| English | 意味 |
|---|---|
| confidence interval/ˈkɒnfɪdəns ˈɪntəvl/ | 信頼区間 |
| standard error/ˈstændəd ˈerə/ | 標準誤差 |
| margin of error/ˈmɑːdʒɪn ɒv ˈerə/ | 許容誤差 |
| confidence level/ˈkɒnfɪdəns ˈlevl/ | 信頼度 |
How precise is sixty out of one hundred?
- In a fictional random sample of 100 customers, 60 support a new collection service. We want a confidence interval 置信区间 for the population support proportion.
- The observed estimate is $\hat p=0.60$. The interval should describe customers in the sampled population, not the proportion in every future sample.
Check before estimating uncertainty
- Check random sampling, $100\le0.10N$ for sampling without replacement, and the observed counts 60 and 40. Each count is at least ten.
- The standard error 标准误 estimates the sampling variability using $\hat p$, since the population proportion is unknown.
Match each sample with the large-counts decision.
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
Both observed counts must be at least ten for the normal interval used here.
Build estimate plus or minus uncertainty
- Use the formula $\hat p\pm z^*SE$, with $SE=\sqrt{\hat p(1-\hat p)/n}$. A 95% normal interval uses $z^*=1.96$.
- Here $SE=\sqrt{0.60(0.40)/100}\approx0.04899$. Keep these extra digits until the endpoints are calculated.
n = 100, p̂ = 0.60. Compute the standard error √(p̂(1−p̂)/n). Two decimals.
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
√(0.6·0.4/100) = √0.0024 ≈ 0.049 ≈ 0.05.
A confidence interval has the form...
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
You must multiply SE by the critical value z*.
For a 95% confidence interval, what critical value z* do you use? (two decimals)
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
z* = 1.96 for 95% confidence.
Calculate the endpoints
- The margin of error 误差幅度 is $ME=z^*SE=1.96(0.04899)\approx0.09602$. The interval is $(0.50398,0.69602)$.
- The confidence level 置信水平 describes the repeated-sampling method. The shaded normal area at $\pm1.96$ is about 0.95; it is not a plot of individual customer answers.
The interval width is about 0.1920, while its margin of error is about 0.0960. Do not report the whole width as the margin.
Normal reference area for 95% intervals
The band from -1.96 to 1.96 contains approximately 95% of a standard normal distribution. This reference area is not the probability assigned to an already observed population interval.
With SE ≈ 0.049 and z* = 1.96, find the margin of error z*·SE. Two decimals.
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
1.96 × 0.049 ≈ 0.096 ≈ 0.10.
Change one factor at a time
- With the same observed proportion and confidence level, quadrupling $n$ halves $SE$ and the margin of error. Raising confidence increases $z^*$ and widens the interval.
- These width comparisons hold the other inputs fixed. A new sample can also change $\hat p$, so its observed width need not follow a sample-size slogan exactly.
Ninety-five percent confidence does not mean that 95% of individual customers lie inside an interval for p.
Raising the confidence level (say 95% → 99%) makes the interval...
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
Higher confidence → larger z* → wider interval.
After calculating this interval, there is a 95% probability that the fixed population proportion lies inside it.
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
Frequentist confidence describes the long-run coverage of the procedure, not a probability assigned to a fixed parameter after observing this interval.
Interpret the estimate in context
- We are 95% confident that about 50.4% to 69.6% of customers in the sampled population support the service, using this checked procedure.
- The interval covers sampling uncertainty, not every source of error. Nonresponse, misleading questions and selection bias need separate attention.
We are 95% confident that about 50.4% to 69.6% of customers in the sampled population support the service, using this checked procedure.
Holding the observed proportion fixed, which changes narrow a normal proportion interval?
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
More observations reduce standard error; lower confidence reduces the critical value. Quadrupling n halves the standard error.