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https://www.youtube.com/watch?v=vzShqT6Eer8&list=PLsri7w6p16vvQCo9pmuRNY_SYoOGB6bWM&index=4
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Case3:
"not enough sample", variance of 2 populations is same
subcase1:
2 populations have normal distribution shape or normal distribution-like shape
population1 follows $$$N(\mu_A,\sigma_A^2)$$$
population2 follows $$$N(\mu_B,\sigma_B^2)$$$
$$$\sigma_A^2 = \sigma_B^2 = \sigma^2$$$
* Since $$$\sigma_A^2 = \sigma_B^2 = \sigma^2$$$
you will predict common variance $$$\sigma_A^2$$$ and will use it
* But you don't know variance of 2 populations,
so you will use variance of 2 samples
$$$s_A^2 = s_B^2 = s^2$$$
* Above is called "pooled variance estimator"
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"pooled variance estimator"
"pooled variance estimator" is common variance of independent populations
in unbiased estimator form
"pooled variance estimator" is notated by $$$s_p^2$$$
$$$s_p^2 = \dfrac{\sum\limits_{i=1}^{N} (n_i-1)\times s_i^2}{\sum\limits_{i=1}^{N} (n_i-1)}$$$
$$$n_i$$$: number of sample from ith population
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Let's apply $$$s_p^2$$$ for population A and B
$$$s_p^2
= \dfrac{(n_A-1)\times s_A^2 + (n_B-1)\times s_B^2}{(n_A-1)+(n_B-1)} \\
= \dfrac{(n_A-1)\times s_A^2 + (n_B-1)\times s_B^2}{n_A+n_B-2}$$$
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Trusted region (lower bound and upper bound)
Trusted region wrt $$$\mu_A- \mu_B$$$ by using $$$100(1-\alpha)\%$$$
2 sided test : $$$\frac{\alpha}{2}$$$
$$$(\bar{x}_A-\bar{x}_B) - t_{(n_A+n_B -2,\frac{\alpha}{2})} \times s_p \sqrt{\frac{1}{n_A} + \frac{1}{n_B}}
\le \mu_A- \mu_B
\le (\bar{x}_A-\bar{x}_B) + t_{(n_A+n_B -2,\frac{\alpha}{2})} \times s_p \sqrt{\frac{1}{n_A} + \frac{1}{n_B}} $$$
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Let's establish hypothesis
2 sided test
$$$H_0:\mu_A-\mu_B=0$$$
$$$H_1:\mu_A-\mu_B\ne 0$$$
left sided test
$$$H_0:\mu_A-\mu_B=0$$$
$$$H_1:\mu_A-\mu_B\lt 0$$$
right sided test
$$$H_0:\mu_A-\mu_B=0$$$
$$$H_1:\mu_A-\mu_B\gt 0$$$
test statistics
$$$t_{n_A+n_B-2, \frac{\alpha}{2}} \\
= \dfrac{(\bar{x}_A-\bar{x}_B)-(\mu_A-\mu_B)}{s_p \times \sqrt{\frac{1}{n_A} + \frac{1}{n_B}}}$$$
Since $$$H_0:\mu_A-\mu_B=0$$$
$$$= \dfrac{(\bar{x}_A-\bar{x}_B)}{s_p \times \sqrt{\frac{1}{n_A} + \frac{1}{n_B}}}$$$
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Case4
"Not enough sample", "identical variance of 2 populations is not known"
$$$\bar{x}_A-\bar{x}_B$$$
$$$E(\bar{x}_A-\bar{x}_B) \\
= E(\bar{x}_A)-E(\bar{x}_B) \\
= \mu_A - \mu_B$$$
$$$s^2(\bar{x}_A-\bar{x}_B)
= s^2(\bar{x}_A) + s^2(\bar{x}_B) - 2Cov(\bar{x}_A-\bar{x}_B) \\
= \frac{s_{x_A}^2}{n_A} + \frac{s_{x_B}^2}{n_B}$$$
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Trusted region wrt $$$\mu_A-\mu_B$$$ by using $$$100(1-\alpha)\%$$$
$$$(\bar{x}_A-\bar{x}_B) - t_{(n_A+n_B-2,\frac{\alpha}{2})} \sqrt{\frac{s_A^2}{n_A} + \frac{s_B^2}{n_B}}
\le \mu_A-\mu_B
\le (\bar{x}_A-\bar{x}_B) + t_{(n_A+n_B-2,\frac{\alpha}{2})} \sqrt{\frac{s_A^2}{n_A} + \frac{s_B^2}{n_B}}$$$
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Let's establish hypothesis
2 sided test
$$$H_0:\mu_A-\mu_B=0$$$
$$$H_1:\mu_A-\mu_B\ne 0$$$
left sided test
$$$H_0:\mu_A-\mu_B=0$$$
$$$H_1:\mu_A-\mu_B\lt 0$$$
right sided test
$$$H_0:\mu_A-\mu_B=0$$$
$$$H_1:\mu_A-\mu_B\gt 0$$$
test statistics
$$$t_{(n_A+n_B-2, \frac{\alpha}{2})} \\
= \dfrac{(\bar{x}_A-\bar{x}_B)-(\mu_A-\mu_B)}{\sqrt{\frac{1}{n_A} + \frac{1}{n_B}}}$$$
Since $$$H_0:\mu_A-\mu_B=0$$$
$$$= \dfrac{(\bar{x}_A-\bar{x}_B)}{\sqrt{\frac{1}{n_A} + \frac{1}{n_B}}}$$$
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