Guessing with Confidence

Point estimators are valuable, but they may give slight errors.

Rather than specify an exact value, we can specify two values we expect flavor duration to lie between.

: As an example, you may want to choose a and b so that there’s a 95% chance of the interval containing the population mean. Finding the exact spot of a and b is the problem we are trying to solve.

The far side of each end, (a, b) is called a confidence interval.

Step 1: Choose your population statistic
If we go back to the work we did in the last chapter, then the sampling distribution of means has the following expectation and variance:

Step 2: Find its sampling distribution
샘플평균들의 분산은 ($Var(\overline{X})$) 모집단의 특성인데 (parameter), 이를 알 수는 없으므로 아래와 같이 샘플의 분산값을 ($s^{2}$) 사용하여 샘플평균들의 분포를 만든다.

위대한 풍선껌은 (Mighty Gumball) 100개의 풍선검을 샘플로 이용하여 단맛의 지속시간을 측정하고, 이 샘플의 분산값으로 (s²) 25를 얻었다. 이를 이용하여 샘플평균들의 (n=100일 때) 분포의 (distribution) 분산값을 예측해보면 0.25를 얻는다.

위를 일반화해서 생각해보면 $X \sim N(\mu, \sigma^{2})$이라고 할 때, 샘플의 숫자가 충분히 크다고 할 때 (n=100과 같이), $E(\overline{X})$ 값과 $Var(\overline{X})$ 값은 아래와 같다.

Step 3: Decide on the level of confidence
Step 4: Find the confidence limits

$$P(z_{a} < Z < z_{b}) = 0.95$$
$$P(Z < z_{a}) = 0.025$$
$$z_{a} = -1.96$$
$$P(Z > z_{b}) = 0.025$$
$$z_{b} = +1.96$$

\begin{eqnarray*} P\left(-1.96 < z < 1.96 \right) = 0.95 \\ P\left(-1.96 < \frac{\overline{X}-\mu}{0.5} < 1.96 \right) = 0.95 \end{eqnarray*}

\begin{eqnarray*} -1.96 < \frac{\overline{X}-\mu}{0.5} < 1.96 \\ \end{eqnarray*}

\begin{eqnarray} -1.96 & < & \frac{\overline{X}-\mu}{0.5} \nonumber \\ -1.96 * 0.5 & < & \overline{X}-\mu \nonumber \\ -0.98 + \mu & < & \overline{X} \nonumber \\ \mu & < & \overline{X} + 0.98 \\ \nonumber \\ \nonumber \\ \frac{\overline{X}-\mu}{0.5} & < & 1.96 \nonumber \\ \overline{X}-\mu & < & 1.96 * 0.5 \nonumber \\ \overline{X} & < & 0.98 + \mu \nonumber \\ \overline{X} - 0.98 & < & \mu \end{eqnarray}

(1)과 (2)에서

\begin{eqnarray*} \;\;\; \overline{X} - 0.98 < \mu < \overline{X} + 0.98 \end{eqnarray*}

\begin{eqnarray*} P(\overline{X} - 0.98 < \mu < \overline{X} + 0.98) = 0.95 \end{eqnarray*}

$\overline{X} = 62.7$ 이었으므로 $62.7 - 0.98$와 $62.7 + 0.98$이 구하는 공간 (interval). 즉,

$(61.72, 63.68)$

$$\left(p_{s} - c \sqrt{\frac{p_{s}*q_{s}}{n}}, \quad p_{s} + c \sqrt{\frac{p_{s}*q_{s}}{n}}\right)$$

\begin{eqnarray*} c & = & 2.58 \\ p_{s} & = & 0.25 \\ q_{s} & = & 0.75 \\ n & = & 50 \end{eqnarray*}

\begin{eqnarray*} \text{CI} & = & \left(p_{s} - c \sqrt{\frac{p_{s}*q_{s}}{n}}, \quad p_{s} + c \sqrt{\frac{p_{s}*q_{s}}{n}}\right) \\ & = & \left(0.25 - 2.58*\sqrt{\frac{0.25 * 0.75}{50}}, \quad 0.25 + 2.58*\sqrt{\frac{0.25 * 0.75}{50}} \right) \\ & = & (0.25 - 2.58 * 0.0612, \; 0.25 + 2.58 * 0.0612) \\ & = & (0.25 - 0.158, \; 0.25 + 0.158) \\ & = & (0.092, \; 0.408) \end{eqnarray*}

Mighty Gumball has taken a representative sample of 10 gumballs and weighed each one. In their sample, x = 0.5 oz and s² = 0.09.

The normal distribution isn't a good approximation for every situation.

When sample sizes are large, the normal distribution is ideal for finding confidence intervals. It gives accurate results, irrespective of how the population itself is distributed.

Here we have a different situation. Even though $X$ itself is distributed normally, $\overline{X}$ isn’t.

Because of small number of sample . . . .

So what sort of distribution does X follow? It actually follows a t-distribution.

v is called the number of degrees of freedom