Chapter 1, Problem 79SE

Let ${\bar{x}}_n$ and $s_n^2$ denote the sample mean and variance for the sample x₁,...., x_n and let ${\bar{x}}_{n\text{ + 1}}$ and $s_{n\text{ + 1}}^2$ denote these quantities when an additional observation x_{n + l} is added to the sample.

1. a. Show how ${\stackrel{-}{x}}_{n\text{ + 1}}$ can be computed from ${\stackrel{-}{x}}_n$ and x_{n + l}.
2. b. Show that $n{s}_{n\text{ + 1}}^2\text{ = (}n\text{ - 1)}{s}_n^2\text{ + }\frac{n}{n\text{ + 1}}{(x_{n+1}-{\stackrel{-}{x}}_n)}^2$ so that $s_{n\text{ + 1}}^2$ can be computed from x_{n + 1} and $s_n^2$.
3. c. Suppose that a sample of 15 strands of drapery yarn has resulted in a sample mean thread elongation of 12.58 mm and a sample standard deviation of .512 mm. A 16^th strand results in an elongation value of

To determine

Prove that ${\overline{x}}_{n+1}$ can be computed with ${\overline{x}}_n$ and $x_{n+1}$.

Answer to Problem 79SE

The formula to compute ${\overline{x}}_{n+1}$ is ${\overline{x}}_{n+1}=\frac{n{\overline{x}}_n+x_{n+1}}{n+1}$.

Explanation of Solution

Given info:

For a sample $x_1,x_2,\mathrm{...},x_n$ the sample mean is ${\overline{x}}_n$ and the sample variance is $s_n{}^2$. When a new observation $x_{n+1}$ is added to the sample then the sample mean is ${\overline{x}}_{n+1}$ and the sample variance is $s_{n+1}{}^2$.

Calculation:

Mean:

The arithmetic mean (also called the mean) is the most commonly used measure of central tendency. It is calculated by summing the observed numerical values of a variable in a set of data and then dividing the total by the number of observations involved.

The general formula to obtain mean is,

$\begin{array}{l}{\overline{x}}_n=\frac{{\displaystyle \sum _{i=1}^{n}x}}{n}\\ \Rightarrow {\displaystyle \sum _{i=1}^{n}x}=n{\overline{x}}_n\end{array}$

From the definition of mean ${\overline{x}}_{n+1}$ is obtained as follows:

$\begin{array}{c}{\overline{x}}_{n+1}=\frac{{\displaystyle \sum _{i=1}^{n+1}x}}{n+1}\\ =\frac{{\displaystyle \sum _{i=1}^{n}x+x}}{n+1}\\ =\frac{n{\overline{x}}_n+x}{n+1}\end{array}$

Thus, the formula to compute ${\overline{x}}_{n+1}$ is ${\overline{x}}_{n+1}=\frac{n{\overline{x}}_n+x_{n+1}}{n+1}$.

To determine

Show that $n{s}_{n+1}^2=\left(n-1\right)s_n^2+\frac{n}{n+1}{\left(x_{n+1}-{\overline{x}}_n\right)}^2$

Answer to Problem 79SE

$n{s}_{n+1}^2=\left(n-1\right)s_n^2+\frac{n}{n+1}{\left(x_{n+1}-{\overline{x}}_n\right)}^2$

Explanation of Solution

Calculation:

Variance:

The variance is based on how much each observation deviates from a central point represented by the mean. In general, the greater the distances between the individual observations and the mean, the greater the variability of the data set.

The general formula to obtain variance is,

$\begin{array}{l}{s}_n^2=\frac{{\displaystyle \sum _{i=1}^n{\left(x_i-{\overline{x}}_n\right)}^2}}{n-1}\\ \Rightarrow {\displaystyle \sum _{i=1}^n{\left(x_i-{\overline{x}}_n\right)}^2}=n{s}_n^2\end{array}$

From the definition of variance $s_n^2$ is obtained as follows:

$\begin{array}{c}{s}_{n+1}^2=\frac{{\displaystyle \sum _{i=1}^{n+1}{\left(x-{\overline{x}}_{n+1}\right)}^2}}{\left(n+1\right)-1}\\ n{s}_{n+1}^2={\displaystyle \sum _{i=1}^{n+1}{\left(x-{\overline{x}}_{n+1}\right)}^2}\\ ={\displaystyle \sum _{i=1}^{n+1}\left(x_i^2-2x\overline{x}{}_{n+1}+{\overline{x}}_{n+1}^2\right)}\\ ={\displaystyle \sum _{i=1}^{n+1}{x}_i^2}-{\displaystyle \sum _{i=1}^{n+1}2x\overline{x}{}_{n+1}+}{\displaystyle \sum _{i=1}^{n+1}{\overline{x}}_{n+1}^2}\end{array}$

$\begin{array}{c}={\displaystyle \sum _{i=1}^{n+1}{x}_i^2}-2{\overline{x}}_{n+1}\times {\displaystyle \sum _{i=1}^{n+1}x+}{\displaystyle \sum _{i=1}^{n+1}{\overline{x}}_{n+1}^2}\\ ={\displaystyle \sum _{i=1}^{n+1}{x}_i^2}-2{\overline{x}}_{n+1}\times \left(n+1\right){\overline{x}}_{n+1}+\left(n+1\right){\overline{x}}_{n+1}^2\\ ={\displaystyle \sum _{i=1}^{n+1}{x}_i^2}-\left(n+1\right){\overline{x}}_{n+1}^2\end{array}$

$\begin{array}{c}={\displaystyle \sum _{i=1}^{n+1}{x}_i^2}-\left(n+1\right){\overline{x}}_{n+1}^2\\ n{s}_{n+1}^2={\displaystyle \sum _{i=1}^n{x}_i^2+x_{n+1}^2}-\left(n+1\right){\overline{x}}_{n+1}^2\end{array}$

Foe easy computation procedure, add and subtract $n{\overline{x}}_n^2$ to the left hand quantity of above equation.

$\begin{array}{c}n{s}_{n+1}^2={\displaystyle \sum _{i=1}^n{x}_i^2+x_{n+1}^2}-\left(n+1\right){\overline{x}}_{n+1}^2\\ =\left({\displaystyle \sum _{i=1}^n{x}_i^2-n{\overline{x}}_n^2}\right)+n{\overline{x}}_n^2+x_{n+1}^2-\left(n+1\right){\overline{x}}_{n+1}^2\\ ={\displaystyle \sum _{i=1}^n{\left(x_i-{\overline{x}}_n\right)}^2}+n{\overline{x}}_n^2+x_{n+1}^2-\left(n+1\right){\overline{x}}_{n+1}^2\\ =\left(n-1\right)s_n^2+\left(n{\overline{x}}_n^2+x_{n+1}^2-\left(n+1\right){\overline{x}}_{n+1}^2\right)\end{array}$

From part (a), the formula for ${\overline{x}}_{n+1}$ is ${\overline{x}}_{n+1}=\frac{n{\overline{x}}_n+x_{n+1}}{n+1}$.

Substituting ${\overline{x}}_{n+1}=\frac{n{\overline{x}}_n+x_{n+1}}{n+1}$ in the quantity $\left(n{\overline{x}}_n^2+x_{n+1}^2-\left(n+1\right){\overline{x}}_{n+1}^2\right)$ of above equation,

$\begin{array}{c}\left(n{\overline{x}}_n^2+x_{n+1}^2-\left(n+1\right){\overline{x}}_{n+1}^2\right)=n{\overline{x}}_n^2+x_{n+1}^2-\left(n+1\right)\times {\left(\frac{n{\overline{x}}_n+x_{n+1}}{n+1}\right)}^2\\ =n{\overline{x}}_n^2+x_{n+1}^2-\frac{{\left(n{\overline{x}}_n+x_{n+1}\right)}^2}{n+1}\\ =\frac{n\left(n+1\right){\overline{x}}_n^2+\left(n+1\right)x_{n+1}^2-{\left(n{\overline{x}}_n+x_{n+1}\right)}^2}{n+1}\\ =\frac{n\left(n+1\right){\overline{x}}_n^2+\left(n+1\right)x_{n+1}^2-{\left(n{\overline{x}}_n\right)}^2+x_{n+1}^2+2n{\overline{x}}_n{x}_{n+1}}{n+1}\end{array}$

$\left(n{\overline{x}}_n^2+x_{n+1}^2-\left(n+1\right){\overline{x}}_{n+1}^2\right)=\frac{n}{n+1}{\left(x_{n+1}-{\overline{x}}_n\right)}^2$

Therefore, the variance equation reduces as follows:

$\begin{array}{c}n{s}_{n+1}^2=\left(n-1\right)s_n^2+\left(n{\overline{x}}_n^2+x_{n+1}^2-\left(n+1\right){\overline{x}}_{n+1}^2\right)\\ =\left(n-1\right)s_n^2+\frac{n}{n+1}{\left(x_{n+1}-{\overline{x}}_n\right)}^2\end{array}$

Thus, $n{s}_{n+1}^2=\left(n-1\right)s_n^2+\frac{n}{n+1}{\left(x_{n+1}-{\overline{x}}_n\right)}^2$

To determine

Find the sample mean of all 16 elongation observations.

Find the sample standard deviation of all 16 elongation observations.

Answer to Problem 79SE

The sample mean of all 16 elongation observations is $\underset{\_}{{\overline{x}}_{16}=12.53\text{mm}}$.

The sample standard deviation of all 16 elongation observations is $\underset{\_}{s_{16}=0.532\text{mm}}$.

Explanation of Solution

Given info:

The mean and the standard deviation for a sample of 15 elongation observations is 12.58mm and 0.512 mm. An additional observation 11.8 is added to the sample.

Calculation:

Sample mean ${\overline{x}}_{16}$:

The mean for the sample of 15 observations is ${\overline{x}}_{15}=12.58$ and the 16^th observation is $x_{16}=11.8\text{mm}$.

From part (a),

The formula to compute ${\overline{x}}_{n+1}$ is ${\overline{x}}_{n+1}=\frac{n{\overline{x}}_n+x_{n+1}}{n+1}$.

For $n=15$, ${\overline{x}}_n=12.58\text{mm}$ and $x_{n+1}=11.8\text{mm}$.

The sample mean ${\overline{x}}_{16}$ is computed as follows:

$\begin{array}{c}{\overline{x}}_{n+1}=\frac{n{\overline{x}}_n+x_{n+1}}{n+1}\\ {\overline{x}}_{15+1}=\frac{15\times {\overline{x}}_{15}+x_{15+1}}{15+1}\\ {\overline{x}}_{16}=\frac{15\times 12.58+11.8}{16}\\ =\frac{200.5}{16}\end{array}$

$=12.53$

Thus, the sample mean of all 16 elongation observations is $\underset{\_}{{\overline{x}}_{16}=12.53\text{mm}}$.

Sample variance $s_{16}^2$:

The sample standard deviation for 15 observations is $s_{15}=0.512\text{mm}$

From part (b), the formula for variance is obtained as follows:

$\begin{array}{c}n{s}_{n+1}^2=\left(n-1\right)s_n^2+\frac{n}{n+1}{\left(x_{n+1}-{\overline{x}}_n\right)}^2\\ s_{n+1}^2=\frac{\left(n-1\right)s_n^2+\frac{n}{n+1}{\left(x_{n+1}-{\overline{x}}_n\right)}^2}{n}\end{array}$

Therefore, the variance of 16 observations is obtained as follows:

$\begin{array}{c}{s}_{n+1}^2=\frac{\left(n-1\right)s_n^2+\frac{n}{n+1}{\left(x_{n+1}-{\overline{x}}_n\right)}^2}{n}\\ s_{15+1}^2=\frac{\left(15-1\right)s_{15}^2+\frac{15}{15+1}{\left(x_{15+1}-{\overline{x}}_{15}\right)}^2}{15}\\ s_{16}^2=\frac{\left(14\right){0.512}^2+\frac{15}{16}{\left(11.8-12.58\right)}^2}{15}\\ =0.238\end{array}$

Thus, the variance of 16 observations is $s_{16}^2=0.238$.

Sample standard deviation $s_{16}$:

The general formula for standard deviation is,

$\text{Standard}\text{deviation}=\sqrt{\text{Variance}}$

The standard deviation of 16 observations is,

$\begin{array}{c}{s}_{16}=\sqrt{s_{16}^2}\\ =\sqrt{0.238}\\ =0.532\end{array}$

Thus, the standard deviation of 16 observations is $\underset{\_}{s_{16}=0.532\text{mm}}$.