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Statistics and Probability - Confidence intervals (HL)

Grade 12IB_AI

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A Confidence Interval (CI) is a range of values, derived from sample data, that is likely to contain the value of an unknown population parameter. The confidence level (e.g., 95%95\%) represents the frequency with which the interval contains the parameter if the experiment is repeated many times.

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For the population mean μ\mu, a zz-interval is used when the population standard deviation σ\sigma is known and the population is normally distributed or the sample size nn is large (n≥30n \ge 30).

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For the population mean μ\mu, a tt-interval is used when the population standard deviation σ\sigma is unknown. In this case, we use the unbiased estimate of the population standard deviation, sn−1s_{n-1}, and the tt-distribution with df=n−1df = n - 1 degrees of freedom.

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A Confidence Interval for a population proportion pp is calculated using the sample proportion p^=xn\hat{p} = \frac{x}{n}, where xx is the number of successes in nn trials.

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The width of a confidence interval is affected by the confidence level (higher confidence leads to a wider interval) and the sample size (larger nn leads to a narrower interval/more precision).

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The Margin of Error (EE) is half the width of the confidence interval, representing the maximum expected difference between the sample statistic and the true population parameter.

📐Formulae

CI for mean μ (σ known): xˉ±z∗(σn)\text{CI for mean } \mu \text{ (} \sigma \text{ known): } \bar{x} \pm z^* \left( \frac{\sigma}{\sqrt{n}} \right) patterns

CI for mean μ (σ unknown): xˉ±t∗(sn−1n)\text{CI for mean } \mu \text{ (} \sigma \text{ unknown): } \bar{x} \pm t^* \left( \frac{s_{n-1}}{\sqrt{n}} \right)

Unbiased variance estimate: sn−12=nn−1sn2=∑(x−xˉ)2n−1\text{Unbiased variance estimate: } s_{n-1}^2 = \frac{n}{n-1}s_n^2 = \frac{\sum (x - \bar{x})^2}{n-1}

CI for proportion p:p^±z∗p^(1−p^)n\text{CI for proportion } p: \hat{p} \pm z^* \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}

Margin of Error (Mean): E=z∗σn or E=t∗sn−1n\text{Margin of Error (Mean): } E = z^* \frac{\sigma}{\sqrt{n}} \text{ or } E = t^* \frac{s_{n-1}}{\sqrt{n}}

💡Examples

Problem 1:

A random sample of 1010 organic apples has a mean weight of xˉ=155\bar{x} = 155 grams and a sample standard deviation of sn−1=12s_{n-1} = 12 grams. Assuming weights are normally distributed, calculate a 95%95\% confidence interval for the population mean weight μ\mu.

Solution:

Given n=10n = 10, xˉ=155\bar{x} = 155, sn−1=12s_{n-1} = 12, and confidence level 0.950.95. Since σ\sigma is unknown and nn is small, we use the tt-distribution with df=10−1=9df = 10 - 1 = 9.

  1. Find critical value t∗t^* for df=9df = 9 at 95%95\% confidence: t∗≈2.262t^* \approx 2.262.
  2. Standard error SE=sn−1n=1210≈3.795SE = \frac{s_{n-1}}{\sqrt{n}} = \frac{12}{\sqrt{10}} \approx 3.795.
  3. Margin of error E=2.262×3.795≈8.584E = 2.262 \times 3.795 \approx 8.584.
  4. Interval: 155±8.584=[146.416,163.584]155 \pm 8.584 = [146.416, 163.584]. Final answer: [146.4,163.6][146.4, 163.6] grams (3 sig figs).

Explanation:

Because the population standard deviation is unknown, we must use the tt-interval. The degrees of freedom is n−1n-1.

Problem 2:

In a survey of 200200 residents, 120120 stated they support the construction of a new park. Calculate the 90%90\% confidence interval for the true proportion of residents who support the park.

Solution:

Given n=200n = 200 and x=120x = 120.

  1. Sample proportion p^=120200=0.6\hat{p} = \frac{120}{200} = 0.6.
  2. For a 90%90\% confidence level, the critical value z∗≈1.645z^* \approx 1.645.
  3. SE=0.6(1−0.6)200=0.6×0.4200=0.0012≈0.03464SE = \sqrt{\frac{0.6(1 - 0.6)}{200}} = \sqrt{\frac{0.6 \times 0.4}{200}} = \sqrt{0.0012} \approx 0.03464.
  4. CI: 0.6±1.645(0.03464)=0.6±0.056980.6 \pm 1.645(0.03464) = 0.6 \pm 0.05698. Interval: [0.543,0.657][0.543, 0.657].

Explanation:

This is a confidence interval for a proportion. We use the zz-distribution because the sample size is large enough to satisfy the normal approximation conditions (np^>5n\hat{p} > 5 and n(1−p^)>5n(1-\hat{p}) > 5).