A test gives the statistic chi-square(1) = 137.38. Find the p-value.

Step-by-step solution

1. 1

   Identify what a p-value is here. The $p$-value is the upper-tail probability $P\!\left(\chi^2_1 \ge 137.38\right)$: the chance of seeing a statistic at least this extreme if the null hypothesis were true. A chi-square test is always one-tailed on the right, because both directions of departure inflate the statistic.

2. 2

   Use the identity that links chi-square with 1 df to the standard normal. If $Z\sim N(0,1)$ then $Z^2 \sim \chi^2_1$. Hence $P\!\left(\chi^2_1 \ge c\right) = P\!\left(|Z| \ge \sqrt{c}\right) = 2\left(1-\Phi\!\left(\sqrt{c}\right)\right).$ This turns an obscure chi-square tail into a familiar normal tail.

3. 3

   Convert the statistic into a z-score. $\sqrt{137.38} = 11.720921\ldots$ so the observed effect is nearly $11.7$ standard errors from the null value.

4. 4

   Evaluate the two-tailed normal tail. In closed form $2\left(1-\Phi(z)\right)=\operatorname{erfc}\!\left(z/\sqrt{2}\right)$, so $p = \operatorname{erfc}\!\left(\sqrt{137.38/2}\right) = \operatorname{erfc}(8.28596) \approx 9.96\times 10^{-32}.$

5. 5

   Confirm with the asymptotic tail bound. For large $z$, $1-\Phi(z)\approx \dfrac{e^{-z^2/2}}{z\sqrt{2\pi}}$, which gives $p \approx \dfrac{2e^{-68.69}}{11.7209\sqrt{2\pi}} \approx 1.00\times10^{-31}$ - the same order of magnitude, so the computed value is trustworthy.

6. 6

   Interpret the result. Since $p \approx 10^{-31}$ is astronomically smaller than any usual threshold such as $0.05$ or $0.001$, the null hypothesis is rejected decisively. Software often prints this as p < 0.001 or even p = 0; report it as $p<0.001$ rather than literally zero.