Moment coefficient of kurtosis for ungrouped data
Let $x_1, x_2,\cdots, x_n$ be $n$ observations. The mean of $X$ is denoted by $\overline{x}$ and is given by
$$ \begin{eqnarray*} \overline{x}& =\frac{1}{n}\sum_{i=1}^{n}x_i \end{eqnarray*} $$
Formula
The moment coefficient of kurtosis $\beta_2$ is defined as
$\beta_2=\dfrac{m_4}{m_2^2}$
The moment coefficient of kurtosis $\gamma_2$ is defined as
$\gamma_2=\beta_2-3$
where
$n$total number of observations$\overline{x}$sample mean$m_2 =\dfrac{1}{n}\sum_{i=1}^n (x_i-\overline{x})^2$is second sample central moment$m_4 =\dfrac{1}{n}\sum_{i=1}^n (x_i-\overline{x})^4$is fourth sample central moment
Example
The hourly earning (in dollars) of sample of 7 workers are :
26, 21, 24, 22, 25, 24, 23.
Compute coefficient of kurtosis based on moments.
Solution
The mean of $X$ is
$$ \begin{aligned} \overline{x} &=\frac{1}{n}\sum_{i=1}^n x_i\\ &=\frac{175}{7}\\ &=25 \text{ dollars} \end{aligned} $$
| $x$ | $(x-xb)$ | $(x-xb)^2$ | $(x-xb)^4$ | |
|---|---|---|---|---|
| 27 | 2 | 4 | 16 | |
| 27 | 2 | 4 | 16 | |
| 24 | -1 | 1 | 1 | |
| 26 | 1 | 1 | 1 | |
| 25 | 0 | 0 | 0 | |
| 24 | -1 | 1 | 1 | |
| 22 | -3 | 9 | 81 | |
| Total | 175 | 0 | 20 | 116 |
Second sample central moment
The second sample central moment is
$$ \begin{aligned} m_2 &=\frac{1}{n}\sum_{i=1}^n (x_i-\overline{x})^2\\ &=\frac{20}{7}\\ &=2.8571 \end{aligned} $$
Fourth sample central moment
The fourth sample central moment is
$$ \begin{aligned} m_4 &=\frac{1}{n}\sum_{i=1}^n (x_i-\overline{x})^4\\ &=\frac{116}{7}\\ &=16.5714 \end{aligned} $$
Coefficient of Kurtosis
The coefficient of kurtosis based on moments ($\beta_2$) is
$$ \begin{aligned} \beta_2 &=\frac{m_4}{m_2^2}\\ &=\frac{(16.5714)}{(2.8571)^2}\\ &=\frac{16.5714}{8.163}\\ &=2.0301 \end{aligned} $$
The coefficient of kurtosis based on moments ($\gamma_2$) is
$$ \begin{aligned} \gamma_2 &=\beta_2-3\\ &=2.0301 -3\\ &=-0.9699 \end{aligned} $$
As the value of $\gamma_2 < 0$, the data is $\text{platy-kurtic}$.
