t-tests

• the population is normally distributed
• the sample size is large enogh (e.g. $n \ge 15$),
• do not use this test if there are outlier or the population is very skewed
  • skewness can be ignored if $n \ge 40$
• the population standard deviation $\sigma$ is unkown

Note

• An estimator is a r.v.
• Variable without subscption such as $X$, $Y$ denote true desitribution

Definition

• \(X, X_{1}, \ldots, X_{N}\),
  • r.v.
• \(Y, Y_{1}, \ldots, Y_{M}\),
  • r.v.
• \(\displaystyle \bar{X}_{N} := \sum_{i=1}^{N} \frac{ X_{i} }{ N }\),
  • sample mean
• \(\displaystyle V_{N}(X_{1}, \ldots, X_{N}) := \frac{ \sum_{i=1}^{N} (X_{i} - \bar{X}_{N})^{2} }{ N - 1 } \label{def_sample_variance}\),
  • unbiased estimator of variance
• \(\displaystyle V_{N}^{S} := \frac{ \sum_{i=1}^{N} (X_{i} - \bar{X}_{N})^{2} }{ N }\),
  • sample variance

If $X_{1}, \ldots, X_{N}$ is I.I.D. sequence of $X$, the follwoing statement holds true.

\[\begin{eqnarray} \mathrm{E}(\bar{X}_{N}) &= & \mathrm{E}(X), \nonumber \\ \mathrm{Var}(\bar{X}_{N}) & = & \frac{ \mathrm{Var}(X) }{ N }, \nonumber \end{eqnarray}\]

$N$ is called sample size. The more sample size, the less sample variance.

one sample t-tests for population mean

Statistical test for mean of one normal r.v.

• True distribution: $X \sim \mathrm{N}(\mu, \sigma)$
• The value of $\sigma$ is unkown
• the sample size is large enough that the mean is normally distributed
  • $n \ge 15$,
• skew can be ignored if $n \ge 40$,
• the population standard deviation $\sigma$ is unkown
• If $n \ge 30$, $t$ distribution can be approximated with the normal distribution
• In practice, $\sigma$ is rarely known

Example

• $\mu_{0} \in \mathbb{R}$
  • mean 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 one-sample t-test for population mean are as follows;

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

\[\begin{eqnarray} t & = & \frac{ \bar{x} - \mu_{0} }{ \frac{s}{\sqrt{n}} } \end{eqnarray}\]

• (3) Compute the $p$ value
  • $T$ follows t distribution with $n - 1$ degree of freedom
  • (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 $H_{0}$,

Lemma 5

\[\sum_{i=1}^{N} (X_{i} - \bar{X}_{N}) = 0\]

proof.

Easy to check.

$\Box$

Theorem 19

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

Then

\[\frac{ (N - 1)V_{N}(X_{1}, \ldots, X_{N}) }{ \sigma^{2} } \sim \chi^{2}(N-1) .\]

proof.

Firstly, we have

\[\frac{ (N - 1)V_{N}(X_{1}, \ldots, X_{N}) }{ \sigma^{2} } \sim \chi^{2}(N-1) = \frac{ 1 }{ \sigma^{2} } \sum_{i=1}^{N} (X_{i} - \bar{X}_{N})^{2} .\]

Hence we will evaluate the distribution of the right hand. We can obvserve that

\[\begin{eqnarray} \frac{1}{\sigma^{2}} \sum_{i=1}^{N} (X_{i} - \mu)^{2} & = & \frac{1}{\sigma^{2}} \sum_{i=1}^{N} \left( (X_{i} - \bar{X}_{N}) + (\bar{X}_{N} - \mu) \right)^{2} \nonumber \\ & = & \frac{1}{\sigma^{2}} \sum_{i=1}^{N} \left( (X_{i} - \bar{X}_{N})^{2} + 2 (X_{i} - \bar{X}_{N}) (\bar{X}_{N} - \mu) + (\bar{X}_{N} - \mu)^{2} \right) \nonumber \\ & = & \frac{1}{\sigma^{2}} \sum_{i=1}^{N} (X_{i} - \bar{X}_{N})^{2} + 2 \frac{1}{\sigma^{2}} (\bar{X}_{N} - \mu) \sum_{i=1}^{N} (X_{i} - \bar{X}_{N}) + \frac{1}{\sigma^{2}} \sum_{i=1}^{N} (\bar{X}_{N} - \mu)^{2} \nonumber \\ & = & \frac{1}{\sigma^{2}} \sum_{i=1}^{N} (X_{i} - \bar{X}_{N})^{2} + \frac{1}{\sigma^{2}} \sum_{i=1}^{N} (\bar{X}_{N} - \mu)^{2} \nonumber \\ & = & \frac{(N-1)}{\sigma^{2}} V_{N}(X_{1}, \ldots, X_{N}) + \frac{N}{\sigma^{2}} (\bar{X}_{N} - \mu)^{2} \nonumber \\ & = & \frac{(N-1)}{\sigma^{2}} V_{N}(X_{1}, \ldots, X_{N}) + \left( \frac{ \bar{X}_{N} - \mu }{ \frac{ \sigma }{ \sqrt{N} } } \right)^{2} \nonumber \end{eqnarray} .\]

where 4th inequality is consequence of lemma 5. Now, let

\[\begin{eqnarray} U & := & \frac{1}{\sigma^{2}} \sum_{i=1}^{N} (X_{i} - \mu)^{2}, \nonumber \\ W & := & \frac{1}{\sigma^{2}} V_{N}(X_{1}, \ldots, X_{N}) \nonumber \\ V & := & \left( \frac{ \bar{X}_{N} - \mu }{ \frac{ \sigma }{ \sqrt{N} } } \right)^{2} . \end{eqnarray}\]

Then $W = U + V$.

We calculate the distribution of $W$. Let

\[Z_{i} := \frac{ X_{i} - \bar{X}_{N} }{ \sigma } .\]

Hence \(Z_{i} \sim \mathrm{N}(0, 1)\) and \(\{Z_{i}\}\) is independent. Since

\[\begin{eqnarray} \frac{ (N - 1)V_{N}(X_{1}, \ldots, X_{N}) }{ \sigma^{2} } & = & \frac{ (N - 1) }{ \sigma^{2} } \frac{ \sum_{i=1}^{N} (X_{i} - \bar{X}_{N})^{2} }{ N - 1 } \nonumber \\ & = & \frac{ \sum_{i=1}^{N} (X_{i} - \bar{X}_{N})^{2} }{ \sigma^{2} } \nonumber \\ & = & \sum_{i=1}^{N} \left( \frac{ (X_{i} - \bar{X}_{N}) }{ \sigma } \right)^{2} \nonumber \\ & = & \sum_{i=1}^{N} Z_{i}^{2}, \end{eqnarray}\]

we have $W \sim \chi^{2}(N)$. By theorem, $\sqrt{V} \sim \mathrm{N}(0, 1)$ and especially $V \sim \chi^{2}(1)$. Moreover, since $W - V = U$, we calculate moment generating function of both the side of the equation;

\[\begin{eqnarray} M_{U}(t) & = & \frac{ M_{W}(t) }{ M_{V}(t) } \nonumber \\ & = & \frac{ (1 - 2t)^{-N/2} }{ (1 - 2t)^{-1/2} } \nonumber \\ & = & (1 - 2t)^{-(n-1)/2} . \end{eqnarray}\]

Therefore, $U \sim \chi^{2}(N-1)$.

$\Box$

Thereom 20

• $X_{1}, \ldots, X_{N}$,
  • independent r.v.s
• \(X_{i} \sim \mathrm{N}(\mu, \sigma^{2})\),

Then

\[\frac{ \bar{X}_{N} - \mu }{ \sqrt{ \frac{ V_{N}(X_{1}, \dots, X_{N}) }{ N } } } \sim t(N-1) .\]

proof.

まず、

\[\begin{eqnarray} U & := & \frac{ (N-1) V_{N}(X_{1}, \dots, X_{N}) }{ \sigma^{2} } \nonumber \\ Z & := & \frac{ \bar{X}_{N} - \mu }{ \sqrt{ \frac{ \sigma^{2} }{ N } } } \nonumber \end{eqnarray}\]

とおくと、定理より$Z \sim \mathrm{N}(0, 1)$かつ$U \sim \chi^{2}(N-1)$である。

\[\begin{eqnarray} \frac{ \bar{X}_{N} - \mu }{ \sqrt{ \frac{ V_{N}(X_{1}, \ldots, X_{N}) }{ N } } } & = & \frac{1}{V_{N}(X_{1}, \ldots, X_{N})} \frac{ \bar{X}_{N} - \mu }{ \frac{ 1 }{ \sqrt{N} } } \sqrt{ \frac{ \sigma^{2} }{ \sigma^{2} } } \nonumber \\ & = & \sqrt{ \frac{\sigma^{2}}{V_{N}} } \frac{ \bar{X}_{N} - \mu }{ \frac{ \sigma^{2} }{ \sqrt{N} } } \nonumber \\ & = & \sqrt{ \frac{(N-1)\sigma^{2}}{(N-1)V_{N}(X_{1}, \ldots, X_{N})} } Z \nonumber \\ & = & \frac{ 1 }{ \sqrt{ \frac{ (N-1)V_{N}(X_{1}, \ldots, X_{N}) }{ (N-1)\sigma^{2} } } } Z \nonumber \\ & = & \frac{ Z }{ \sqrt{ \frac{ U }{ (N-1) } } } \nonumber \end{eqnarray}\]

となる。

$\Box$

paired samples t-test for the population mean of paired samples

A test for the difference of two random variables with normal distribution whose variance is equal.

• True distribution: $X \sim \mathrm{N}(\mu_{X}, \sigma)$
• True distribution: $Y \sim \mathrm{N}(\mu_{Y}, \sigma)$
• two samples are paired
  • Here paired means that $X$ and $Y$ has same sample size $n$
• The value of $\sigma$ is uknown but $X$ and $Y$ has same $\sigma$
• both normal distribution
• both simze sizes are large enough
  • $n \ge 15$,

Example

• $D \in \mathbb{R}$,
  • the difference mean
• $n \in \mathbb{N}$,
  • the sample size
• $X_{1}, \ldots, X_{n}$,
  • i.i.d. of $X$
• $Y_{1}, \ldots, Y_{n}$,
  • i.i.d. of $Y$
• $D_{1} := X_{1} - Y_{1}, \ldots, D_{n} := X_{n} - Y_{n}$,
  • i.i.d. of $D := X - Y$,
• \(s_{d} := \mathrm{Var}_{n}(D_{1}, \ldots, D_{N})(\omega)\),
  • sample standard deviation of $D$
• $\alpha \in (0, 1)$,
  • significance level
• (1) State the hypothesis
  • null hypothesis
    • $H_{0}:\mu_{D} = D $
  • alternative hypothesis
    • (a) $H_{0}:\mu_{D} \neq D $
    • (b) $H_{0}:\mu_{D} > D $
    • (c) $H_{0}:\mu_{D} < D $
• (2) compute the test statistic

\[\begin{eqnarray} t = \frac{ \bar{d} - D }{ \frac{s_{d}}{\sqrt{n}} } \end{eqnarray}\]

• (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 $H_{0}$,

Theorem21

• \(X_{1}, \ldots, X_{N}\),
  • normal distribution
• \(Y_{1}, \ldots, Y_{N}\),
  • normal distribution

Moreover, for each $i$, $X_{i}$ and $Y_{i}$ are independent. Then

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

where

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

proof.

Since $D_{i}$ follows normal distribution, we can apply theorem 20 directly.

$\Box$

Two sample t-test for population means (equal variances)

Test for means of two random variables with normal distribution whose variances are equal.

• True distribution: $X \sim \mathrm{N}(\mu_{X}, \sigma)$
• True distribution: $Y \sim \mathrm{N}(\mu_{Y}, \sigma)$
• $X$, $Y$ are independent
• $n$ is the sample size of $X$
• $m$ is the sample size of $Y$
• Two sample means that the sample size can be different
• $\sigma$ are unkown but euqal to the other
  • This assumption is not realistic.

In practical, $D = 0$.

Example

• $D \in \mathbb{R}$,
  • the difference of means to be tested
• $n \in \mathbb{N}$,
  • the sample size of $X$
• $m \in \mathbb{N}$,
  • the sample size of $Y$
• $X_{1}, \ldots, X_{N}$,
  • i.i.d. of $X$
• $Y_{1}, \ldots, Y_{M}$,
  • i.i.d. of $Y$
• $\bar{X}_{N}$,
  • estimator of mean of $X$,
• $\bar{Y}_{M}$,
  • estimator of mean of $Y$,
• $\mu_{X} := \bar{X}_{N}(\omega)$,
• $\mu_{Y} := \bar{Y}_{M}(\omega)$,
• \(S_{X} := \mathrm{Var}_{N}(X_{1}, \ldots, X_{N})\),
  • sample standard deviation of $X$
• \(S_{Y} := \mathrm{Var}_{M}(Y_{1}, \ldots, Y_{M})\),
  • sample standard deviation of $Y$
• \(s_{X} := S_{X}(\omega)\),
• \(s_{Y} := S_{Y}(\omega)\),
• $\alpha \in (0, 1)$,
  • significance level

The steps for two-sample t-tests for population mean are as follows;

• (1) State the hypothesis
  • null hypothesis
    • $H_{0}:\mu_{X} - \mu_{Y} = D$,
  • alternative hypothesis
    • (a) $H_{A}:\mu_{X} - \mu_{Y} \neq D $
    • (b) $H_{A}:\mu_{X} - \mu_{Y} > D $
    • (c) $H_{A}:\mu_{X} - \mu_{Y} < D $
• (2) compute the test statistic

\[\begin{eqnarray} t & = & \frac{ (\mu_{X} - \mu_{Y}) - D }{ \sqrt{ s_{P}^{2} \left( \frac{1}{N} + \frac{1}{M} \right) } } \nonumber \\ s_{P}^{2} & := & \frac{ (N - 1)^{2}s_{X}^{2} + (M - 1)^{2}s_{Y}^{2} }{ N + M - 2 } \nonumber \end{eqnarray}\]

• (3) Compute $p$ value
  • $T \sim \mathrm{t}(n m - 2)$,
  • (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 $H_{0}$,

Lemma 6

Suppose that \(\{ X_{i} \}_{i=1}^{N}\) and \(\{ Y_{i}\}_{i=1}^{M}\) are independent.

• \(X_{i} \sim \mathrm{N}(\mu_{X}, \sigma^{2})\),
  • i.i.d.
• \(Y_{i} \sim \mathrm{N}(\mu_{Y}, \sigma^{2})\),
  • i.i.d.

Then

\[\bar{X}_{N} - \bar{Y}_{M} \sim \mathrm{N} \left( \mu_{X} - \mu_{Y}, \sigma^{2} \left( \frac{1}{N} + \frac{1}{M} \right) \right).\]

proof.

By Lemma3, it follows that

\[\begin{eqnarray} \bar{X}_{N} & \sim & \mathrm{N}(\mu_{X}, \frac{\sigma^{2}}{N}), \nonumber \\ \bar{Y}_{M} & \sim & \mathrm{N}(\mu_{Y}, \frac{\sigma^{2}}{M}), \nonumber \end{eqnarray} .\]

Since the sum of the independet normally distributed r.v.s follows normal distribution, we have

\[\begin{eqnarray} \bar{X}_{N} - \bar{Y}_{M} & = & \bar{X}_{N} + (-1)\bar{Y}_{M} \nonumber \\ & \sim & \mathrm{N} \left( \mu_{X} - \mu_{Y}, \frac{ \sigma^{2} }{ N } + (-1)^{2} \frac{ \sigma^{2} }{ M } \right) \nonumber \end{eqnarray} .\]

$\Box$

Lemma 7

• \(X \sim \mathrm{N}(\mu_{X}, \sigma^{2})\),
• \(Y \sim \mathrm{N}(\mu_{Y}, \sigma^{2})\),
• \(X_{1}, \ldots, X_{N}\),
  • i.i.d. of $X$
• \(Y_{1}, \ldots, Y_{M}\),
  • i.i.d. of $Y$
• \(\{X_{i} \}\) and \(\{Y_{i}\}\) are independent

Then

\[\frac{ \bar{X}_{N} - \bar{Y}_{M} - (\mu_{X} - \mu_{Y}) }{ \sigma \sqrt{ \frac{ \frac{ 1 }{ N } }{ \frac{1}{M} } } } \sim \mathrm{N}(0, 1)\]

proof.

By lemma6, it follows that

\[\bar{X}_{N} - \bar{X}_{M} \sim \mathrm{N} \left( \mu_{X} - \mu_{Y}, \sigma^{2} \left( \frac{1}{N} + \frac{1}{M} \right) \right) .\]

Thus, the statement holds.

$\Box$

Theorem 22

• \(X \sim \mathrm{N}(\mu_{X}, \sigma)\),
• \(Y \sim \mathrm{N}(\mu_{Y}, \sigma)\),
• \(X_{1}, \ldots, X_{N}\),
  • i.i.d. of $X$
• \(Y_{1}, \ldots, Y_{M}\),
  • i.i.d. of $Y$
• \(\{X_{i} \}\) and \(\{Y_{i}\}\) are independent

Then

\[T := \frac{ (\bar{X}_{N} - \bar{Y}_{M}) - (\mu_{X} - \mu_{Y}) }{ \sqrt{ V_{N}^{P} \left( \frac{1}{N} + \frac{1}{M} \right) } } \sim t(N + M - 2)\]

where

\[\begin{eqnarray} V^{P} & := & \frac{ (N-1)V_{N}(X_{1}, \ldots, X_{N}) + (M - 1)V_{M}(Y_{1}, \ldots, Y_{M}) }{ N + M - 2 } . \nonumber \end{eqnarray}\]

proof.

By the lemma 7, we have

\[\begin{eqnarray} Z & := & \frac{ (\bar{X}_{N} - \bar{Y}_{M}) - (\mu_{X} - \mu_{Y}) }{ \sigma \sqrt{ \frac{1}{N} + \frac{1}{M} } }, \nonumber \\ & \sim & \mathrm{N}(0, 1) \nonumber \\ U_{1} & := & \frac{ (N - 1) V_{N}(X_{1}, \ldots, X_{N}) }{ \sigma^{2} } \nonumber \\ & \sim & \chi^{2}(N - 1) \nonumber \\ U_{2} & := & \frac{ (M-1) V_{M}(Y_{1}, \ldots, Y_{M}) }{ \sigma^{2} } \nonumber \\ & \sim & \chi^{2}(M - 1) \nonumber \\ U & := & U_{1} + U_{2} \nonumber \\ & \sim & \chi^{2}(N + M - 2) . \nonumber \end{eqnarray}\]

$U_{i}$ follows $\chi^{2}$ distribution. Hence the sum of the$\chi^{2}$ distributions also follows $\chi^{2}$ distribution.

$T$ can be represented with $U$ and $Z$.

\[\begin{eqnarray} T & = & \frac{ (\bar{X}_{N} - \bar{Y}_{M}) - (\mu_{X} - \mu_{Y}) }{ \sqrt{ \frac{ (N-1)V_{N}(X_{1}, \ldots, X_{N}) + (M - 1)V_{M}(Y_{1}, \ldots, Y_{M}) }{ N + M - 2 } \left( \frac{1}{N} + \frac{1}{M} \right) } } \frac{ \sigma }{ \sigma } \nonumber \\ & = & \frac{ (\bar{X}_{N} - \bar{Y}_{M}) - (\mu_{X} - \mu_{Y}) }{ \sigma \sqrt{ \left( \frac{1}{N} + \frac{1}{M} \right) } } \frac{ \sigma }{ \sqrt{ \frac{ (N-1)V_{N}(X_{1}, \ldots, X_{N}) + (M - 1)V_{M}(Y_{1}, \ldots, Y_{M}) }{ N + M - 2 } } } \nonumber \\ & = & Z \sqrt{ \frac{ 1 }{ \frac{ \frac{ (N-1)V_{N}(X_{1}, \ldots, X_{N}) }{ \sigma^{2} } + \frac{ (M - 1)V_{M}(Y_{1}, \ldots, Y_{M}) }{ \sigma^{2} } }{ N + M - 2 } } } \nonumber \\ & = & \frac{ Z }{ \sqrt{ \frac{ U }{ N + M - 2 } } } \nonumber \end{eqnarray}\]

$\Box$

Two sample t-test for population means (unequal variances)

A test for two random variables with Normal distirbution whose variance is unequal. This test uses the approximated distirbution for test statistics. Null-hypothesis can be rejected in terms of the approximated distribution.

• True distribution: $X \sim \mathrm{N}(\mu_{X}, \sigma_{X}^{2})$
• True distribution: $Y \sim \mathrm{N}(\mu_{Y}, \sigma_{Y}^{2})$
• $N$ is the sample size of $X$
• $M$ is the sample size of $Y$
• Two sample means that the sample size can be different
• $\sigma_{X}$ and $\sigma_{Y}$ are unkown but are not equal.
• $\mu_{X} - \mu_{Y} = D$ is null hypothesis

Example

• $D \in \mathbb{R}$,
  • the difference of means
• $N \in \mathbb{N}$,
  • the sample size of $X$,
• $M \in \mathbb{N}$,
  • the sample size of $Y$,
• $X_{1}, \ldots, X_{N}$,
  • i.i.d. of $X$
• $Y_{1}, \ldots, Y_{M}$,
  • i.i.d. of $Y$
• \(\bar{x}_{N} := \bar{X}_{N}(\omega)\),
• \(\bar{y}_{M} := \bar{Y}_{M}(\omega)\),
• \(S_{X} := \mathrm{Var}_{n}(X_{1}, \ldots, X_{N})\),
  • sample standard deviation of $X$
• \(S_{Y} := \mathrm{Var}_{m}(Y_{1}, \ldots, Y_{M})\),
  • sample standard deviation of $Y$
• \(s_{X} := S_{X}(\omega)\),
• \(s_{Y} := S_{Y}(\omega)\),
• $\alpha \in (0, 1)$,
  • significance level

The steps for two sample t-tests are as follows;

• (1) State the hypothesis
  • null hypothesis
    • \(H_{0}:\bar{x}_{N} - \bar{y}_{M} = D\),
  • alternative hypothesis
    • (a) \(H_{A}:\bar{x}_{N} - \bar{y}_{M} \neq D\),
    • (b) \(H_{A}:\bar{x}_{N} - \bar{y}_{M} > D\),
    • (c) \(H_{A}:\bar{x}_{N} - \bar{y}_{M} < D\),
• (2) compute the test statistic

\[\begin{eqnarray} t & = & \frac{ (\bar{x}_{N} - \bar{y}_{M}) - D }{ \sqrt{ \frac{s_{X}^{2}}{N} + \frac{s_{Y}^{2}}{M} } } \nonumber \\ df & := & \frac{ \left( \frac{s_{X}^{2}}{N} + \frac{s_{y}^{2}}{M} \right)^{2} }{ \frac{ s_{X}^{4} }{ N^{2}(N - 1) } + \frac{ s_{Y}^{4} }{ M^{2}(M - 1) } } \nonumber \end{eqnarray}\]

• (3) Compute $p$ value
  • $T \sim \mathrm{t}(df)$,
  • (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

Lemma 8

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

Then

\[\begin{equation} \mathrm{Var} \left[ V_{N}(X_{1}, \ldots, X_{N}) \right] = \frac{ 2 \sigma^{4} }{ N - 1 } \end{equation}\]

proof.

By Theorem 19, it follows that

\[\begin{equation} \frac{ (N - 1) V_{N}(X_{1} ,\ldots, X_{N}) }{ \sigma^{2} } \sim \chi_{N-1}^{2} \end{equation}\]

Hence

\[\mathrm{Var} \left[ \frac{ (N - 1) V_{N}(X_{1} ,\ldots, X_{N}) }{ \sigma^{2} } \right] = 2 (N - 1) .\]

On the other hand,

\[\begin{eqnarray} \mathrm{Var} \left[ \frac{ (N - 1) V_{N}(X_{1} ,\ldots, X_{N}) }{ \sigma^{2} } \right] & = & \frac{ (N - 1)^{2} }{ \sigma^{4} } \mathrm{Var} \left[ V_{N}(X_{1} ,\ldots, X_{N}) \right] \end{eqnarray}\]

Thus,

\[\begin{eqnarray} & & \frac{ (N - 1)^{2} }{ \sigma^{4} } \mathrm{Var} \left[ V_{N}(X_{1} ,\ldots, X_{N}) \right] = 2 (N - 1) \nonumber \\ & \Leftrightarrow & \mathrm{Var} \left[ V_{N}(X_{1} ,\ldots, X_{N}) \right] = \frac{ 2 \sigma^{4} }{ (N - 1) } \nonumber \end{eqnarray}\]

$\Box$

Theorem 23

\[\hat{\beta} \frac{ \frac{ V_{N}(X_{1}, \ldots, X_{N}) }{ N } + \frac{ V_{M}(Y_{1}, \ldots, Y_{M}) }{ M } }{ \frac{ \sigma^{2}_{X} }{ N } + \frac{ \sigma_{Y}^{2} }{ M } } \overset{\mathrm{approx}}{\sim} \chi_{\hat{\beta}}^{2},\]

where

\[\begin{equation} \hat{\beta} := \frac{ \left( \frac{ V_{N}(X_{1}, \ldots, X_{N}) }{ N } + \frac{ V_{M}(Y_{1}, \ldots, Y_{M}) }{ M } \right)^{2} }{ \frac{ V_{N}(X_{1}, \ldots, X_{N})^{2} }{ N(N-1) } + \frac{ V_{M}(Y_{1}, \ldots, Y_{N})^{2} }{ M(M-1) } } . \end{equation}\]

proof.

$\Box$

Theorem 24

• $X \sim \mathrm{N}(\mu_{X}, \sigma_{X}^{2})$,
• $Y \sim \mathrm{N}(\mu_{Y}, \sigma_{Y}^{2})$,
• \(X_{1}, \ldots, X_{N}\),
  • i.i.d. of $X$
• \(Y_{1}, \ldots, Y_{M}\),
  • i.i.d. of $y$

Then

\[T := \frac{ (\bar{X}_{N} - \bar{Y}_{M}) - (\mu_{X} - \mu_{Y}) }{ \sqrt{ \frac{ V_{N}(X_{1}, \ldots, X_{N}) }{ N } + \frac{ V_{M}(Y_{1}, \ldots, Y_{M}) }{ M } } } \overset{\mathrm{approx}}{\sim} t(v)\]

where

\[v := \frac{ \left( \frac{ V_{N}(X_{1}, \ldots, X_{N}) }{ N } + \frac{ V_{M}(Y_{1}, \ldots, Y_{M}) }{ M } \right)^{2} }{ \frac{ V_{N}(X_{1}, \ldots, X_{N})^{2} }{ N^{2}(N - 1) } + \frac{ V_{M}(Y_{1}, \ldots, Y_{M})^{2} }{ M^{2}(M - 1) } } .\]

proof.

First, let

\[Z := \frac{ (\bar{X}_{N} - \bar{Y}_{M}) - (\mu_{X} - \mu_{Y}) }{ \sqrt{ \frac{ \sigma_{X}^{2} }{ N } + \frac{ \sigma_{Y}^{2} }{ M } } } .\]

$Z \sim \mathrm{N}(0, 1)$. By the previous theorem,

\[V := \hat{\beta} \frac{ \frac{ V_{N}(X_{1}, \ldots, X_{N}) }{ N } + \frac{ V_{M}(Y_{1}, \ldots, Y_{M}) }{ M } }{ \frac{ \sigma_{X}^{2} }{ N } + \frac{ \sigma_{Y}^{2} }{ M } }\]

follows \(V \overset{\mathrm{approx}}{\sim} \chi_{\hat{\beta}}^{2}\). Thus,

\[\begin{eqnarray} T & = & \frac{ (\bar{X}_{N} - \bar{Y}_{M}) - (\mu_{X} - \mu_{Y}) }{ \sqrt{ \frac{ V_{N}(X_{1}, \ldots, X_{N}) }{ N } + \frac{ V_{M}(Y_{1}, \ldots, Y_{M}) }{ M } } } \frac{ \hat{\beta} \sqrt{ \frac{ \sigma_{X}^{2} }{ N } + \frac{ \sigma_{Y}^{2} }{ M } } }{ \hat{\beta} \sqrt{ \frac{ \sigma_{X}^{2} }{ N } + \frac{ \sigma_{Y}^{2} }{ M } } } \nonumber \\ & = & \frac{ (\bar{X}_{N} - \bar{Y}_{M}) - (\mu_{X} - \mu_{Y}) }{ \sqrt{ \frac{ \sigma_{X}^{2} }{ N } + \frac{ \sigma_{Y}^{2} }{ M } } } \left( \frac{ \hat{\beta} \sqrt{ \frac{ V_{N}(X_{1}, \ldots, X_{N}) }{ N } + \frac{ V_{M}(Y_{1}, \ldots, Y_{M}) }{ M } } }{ \hat{\beta} \sqrt{ \frac{ \sigma_{X}^{2} }{ N } + \frac{ \sigma_{Y}^{2} }{ M } } } \right)^{-1} \nonumber \\ & = & \hat{\beta} \frac{ Z }{ \sqrt{ \frac{ V }{ \hat{\beta} } } } . \end{eqnarray}\]

$\Box$

Reference

• Student’s t-test - Wikipedia