β€œν™•λ₯ κ³Ό 톡계(MATH230)” μˆ˜μ—…μ—μ„œ 배운 것과 κ³΅λΆ€ν•œ 것을 μ •λ¦¬ν•œ ν¬μŠ€νŠΈμž…λ‹ˆλ‹€. 전체 ν¬μŠ€νŠΈλŠ” Probability and Statisticsμ—μ„œ ν™•μΈν•˜μ‹€ 수 μžˆμŠ΅λ‹ˆλ‹€ 🎲

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β€œν™•λ₯ κ³Ό 톡계(MATH230)” μˆ˜μ—…μ—μ„œ 배운 것과 κ³΅λΆ€ν•œ 것을 μ •λ¦¬ν•œ ν¬μŠ€νŠΈμž…λ‹ˆλ‹€. 전체 ν¬μŠ€νŠΈλŠ” Probability and Statisticsμ—μ„œ ν™•μΈν•˜μ‹€ 수 μžˆμŠ΅λ‹ˆλ‹€ 🎲

Choice of Sample Size

μ‹€μ „μ—μ„œλŠ” μ‹€ν—˜(experiment)λ₯Ό μˆ˜ν–‰ν•˜κΈ° 전에 주어진 significance level $\alpha$ μ•„λž˜μ—μ„œ μ μ ˆν•œ κ²€μ •λ ₯을 κ°–λŠ” sample sizeλ₯Ό 미리 μ„€μ •ν•œ 후에 μ‹€ν—˜μ„ μˆ˜ν–‰ν•œλ‹€! 이 과정은 data-taking process 이전이라면, 반.λ“œ.μ‹œ. μˆ˜ν–‰ν•΄μ•Ό ν•˜λŠ” 과정이닀!

쒋은 κ²€μ •λ ₯을 μ–»κΈ° μœ„ν•΄ μˆ˜ν–‰ν•˜λŠ” β€œμƒ˜ν”Œμ˜ μˆ˜β€λ₯Ό κ²°μ •ν•˜λŠ” 과정은 $\alpha$ κ°’κ³Ό $H_1: \mu = \mu_1$의 값을 κ³ μ •ν•˜κ³  μˆ˜ν–‰ν•œλ‹€.

μ΄λ•Œ, <κ²€μ •λ ₯>은 μ•„λž˜μ™€ κ°™λ‹€.

\[1 - \beta = P(\text{rejeect} \; H_0 \mid H_1 \; \text{is true})= P(\bar{X} > a \;\; \text{when} \;\; \mu = \mu_0 + \delta)\]

μ΄λ•Œ, $\beta$λŠ” T2 Errorλ‹€!!

1. Set Hypothesis

  • $H_0$: $\mu=190$ (cm)
  • $H_1$: $\mu=195$ (cm)

2. we want

  • $\alpha = 0.05$
  • $1 - \beta \ge 0.9$

3. Evaluate T1 Error

\[\begin{aligned} \alpha &= P(\text{rejeect} \; H_0 \mid \mu = 190) \\ &= P \left( \frac{\bar{X} - \mu_0}{\sigma/\sqrt{n}} > z_{\alpha} \right) \\ \end{aligned}\]

μœ„μ˜ 식은 μ–΄λ–€ $n$을 μ„ νƒν•˜λ”λΌλ„ 항상 참인 λͺ…μ œλ‹€!

4. Evaluate T2 Error

$1 - \beta$ = (power at $\mu = \mu_1$) $\ge 0.9$.

\[1 - \beta = P \left( \text{reject}\; H_0 \mid \mu = \mu_1 \right) = P \left( \frac{\bar{X} - \mu_0}{\sigma/\sqrt{n}} > z_{\alpha} \mid \mu = \mu_1 \right) \ge 0.9\]

Now, let’s find $n$ which guarantees the eq. of (3) and (4).

\[\begin{aligned} P \left( \frac{\bar{X} - \mu_0}{\sigma/\sqrt{n}} < z_{\alpha} \mid \mu = \mu_1 \right) &\le \beta \\ P \left( \frac{\bar{X} - \mu_1 + \mu_1 - \mu_0}{\sigma/\sqrt{n}} < z_{\alpha} \mid \mu = \mu_1 \right) &\le \beta \\ P \left( z < z_{\alpha} - \frac{\mu_1 - \mu_0}{\sigma/\sqrt{n}} \right) &\le \beta \end{aligned}\]

μ΄λ•Œ, $\mu_1 > \mu_0$ and $n$ is large,

\[z_{\alpha} - \frac{\mu_1 - \mu_0}{\sigma/\sqrt{n}} < 0\]

More specifically,

\[z_{\alpha} - \frac{\mu_1 - \mu_0}{\sigma/\sqrt{n}} = - z_{\beta}\]

Then, if we solve the above inequality, then we get a inequality for sample size $n$!

\[n \ge \left( \frac{(z_\alpha + z_\beta) \sigma }{\mu_1 - \mu_0} \right)^2\]

κ΅μž¬μ—μ„œλŠ” μœ„μ˜ 상황을 μ•„λž˜μ˜ 그림처럼 ν‘œν˜„ν•˜κ³  μžˆλ‹€!


πŸ’₯ (two-sided case) If $H_1$ is a form of $H_1: \mu \ne \mu_0$ at the level $\alpha$, and we want the power at $\mu = \mu_1$ to be at least $1 - \beta$?

이 κ²½μš°μ—λŠ” 식이

\[n \ge \left( \frac{(z_{\alpha/2} + z_\beta) \sigma }{\mu_1 - \mu_0} \right)^2\]

κ°€ λœλ‹€!


μ΄μ–΄μ§€λŠ” ν¬μŠ€νŠΈμ—μ„œλŠ” <Proportion>κ³Ό <Variance>의 검정에 λŒ€ν•΄ μ‚΄νŽ΄λ³Έλ‹€!! πŸ˜†

πŸ‘‰ Proportion Test πŸ‘‰ Variance Test