Interval Estimation — Part 2

Topics covered

2’) Confidence Interval for the Population Mean, \(\mu\), of Non-Normal Populations

3. Confidence Interval for the Proportion, \(p\), of a Bernoulli Population

4. Confidence Interval for the Variance, \(\sigma^2\), and Standard Deviation, \(\sigma\), of a Normal Population

Case 3 — Non-Normal Population and \(n > 30\)

Consider a random sample \(X_1, X_2, \ldots, X_n\), \(n \in \mathbb{N}\) and \(n > 30\), drawn from a population \(X\) of unknown or non-Normal distribution.

Regardless of whether \(\sigma\) is known or not, the pivot statistic is always approximately \(N(0,1)\) (by the Central Limit Theorem):

Case 3 — Confidence Intervals for \(\mu\)

The approximate \((1-\alpha)\times 100\%\) confidence intervals for \(\mu\) are:

If \(\sigma\) known:

\[IC_{(1-\alpha)\times100\%}(\mu) \approx \left(\bar{x} - z_{\alpha/2}\frac{\sigma}{\sqrt{n}},\; \bar{x} + z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\right)\]

If \(\sigma\) unknown:

\[IC_{(1-\alpha)\times100\%}(\mu) \approx \left(\bar{x} - z_{\alpha/2}\frac{s'}{\sqrt{n}},\; \bar{x} + z_{\alpha/2}\frac{s'}{\sqrt{n}}\right)\]

Application Exercise — Setup

An insurance company has budgeted an average daily payout of no more than 5,000 monetary units (m.u.) per day to cover its policyholders’ claims. To validate this estimate, 100 days were randomly selected from the company’s historical records, yielding:

\[\bar{x} = 5{,}625 \text{ m.u.} \qquad s' = 2{,}500 \text{ m.u.}\]

Find a 95% CI for the mean daily payout and determine whether 5,000 m.u./day is sufficient.

Application Exercise — Solution

Population distribution: unknown; \(\sigma\): unknown.

Sample: \(n = 100\), \(\bar{x} = 5{,}625\), \(s' = 2{,}500\).

Pivot statistic: \(\;Z = \dfrac{\bar{X} - \mu}{S'/\sqrt{n}} \;\dot{\sim}\; N(0,1)\)

\[1-\alpha = 0.95 \;\Leftrightarrow\; \alpha = 0.05 \;\Rightarrow\; z_{\alpha/2} = 1.96\]

\[IC_{95\%}(\mu) \approx \left(5{,}625 - 1.96\times\frac{2{,}500}{\sqrt{100}}\;;\; 5{,}625 + 1.96\times\frac{2{,}500}{\sqrt{100}}\right)\]

\[\boxed{IC_{95\%}(\mu) \approx (5{,}135\;;\; 6{,}115)}\]

Application Exercise — Conclusion

\[IC_{95\%}(\mu) \approx (5{,}135\;;\; 6{,}115)\]

CI for the Proportion — Setup

Consider a random sample \(X_1, X_2, \ldots, X_n\), \(n \in \mathbb{N}\), \(n > 30\), drawn from a Bernoulli population:

\[X \sim \mathrm{Bernoulli}(p), \quad 0 < p < 1\]

(equivalently, \(X \sim \mathrm{Binomial}(1, p)\))

\[\underbrace{p}_{\text{parameter}} \;\longrightarrow\; \underbrace{\frac{Y}{n}}_{\substack{\text{point} \\ \text{estimator}}} \;\longrightarrow\; \underbrace{Z = \dfrac{\hat{p} - p}{\sqrt{\hat{p}\hat{q}/n}}}_{\text{pivot statistic}} \;\dot{\sim}\; N(0,1)\]

where \(Y = \displaystyle\sum_{i=1}^{n} X_i\), \(\;\hat{p} = \bar{X} = \dfrac{Y}{n}\), and \(\hat{q} = 1 - \hat{p}\).

CI for the Proportion — Formula

The approximate \((1-\alpha)\times 100\%\) confidence interval for \(p\) is:

\[\boxed{IC_{(1-\alpha)\times100\%}(p) \approx \left(\hat{p} - z_{\alpha/2}\sqrt{\frac{\hat{p}\,\hat{q}}{n}},\; \hat{p} + z_{\alpha/2}\sqrt{\frac{\hat{p}\,\hat{q}}{n}}\right)}\]

Application Exercise — Setup

In a given district, 840 out of 2,000 voters surveyed in a poll declared they would vote for Party A.

Construct a 95% confidence interval for the proportion of voters supporting Party A.

Population: \(X \sim \mathrm{Bernoulli}(p)\), where \(p\) = proportion of Party A voters.

Sample: \(n = 2{,}000\); \(x = 840\) (observed Party A voters in the sample).

\[\hat{p} = \frac{x}{n} = \frac{840}{2{,}000} = 0.42 \qquad \hat{q} = 1 - \hat{p} = 0.58\]

Application Exercise — Solution

Pivot statistic: \(\;Z = \dfrac{\hat{p} - p}{\sqrt{\hat{p}\hat{q}/n}} \;\dot{\sim}\; N(0,1)\)

\[1-\alpha = 0.95 \;\Leftrightarrow\; \alpha = 0.05 \;\Rightarrow\; z_{\alpha/2} = 1.96\]

\[IC_{95\%}(p) \approx \left(0.42 - 1.96\sqrt{\frac{0.42\times 0.58}{2{,}000}}\;;\; 0.42 + 1.96\sqrt{\frac{0.42\times 0.58}{2{,}000}}\right)\]

With 95% confidence, the proportion of Party A voters in the district is between 39.83% and 44.16%.

CI for the Variance — Setup

Consider a random sample \(X_1, X_2, \ldots, X_n\), \(n \in \mathbb{N}\), drawn from a Normal population \(X \sim N(\mu, \sigma)\), with both \(\mu\) and \(\sigma\) unknown.

\[\underbrace{\sigma^2}_{\text{parameter}} \;\longrightarrow\; \underbrace{S'^{\,2}}_{\substack{\text{point} \\ \text{estimator}}} \;\longrightarrow\; \underbrace{Q = \dfrac{(n-1)S'^{\,2}}{\sigma^2}}_{\text{pivot statistic}} \;\sim\; \chi^2_{(n-1)}\]

The pivot statistic \(Q\) follows exactly a chi-squared distribution with \(n-1\) degrees of freedom.

CI for \(\sigma^2\) — Probability Statement

\[P(Q < q_{\inf}) = P(Q > q_{\sup}) = \frac{\alpha}{2}\]

CI for \(\sigma^2\) and \(\sigma\) — Formulas

The \((1-\alpha)\times100\%\) confidence interval for \(\sigma^2\) is:

\[\boxed{IC_{(1-\alpha)\times100\%}(\sigma^2) = \left(\frac{(n-1)\,s'^{\,2}}{q_{\sup}}\;,\; \frac{(n-1)\,s'^{\,2}}{q_{\inf}}\right)}\]

If a CI for \(\sigma\) is required, take the square root of each bound:

\[\boxed{IC_{(1-\alpha)\times100\%}(\sigma) = \left(\sqrt{\frac{(n-1)\,s'^{\,2}}{q_{\sup}}}\;,\; \sqrt{\frac{(n-1)\,s'^{\,2}}{q_{\inf}}}\right)}\]

where \(q_{\inf}\) and \(q_{\sup}\) are the lower and upper critical values of the \(\chi^2_{(n-1)}\) distribution.

Application Exercise — Setup

Based on a random sample of \(n = 16\) observations drawn from a Normal population, a corrected sample standard deviation of \(s' = 3.872\) was obtained.

Construct a 95% confidence interval for the population variance.

Population: \(X \sim N(\mu, \sigma)\), both parameters unknown.

Sample: \(n = 16\), \(s' = 3.872\).

Pivot statistic:

\[Q = \frac{(n-1)\,S'^{\,2}}{\sigma^2} \sim \chi^2_{(n-1)} \equiv \chi^2_{(15)}\]

Application Exercise — Finding the Critical Values

We need \(q_{\inf}\) and \(q_{\sup}\) from the \(\chi^2_{(15)}\) distribution (Table 6, row 15):

\[1-\alpha = 0.95 \;\Leftrightarrow\; \alpha = 0.05 \;\Rightarrow\; \frac{\alpha}{2} = 0.025\]

\[P(Q > q_{\sup}) = \frac{\alpha}{2} = 0.025 \;\underset{\substack{\text{Table 6} \\ \text{row 15},\; \varepsilon=0.025}}{\Longrightarrow}\; q_{\sup} = 27.488\]

\[P(Q < q_{\inf}) = \frac{\alpha}{2} \;\Leftrightarrow\; P(Q > q_{\inf}) = 1 - \frac{\alpha}{2} = 0.975 \;\underset{\substack{\text{Table 6} \\ \text{row 15},\; \varepsilon=0.975}}{\Longrightarrow}\; q_{\inf} = 6.262\]

Application Exercise — Result for \(\sigma^2\)

\[IC_{95\%}(\sigma^2) = \left(\frac{(16-1)\times 3.872^2}{27.488}\;;\; \frac{(16-1)\times 3.872^2}{6.262}\right)\]

Application Exercise — Result for \(\sigma\)

If a confidence interval for \(\sigma\) is required, take the square root of each bound:

\[IC_{95\%}(\sigma) = \left(\sqrt{\frac{(16-1)\times 3.872^2}{27.488}}\;;\; \sqrt{\frac{(16-1)\times 3.872^2}{6.262}}\right)\]

Summary — All Cases