Chapter 1, Problem 1.76QP

Interpretation Introduction

Interpretation: Whether or not any one of the values in the given set of replicate measurement is an outlier is to be decided using Grubb’s test.

Concept introduction: The outlier value is the suspected value that has $G$ greater than critical $G$ value.

The value of $G$ is calculated by the formula,

$G=\frac{\text{Measurement}\text{of}\text{suspected}\text{value}-\text{Mean}\text{of}\text{all}\text{measurements}}{\text{Standard}\text{deviation}}$

To determine: If there is any outlier in the given set of replicated measurements.

Answer to Problem 1.76QP

Solution:

The measurement $75$ is an outlier.

Explanation of Solution

Given

The given data of the replicated measurements is,

$\begin{array}{cc}\text{Samples}& \text{Measurements}\\ 1& 61\\ 2& 75\\ 3& 64\\ 4& 65\\ 5& 64\\ 6& 66\end{array}$

The outlier in the given set of replicated measurement is identified for the value that is exceptionally higher or lower than the other values and it is known as suspected value.

In the given data, sample number $2$ has exceptional value of $75$. Therefore, this becomes out suspected value.

Therefore, the table for the calculation of the standard deviation is,

$\begin{array}{cccc}\text{Samples}& x\left(\text{Measurements}\right)& x-\overline{x}& {\left(x-\overline{x}\right)}^2\\ 1& 61& 4.83& 23.33\\ 2& 75& -9.17& 84.089\\ 3& 64& 1.83& 3.349\\ 4& 65& 0.83& 0.689\\ 5& 64& 1.83& 3.349\\ 6& 66& -0.17& 0.029\\ \text{Mean}\left(\overline{x}\right)& 65.83& & \end{array}$

The mean of the measurement is given by the formula,

$\text{Mean}=\frac{\text{Sum}\text{of}\text{all}\text{the}\text{percentages}}{\text{total}\text{number}\text{of}\text{samples}}$

Substitute the values of all the measurements and the number of samples of peanuts in the above equation.

$\begin{array}{c}\text{Mean}=\frac{61\text{+75+64+65+64+66}}{6}\\ =65.83\end{array}$

The standard deviation is calculated as,

$\text{Standard}\text{deviation}=\sqrt{\frac{{\displaystyle \sum {\left(x-\overline{x}\right)}^2}}{\text{Number}\text{of}\text{samples}}}$

Substitute the value of ${\displaystyle \sum {\left(x-\overline{x}\right)}^2}$ and the number of samples in the above equation.

$\begin{array}{c}\text{Standard}\text{deviation}=\sqrt{\frac{23.33+84.089+3.349+0.689+3.349+0.029}{\text{6}}}\\ =\sqrt{\frac{114.835}{6}}\\ =\sqrt{19.1392}\\ =4.375\end{array}$

The outlier value is the suspected value that has $G$ greater than critical $G$ value.

The value of $G$ is calculated by the formula,

$G=\frac{\text{Measurement}\text{of}\text{suspected}\text{value}-\text{Mean}\text{of}\text{all}\text{measurements}}{\text{Standard}\text{deviation}}$

Substitute the measurement of suspected value, mean of all measurements and the standard deviation in the above equation.

$\begin{array}{c}G=\frac{\text{75}-\text{65}\text{.83}}{\text{4}\text{.375}}\\ =2.096\end{array}$

The critical value of $G$ for $95\%$ confidence and sample number of six is $1.822$.

Therefore, the calculated value of $G$ is more than its critical value. Therefore, the measurement $75$ is an outlier.

Conclusion:

The measurement $75$ is an outlier