b:head_first_statistics:constructing_confidence_intervals
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b:head_first_statistics:constructing_confidence_intervals [2019/12/09 10:40] – [Four steps for finding confidence intervals] hkimscil | b:head_first_statistics:constructing_confidence_intervals [2023/11/15 08:24] (current) – [Four steps for finding confidence intervals] hkimscil | ||
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Rather than specify an exact value, we can specify two values we expect flavor duration to lie between. | Rather than specify an exact value, we can specify two values we expect flavor duration to lie between. | ||
- | [{{:b: | + | <WRAP group> |
+ | <WRAP 25% column> | ||
+ | $\Large{P(a < \mu < b) = 0.95} $ | ||
+ | </ | ||
+ | <WRAP 50% column> | ||
+ | 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 **// | The far side of each end, (a, b) is called a **// | ||
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$(61.72, 63.68)$ 을 전체 population의 단맛의 지속시간으로 삼는다. | $(61.72, 63.68)$ 을 전체 population의 단맛의 지속시간으로 삼는다. | ||
+ | |||
<WRAP box> | <WRAP box> | ||
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* 따라서 위의 경우는 95%에 해당하는 probability는 | * 따라서 위의 경우는 95%에 해당하는 probability는 | ||
* $P(-2 < z < 2) = .95$ | * $P(-2 < z < 2) = .95$ | ||
- | * $P(-2 < \dfrac {X - \overline{X}}{sd} < 2) = .95$ | + | * $P(-2 < \dfrac {\overline{X} |
* 이렇게 계산을 하면 | * 이렇게 계산을 하면 | ||
* $P(\overline{X} -1 < \mu < \overline{X} + 1) = .95 $ | * $P(\overline{X} -1 < \mu < \overline{X} + 1) = .95 $ | ||
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{{: | {{: | ||
- | v is called the **<fc # | + | |
+ | v is called the number of **<fc # | ||
{{: | {{: | ||
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==== Step 4: Find the confidence limits ==== | ==== Step 4: Find the confidence limits ==== | ||
{{: | {{: | ||
+ | Use degrees of freedom with alpha (p-level) | ||
===== The t-distribution vs. the normal distribution ===== | ===== The t-distribution vs. the normal distribution ===== | ||
{{: | {{: | ||
+ | |||
+ | ===== Exercise ===== | ||
+ | <WRAP help> | ||
+ | Mighty Gumball has noticed a problem with their gumball dispensers. They have taken a sample of 30 machines, and found that the mean number of malfunctions is 15. Construct a 99% confidence interval for the number of malfunctions per month. | ||
+ | </ | ||
+ | |||
+ | 위는 Poisson distribution이므로 $X \sim Po(15)$ 이고 $E(X) = \lambda$이고 $Var(X) = \lambda$이다. 따라서 | ||
+ | |||
+ | $$\text {confidence interval} = (\overline{X} - c * se, \;\; \overline{X} + c * se)$$ | ||
+ | $$\text{se} = \sqrt{(15/ | ||
+ | $$\text{c} = 2.58 (3) $$ 이므로 | ||
+ | |||
+ | \begin{eqnarray*} | ||
+ | \text {confidence interval} & = & (\overline{X} - c * se, \;\; \overline{X} + c * se) \\ | ||
+ | & = & (15 - 3 * \sqrt{(15/ | ||
+ | & = & (15 - 2.58 * \sqrt{(15/ | ||
+ | & = & (15 - 2.58 * 0.707, \;\; 15 + 2.58 * 0.707) \\ | ||
+ | & = & (15 - 1.824, \;\; 15 + 1.824) \\ | ||
+ | & = & (13.176, \;\; 16.824) | ||
+ | \end{eqnarray*} |
b/head_first_statistics/constructing_confidence_intervals.1575855619.txt.gz · Last modified: 2019/12/09 10:40 by hkimscil