Chi-square tests

Chi-Square test for the Population Variance

• True ditribution: $X \sim \mathrm{N}(\mu, \sigma^{2})$,
• $X_{1}, \ldots, X_{N}$,
  • i.i.d. of $X$
• $\mu$ is unknown

Example

• $\sigma_{0} \in \mathbb{R}_{>0}$
  • std deviation for null hypothesis
• $N \in \mathbb{N}$,
  • sample size
• $x_{i} := X_{i}(\omega)$,
• $\alpha \in (0, 1)$,
  • significance level
  • 0.05, 0.01 are commonly used

\[\begin{eqnarray} \bar{x} & := & \bar{X}_{N}(\omega) \nonumber \\ s & := & \sqrt{V_{n}(X_{1}, \ldots, X_{N})(\omega)} \nonumber \end{eqnarray}\]

The steps for chi-square test for population variance are as follos;

• (1) State the hypotheses:
  • null hypothesis
    • $H_{0}$: $\sigma^{2} = \sigma_{0}^{2}$,
  • alternative hypothesis
    • (a) $H_{A}$: $\sigma^{2} \neq \sigma_{0}^{2}$,
    • (b) $H_{A}$: $\sigma^{2} > \sigma_{0}^{2}$,
    • (c) $H_{A}$: $\sigma^{2} < \sigma_{0}^{2}$,
• (2) Compute the test statistics
  • $\bar{x}$
    • the sample mean
  • $s$
    • the sample standard deviation

\[\begin{eqnarray} y & = & \frac{ (N - 1)s^{2} }{ \sigma_{0}^{2} } \end{eqnarray}\]

• (3) Compute the $p$ value
  • $Y$: $\chi^{2}$ distribution with $N - 1$ degree of freedom
  • (a)
    • If $y$ is less than the median, \(p := 2P(Y \le y)\),
    • If $y$ is greater than the median, \(p := 2P(Y \ge y)\),
  • (b) $p := P(Y > y)$,
  • (c) $p := P(Y < y)$,
• (4)
  • If $p < \alpha$, reject $H_{0}$,
  • otherwise, fail to reject $H_{0}$,

Theory

Corollary 9.

• $X \sim \mathrm{N}(\mu, \sigma^{2})$,
• $X_{1}, \ldots, X_{N}$,
  • i.i.d. of $X$

Then

\[\begin{eqnarray} \frac{ (N - 1) V_{N}(X_{1}, \ldots, X_{N}) }{ \sigma^{2} } & \sim & \chi(N - 1) \nonumber \end{eqnarray}\]

proof.

By Theorem 19.

$\Box$

Chi-square test for goodness of fit

The test is also known as Pearson’s chi-squared test. This approximation known as Peason’s approximation.

• True ditribution: $X \sim f(p_{1}, \ldots, p_{m})$,
  • multinomial distribution
  • $f$: p.d.f. of multinomial distribution
  • $\sum_{j=1}^{m} p_{j} = 1$,
  • \(X \in \{1, \ldots, m\}\),
• $N \in \mathbb{N}$,
  • sample size
• $X_{1}, \ldots, X_{n}$,
  • $X$のi.i.d
• $x_{i} := X_{i}(\omega)$,
• All expected values are at least 5

Example

The steps for chi-square test for goodness of fit are as follows;

• (1) State the hypothesis
  • null hypothesis
    • $H_{0}:$ The data fits the proposed distribution
  • alternative hypothesis
    • $H_{A}:$ The data doesn’t fit the proposed distribution
• (2) compute the test statistic

\[\begin{eqnarray} j \in \{1, \ldots, m\}, \ u_{j} & = & \sum_{i :x_{i} = j} x_{i} \nonumber \\ w & = & \sum_{i=1}^{m} \frac{ (u_{j} - n p_{j})^{2} }{ Np_{j} } \nonumber \end{eqnarray}\]

• (3) Compute $p$ value
  • $T \sim \mathrm{t}(N-1)$,
  • (a) \(p := P(T \le |t| \cup T \ge |t|)\)
  • (b) $p := P(T > t)$
  • (c) $p := P(T < t)$
• (4)
  • If $p < \alpha$, reject $H_{0}$,
  • otherwise, fail to reject $H_{0}$,

Theory

Theorem21

\(X_{1}, \ldots, X_{N}\), \(Y_{1}, \ldots, Y_{N}\)が正規分布に従っているとする。 また、$X_{i}$と$Y_{i}$が各$i$について独立とする。 このとき、

\[Z := \frac{ \bar{D}_{N} - \delta }{ \sqrt{ \frac{ V_{N}(D_{1}, \ldots, D_{N}) }{ N } } } \sim t(N - 1)\]

は、 ここで、

• \(D_{i} := X_{i} - Y_{i}\),
• \(\bar{D}_{N} := \frac{\sum_{i=1}^{N} D_{i}}{N}\),
• \(\displaystyle V_{N}(D_{1}, \ldots, D_{N}) := \frac{ \sum_{i=1}^{N} (D_{i} - \bar{D}_{N})^{2} }{ N - 1 }\),

proof.

$\Box$

Proposition 3

• $X \sim f(p_{1}, \ldots, p_{m})$,
  • $f$ is multinomial distribution
  • $\sum_{j=1}^{m}p_{j} = 1$,
  • \(X \in \{1, \ldots, m\}\),
• $X_{1}, \ldots, X_{n}$,
  • I.I.D. of $X$
• $n \in \mathbb{N}$,
  • sample size
• $x_{j} := X_{j}(\omega)$

\[\begin{eqnarray} j = 1, \ldots, m, \ \bar{p}_{j} & := & \sum_{i:x_{i}=j} \frac{ 1 }{ n } \nonumber \\ W & := & \sum_{i=1}^{n} \frac{ (X_{i} - n \bar{p}_{i})^{2} }{ n \bar{p}_{i} } \nonumber \end{eqnarray}\]

Then $W$ has an approximate $\chi^{2}$ distribution with

• $r - m$ degree of freedom

proof.

$\Box$

Chi square tests for Independence

Chi square tests for Homegenity

Reference

• Pearson’s chi-squared test - Wikipedia