Question

Let \({\bar x_n}\) and \(s_n^2\) denote the sample mean and variance for the sample \({x_1},{x_2},...,{x_n}\) and let \({\bar x_{n + 1}}\) and \(s_{n + 1}^2\) denote these quantities when an additional observation \({x_{n + 1}}\) is added to the sample.

a. Show how\({\bar x_{n + 1}}\)can be computed from\({\bar x_n}\)and\({x_{n + 1}}\).

b. Show that

\(ns_{n + 1}^2 = \left( {n - 1} \right)s_n^2 + \frac{n}{{n + 1}}{\left( {{x_{n + 1}} - {{\bar x}_n}} \right)^2}\)

so that\(s_{n + 1}^2\)can be computed from\({x_{n + 1}}\),\({\bar x_n}\), and\(s_n^2\).

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 16th strand results in an elongation value of 11.8. What are the values of the sample mean and sample standard deviation for all 16 elongation observations?

Short answer

a.\({\bar x_{n + 1}}\)can be computed from\({\bar x_n}\)and\({x_{n + 1}}\).

b.\(ns_{n + 1}^2 = \left( {n - 1} \right)s_n^2 + \frac{n}{{n + 1}}{\left( {{x_{n + 1}} - \bar x} \right)^2}\).

c.

The sample mean for all 16 elongation observations is 12.53 mm.

The sample standard deviation for all 16 elongation observations is 0.532mm.

Step-by-step solution

Given information

A sample of size n is \({x_1},{x_2},...,{x_n}\).

\({\bar x_n}\) and \(s_n^2\) denote the sample mean and variance.

\({\bar x_{n + 1}}\) and \(s_{n + 1}^2\) represents these quantities when an additional observation \({x_{n + 1}}\) is added to the sample.

Compute the sample mean of additional observation

a.

We know,

\(\begin{array}{c}\sum\limits_{i = 1}^{n + 1} {{x_i}} &=& \sum\limits_{i = 1}^n {{x_i}} + {x_{n + 1}}\\ &=& n{{\bar x}_n} + {x_{n + 1}}\\\frac{{\sum\limits_{i = 1}^{n + 1} {{x_i}} }}{{n + 1}} &=& \frac{{n{{\bar x}_n} + {x_{n + 1}}}}{{n + 1}}\;\;\;\;\;\;\left( {{\rm{Dividing}}\;{\rm{by}}\;n + 1} \right)\\{{\bar x}_{n + 1}} &=& \frac{{n{{\bar x}_n} + {x_{n + 1}}}}{{n + 1}}\end{array}\)

Hence, showed.

Computing the sample variance of additional observation

b.

It is known that,

\(\begin{array}{c}ns_{n + 1}^2 &=& \sum\limits_{i = 1}^{n + 1} {{{\left( {{x_i} - {{\bar x}_{n + 1}}} \right)}^2}} \\ &=& \sum\limits_{i = 1}^{n + 1} {{{\left( {{x_i} - \frac{{n{{\bar x}_n} + {x_{n + 1}}}}{{n + 1}}} \right)}^2}} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left( {{\rm{Referring}}\;{\rm{to}}\;{\rm{part}}\;{\rm{a}}} \right)\\ &=& \sum\limits_{i = 1}^{n + 1} {x_i^2 - } {\left( {n + 1} \right)^{ - 2}}{x_{n + 1}}\\ &=& \sum\limits_{i = 1}^{n + 1} {x_i^2 - } nx_n^{ - 2} + x_{n + 1}^2 + nx_n^{ - 2} - \left( {n + 1} \right)x_{n + 1}^{ - 2}\\ &=& \left( {n - 1} \right)s_n^2 + \left\{ {x_{n + 1}^2 + nx_n^{ - 2} - \left( {n + 1} \right)x_{n + 1}^{ - 2}} \right\}\\ &=& \left( {n - 1} \right)s_n^2 + \left\{ {x_{n + 1}^2 + nx_n^{ - 2} - nx_{n + 1}^{ - 2} - x_{n + 1}^{ - 2}} \right\}\\ &=& \left( {n - 1} \right)s_n^2 + {\left( {{x_{n + 1}} + nx_n^{ - 1} - nx_{n + 1}^{ - 1} - x_{n + 1}^{ - 1}} \right)^2}\\ &=& \left( {n - 1} \right)s_n^2 + \frac{n}{{n + 1}}{\left( {{x_{n + 1}} - \bar x} \right)^2}\end{array}\)

Therefore,

\(ns_{n + 1}^2 = \left( {n - 1} \right)s_n^2 + \frac{n}{{n + 1}}{\left( {{x_{n + 1}} - \bar x} \right)^2}\).

Computing the sample mean and sample standard deviation

c.

The sample mean thread elongation of 15 strands is 12.58 mm.

The sample standard deviation of 15 strands is 0.512 mm.

The value of 16^th strand elongation value is 11.8.

Referring to parts (a) and (b),

The sample mean for all 16 elongation observations is computed as,

\(\begin{array}{c}{{\bar x}_{15 + 1}} &=& \frac{{15\left( {12.58} \right) + 11.8}}{{16}}\\ &=& \frac{{200.5}}{{16}}\\ &=& 12.53\end{array}\)

Thus, the sample mean for all 16 elongation observations is 12.53 mm.

The sample variance for all 16 elongation observations is computed as,

\(\begin{array}{c}s_{n + 1}^2 &=& \frac{{14}}{{15}}{\left( {0.512} \right)^2} + \frac{{{{\left( {11.8 - 12.58} \right)}^2}}}{{16}}\\ &=& 0.245 + 0.88\\ &=& 0.238\end{array}\)

The sample standard deviation for all 16 elongation observations is computed as,

\(\begin{array}{c}{s_{n + 1}} &=& \sqrt {s_{n + 1}^2} \\ &=& \sqrt {0.238} \\ &=& 0.532\end{array}\)

Therefore, the sample standard deviation for all 16 elongation observations is 0.532mm.