Confidence Interval

Study guides: Exam SRM, Exam MAS-I

A Confidence Interval is a range computed from a sample that covers the unknown parameter with a stated long-run probability $1 - \alpha$ . It reports the same information as a hypothesis test but on the parameter's own scale — the interval contains exactly those null values the test would not reject.

$\bar{X} \pm z_{\alpha/2}\,\frac{\sigma}{\sqrt{n}} \qquad (\sigma \text{ known})$

$\bar{X} \pm t_{\alpha/2,\,n-1}\,\frac{S}{\sqrt{n}} \qquad (\sigma \text{ estimated})$

Both intervals assume a random sample from a normal population. The $z$ interval is then exact, and so is the $t$ interval, because $(\bar{X} - \mu)/(S/\sqrt{n})$ has a $t$ distribution with $n - 1$ degrees of freedom. For a non-normal population both are only approximate, through the CLT, and need a larger $n$ the further the population is from normal
The general form is $\hat\theta \pm (\text{critical value}) \times \text{SE}(\hat\theta)$ , where the critical value comes from the estimator's Sampling Distribution
The coverage statement is about the procedure, not one interval: $95\%$ of intervals built this way contain $\theta$ . Once computed, a given interval either contains $\theta$ or does not
Width shrinks like $1/\sqrt{n}$ and grows with confidence — going from $95\%$ to $99\%$ multiplies the half-width by $2.576/1.96 = 1.31$
Use $t_{n-1}$ whenever $\sigma$ is replaced by $S$ ; the extra width is the price of estimating the spread, and it matters most for small $n$ . Here $z_{\alpha/2}$ and $t_{\alpha/2,\,n-1}$ are upper-tail points (area $\alpha/2$ above them). The CAS MAS-I $t$ table is headed by two-sided tail areas, so $t_{\alpha/2,\,n-1}$ is in the column headed $\alpha$ : $t_{0.025,\,15} = 2.131$ is under $0.05$
For a variance, the interval is built from the $\\chi^2$ distribution, since $(n-1)S^2/\sigma^2 \sim \chi^2_{n-1}$ for a normal sample, and is asymmetric: $\left(\frac{(n-1)S^2}{\chi^2_{n-1}(1-\alpha/2)},\ \frac{(n-1)S^2}{\chi^2_{n-1}(\alpha/2)}\right)$ . Here $\chi^2_{n-1}(p)$ is the point with lower-tail area $p$ , which is the CAS MAS-I chi-square table's column $p$ , so the larger point gives the lower endpoint. For $n = 16$ at $95\%$ the df $15$ row gives $\left(\frac{15S^2}{27.49},\ \frac{15S^2}{6.26}\right)$ . This interval needs the normal population at every $n$ : no CLT rescues it
For an MLE, the large-sample interval $\hat\theta \pm z_{\alpha/2}/\sqrt{n I(\hat\theta)}$ uses the Fisher Information; for a GLM coefficient it is $\hat\beta_j \pm z_{\alpha/2}\text{SE}(\hat\beta_j)$ , and exponentiating both endpoints gives an interval for the multiplicative rating factor

Interval for a Mean with Unknown Variance A sample of $n = 16$ claims, assumed drawn from a normal population, has $\bar{x} = 8{,}400$ and $s = 1{,}200$ . Build a $95\%$ confidence interval for $\mu$ , given $t_{0.025, 15} = 2.131$ .

Answer $\text{SE} = \frac{s}{\sqrt{n}} = \frac{1{,}200}{4} = 300$ $8{,}400 \pm 2.131(300) = 8{,}400 \pm 639 = (7{,}761,\ 9{,}039)$ Because $8{,}000$ lies inside the interval, a two-sided test of $H_0: \mu = 8{,}000$ at $\alpha = 0.05$ would not reject.

From a Coefficient to a Rating Factor A log-link GLM gives $\hat\beta = 0.35$ for "urban territory" with $\text{SE} = 0.12$ . Give a $95\%$ interval for the multiplicative effect on expected claim frequency.

Answer On the linear-predictor scale: $0.35 \pm 1.96(0.12) = (0.115,\ 0.585)$ . Exponentiating the endpoints: $\left(e^{0.115},\ e^{0.585}\right) = (1.12,\ 1.80)$ Urban risks are estimated to have $12\%$ to $80\%$ higher frequency. The interval excludes $1.00$ , so the effect is significant at $5\%$ — but it is wide enough that the indicated relativity is far from precisely known.