Chapter 1, Problem 1.73QP

(a)

Interpretation Introduction

Interpretation: The mean, standard deviation, the values inside $95\%$ confidence interval and facts related to accuracy and precision is to be determined.

Concept introduction: The precision means that the results of repeated measurements agree with each other. But the accuracy means that the observed value is close to the accurate value or not.

To determine: The mean and standard deviation of the given sets of data for three manufacturers.

Answer to Problem 1.73QP

Solution

The mean and standard deviation for the given sets of data is given below.

Explanation of Solution

Explanation

Given

The data provided by three manufacturers is:

Manufacturer 1 $\left(\text{μm}\right)$	Manufacturer 2 $\left(\text{μm}\right)$	Manufacturer 3 $\left(\text{μm}\right)$
0.512	0.514	0.500
0.508	0.513	0.501
0.516	0.514	0.502
0.504	0.514	0.502
0.513	0.512	0.501

The mean of the given set of data is calculated by using the formula,

$\overline{x}=\frac{{\displaystyle \sum x_{\text{i}}}}{n}$

Where,

• $\overline{x}$ is the mean of the given set of data.
• $x_{\text{i}}$ is the i^th value of the distribution.
• $n$ is the number of values in the data.

Substitute the values of $x_{\text{i}}$ and $n$ for the manufacturer 1 in the above equation.

$\begin{array}{c}\overline{x}=\frac{0.512+0.508+0.516+0.504+0.513}{5}\\ =\underset{\_}{0.5106}\end{array}$

Substitute the values of $x_{\text{i}}$ and $n$ for the manufacturer 2 in the above equation.

$\begin{array}{c}\overline{x}=\frac{0.514+0.513+0.514+0.514+0.512}{5}\\ =\underset{\_}{0.5134}\end{array}$

Substitute the values of $x_{\text{i}}$ and $n$ for the manufacturer 3 in the above equation.

$\begin{array}{c}\overline{x}=\frac{0.500+0.501+0.502+0.502+0.501}{5}\\ =\underset{\_}{0.5012}\end{array}$

The standard deviation for the given sets of data is calculated by using the formula,

$s=\sqrt{\frac{{\displaystyle \sum {\left(x_{\text{i}}-\overline{x}\right)}^2}}{n-1}}$

Where,

• $\overline{x}$ is the mean of the given set of data.
• $x_{\text{i}}$ is the i^th value of the distribution.
• $n$ is the number of values in the data.
• $s$ is the standard deviation of the data.

Substitute the values of $x_{\text{i}}$, $\overline{x}$ and $n$ for manufacturer 1 in the above equation.

$\begin{array}{c}s=\sqrt{\frac{\begin{array}{l}{\left(0.512-0.5106\right)}^2+{\left(0.508-0.5106\right)}^2\\ +{\left(0.516-0.5106\right)}^2+{\left(0.504-0.5106\right)}^2+{\left(0.513-0.5106\right)}^2\end{array}}{5-1}}\\ =\sqrt{\frac{0.00000196+0.00000676+0.00002916+0.00004356+0.00000576}{4}}\\ =\underset{\_}{4.67\times 10^{-3}}\end{array}$

Substitute the values of $x_{\text{i}}$, $\overline{x}$ and $n$ for manufacturer 2 in the above equation.

$\begin{array}{c}s=\sqrt{\frac{\begin{array}{l}{\left(0.514-0.5134\right)}^2+{\left(0.513-0.5134\right)}^2\\ +{\left(0.514-0.5134\right)}^2+{\left(0.513-0.5134\right)}^2+{\left(0.512-0.5134\right)}^2\end{array}}{5-1}}\\ =\sqrt{\frac{0.00000036+0.00000016+0.00000036+0.00000036+0.00000196}{4}}\\ =\underset{\_}{8.94\times 10^{-4}}\end{array}$

Substitute the values of $x_{\text{i}}$, $\overline{x}$ and $n$ for manufacturer 3 in the above equation.

$\begin{array}{c}s=\sqrt{\frac{\begin{array}{l}{\left(0.500-0.5012\right)}^2+{\left(0.501-0.5012\right)}^2\\ +{\left(0.502-0.5012\right)}^2+{\left(0.502-0.5012\right)}^2+{\left(0.501-0.5012\right)}^2\end{array}}{5-1}}\\ =\sqrt{\frac{0.00000144+0.00000004+0.00000064+0.00000064+0.00000004}{4}}\\ =\underset{\_}{8.37\times 10^{-4}}\end{array}$

(b)

Interpretation Introduction

To determine: The set of data which include confidence interval $0.500\text{μm}$ out of the given sets of data.

Answer to Problem 1.73QP

Solution

The set of data for manufacturer 3 includes confidence interval $0.500\text{μm}$ out of the given sets of data.

Explanation of Solution

Explanation

The confidence interval is the interval in which the value observed for the given data is considered to be the true value.

The value $0.500\text{μm}$ is present in only one set of data which is for manufacturer 3. Rest of the sets of data does not have this value in the distribution. Therefore, the value $0.500\text{μm}$ comes under the confidence interval of manufacturer 3.

(c)

Interpretation Introduction

To determine: The value which is precise and accurate and precise but not accurate out of the given sets of data.

Answer to Problem 1.73QP

Solution

The value $0.500\mu m$ is precise and accurate but the value $0.501\mu m$ is precise but not accurate.

Explanation of Solution

Explanation

The value of the repeated results of the given measurements is close to each other is known as precision.

The value of repeated results of the given measurement is close to the true value is known as the accuracy.

Since, the value $0.500\text{μm}$ is close to the true value of the measurement. Therefore, it is precise and accurate; whereas, the value $0.501\text{μm}$ is precise as per the data but not accurate as per the true value.

Conclusion

The mean of the data coming of manufacturer 1 is $0.5106$.

The standard deviation of the data coming from manufacture 1 is $4.67\times 10^{-3}$.

The mean of the data coming of manufacturer 2 is $0.5134$.

The standard deviation of the data coming from manufacture 2 is $8.94\times 10^{-4}$.

The mean of the data coming of manufacturer 3 is $0.5012$.

The standard deviation of the data coming from manufacture 3 is $8.37\times 10^{-4}$.

The value $0.500\mu m$ is precise and accurate but the value $0.501\mu m$ is precise but not accurate