be employed.

\({{\rm{\hat \theta }}_{{{\rm{T}}_{\rm{2}}}}}{\rm{ = T - N \times \bar D}}\)

When the total sample mean error is deducted from the total book value, the total book value is the result.

Third, using \(T\)and \(\bar X/\bar Y\),the following statistic can be employed.

This is the quotient of the sample mean audited value and the sample mean book average multiplied by the total book value: \({{\rm{\hat \theta }}_{{{\rm{T}}_{\rm{3}}}}}{\rm{ = T \times }}\frac{{{\rm{\bar X}}}}{{{\rm{\bar Y}}}}\)

Consider the given,

\(\begin{array}{l}{\rm{N = 5000 }}\\{\rm{T = 1,761,300}}\end{array}\)

The point estimations can be determined using the invoice table and the invoice table

\(\begin{array}{c}{\rm{\bar y = }}\frac{{\rm{1}}}{{\rm{5}}}{\rm{(300 + 720 + 526 + 200 + 127)}}\\{\rm{ = 374}}{\rm{.6}}\end{array}\)

The average audited value in the sample is

\(\begin{array}{c}{\rm{\bar x = }}\frac{{\rm{1}}}{{\rm{5}}}{\rm{(300 + 520 + 526 + 200 + 157)}}\\{\rm{ = 340}}{\rm{.6,}}\end{array}\)

Where the standard deviation of the sample mean error is

\(\begin{array}{c}{\rm{\bar d = }}\frac{{\rm{1}}}{{\rm{5}}}{\rm{(0 + 200 + 0 + 0 - 30)}}\\{\rm{ = 34}}\end{array}\)

As a result, the point estimations

\(\begin{array}{c}{\rm{\& }}{{{\rm{\hat \theta }}}_{{{\rm{T}}_{\rm{1}}}}}{\rm{ = N \times \bar x = 5000 \times 340}}{\rm{.6}}\\{\rm{ = 1,704,000,}}\\{\rm{n\& }}{{{\rm{\hat \theta }}}_{{{\rm{T}}_{\rm{2}}}}}{\rm{ = T - N \times \bar d = 1,761,300 - 5,000 \times 34}}{\rm{.0}}\\{\rm{ = 1,591,300,}}\\{{{\rm{\hat \theta }}}_{{{\rm{T}}_{\rm{3}}}}}{\rm{ = T \times }}\frac{{{\rm{\bar x}}}}{{{\rm{\bar y}}}}\\{\rm{ = 1,761,300 \times }}\frac{{{\rm{340}}{\rm{.6}}}}{{{\rm{374}}{\rm{.6}}}}\\{\rm{ = 1,601,438}}{\rm{.281}}\end{array}\)

Therefore, the required estimated value is\({\rm{1,601,438}}{\rm{.281}}\).