Critical Value

Study guide: Exam MAS-I

A critical value $c$ is the boundary of the rejection region in a hypothesis test: the percentile of the test statistic's sampling distribution under $H_0$ chosen so that, if $H_0$ is true, the statistic lands beyond it with probability equal to the significance level $\alpha$ .

$P\left(T \ge c \mid H_0\right) = \alpha$

The block is for an upper-tailed test; a lower-tailed test rejects when $T \le c$ with $P(T \le c \mid H_0) = \alpha$ , and a two-sided test splits $\alpha/2$ into each tail. Rejecting a true $H_0$ is a Type I Error, so $\alpha$ is its probability by construction.
Notation. Here $z_\alpha$ , $t_{\alpha,\nu}$ , $\chi^2_{\alpha,\nu}$ and $F_{\alpha,\nu_1,\nu_2}$ are upper $\alpha$ points — the value with probability $\alpha$ above it. Tables differ, so check which tail yours tabulates.
Tests on a mean. Use $Z = \dfrac{\bar X - \mu_0}{\sigma/\sqrt n}$ against $z_\alpha$ when $\sigma$ is known ( $z_{0.05} = 1.645$ , $z_{0.025} = 1.960$ ). Use $T = \dfrac{\bar X - \mu_0}{S/\sqrt n}$ against $t_{\alpha,\,n-1}$ when $\sigma$ is estimated from a normal sample.
Tests on a variance. Use $\chi^2 = (n-1)S^2/\sigma_0^2$ against $\chi^2_{\alpha,\,n-1}$ . The chi-square is skewed, so a two-sided test needs two different critical values, $\chi^2_{1-\alpha/2,\,n-1}$ and $\chi^2_{\alpha/2,\,n-1}$ , not $\pm$ one number. For two variances, $F = S_1^2/S_2^2$ is compared with $F_{\alpha,\,n_1-1,\,n_2-1}$ . The lower point comes from $F_{1-\alpha,\,\nu_1,\,\nu_2} = 1/F_{\alpha,\,\nu_2,\,\nu_1}$ .
Same decision as the p-Value. The statistic is beyond the critical value exactly when the p-value is below $\alpha$ . A two-sided test at level $\alpha$ rejects $\mu_0$ exactly when $\mu_0$ lies outside the $1-\alpha$ Confidence Interval. Moving the critical value trades Type II Error against Type I error, which is what sets the power.

t-Test on Mean Claim Severity After a change in repair costs, a sample of $n = 16$ auto physical damage claims has mean $5,400 and sample standard deviation $800. Assuming severities are normal, test $H_0: \mu = 5{,}000$ against $H_1: \mu > 5{,}000$ at $\alpha = 0.05$ and at $\alpha = 0.01$ .
Answer $\sigma$ is unknown, so use the $t$ distribution with $15$ degrees of freedom.
 $\begin{align*} T &= \frac{5{,}400 - 5{,}000}{800/\sqrt{16}} \\ &= \frac{400}{200} \\ &= 2.00 \end{align*}$ 
At $\alpha = 0.05$ : $t_{0.05,15} = 1.753$ . Since $2.00 > 1.753$ , reject $H_0$ .

At $\alpha = 0.01$ : $t_{0.01,15} = 2.602$ . Since $2.00 < 2.602$ , do not reject.

The evidence of higher severity is significant at 5% but not at 1%, which places the p-value between them (it is about $0.032$ ).

Chi-Square Test for Increased Volatility A pricing model assumes claim severities have standard deviation $1,000. A sample of $n = 25$ recent claims has $s = 1{,}300$ . Assuming normality, test $H_0: \sigma^2 = 1{,}000^2$ against $H_1: \sigma^2 > 1{,}000^2$ at $\alpha = 0.05$ and $\alpha = 0.01$ .
Answer
 $\begin{align*} \chi^2 &= \frac{(n-1)s^2}{\sigma_0^2} \\ &= \frac{24 \times 1{,}300^2}{1{,}000^2} \\ &= 40.56 \end{align*}$ 
With $24$ degrees of freedom, $\chi^2_{0.05,24} = 36.415$ and $\chi^2_{0.01,24} = 42.980$ .

At $5\%$ : $40.56 > 36.415$ , so reject $H_0$ .

At $1\%$ : $40.56 < 42.980$ , so do not reject.

At the usual 5% level the data say severity is more volatile than the model assumes — a reason to revisit any risk load built on the old standard deviation.