An IQ threshold

Topic: Bayesian hypotheses and model comparison · more questions on this topic
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Model population IQ by \mu\sim\operatorname{N}(100,15^2). Conditional on a person’s true IQ \mu, a standardised test score satisfies X\mid\mu\sim\operatorname{N}(\mu,10^2). A candidate obtains x=135, and the classification threshold for “gifted” is 130.

  1. Find an equal-tailed 95\% posterior credible interval for \mu.

  2. For H_0:\mu<130 and H_1:\mu\geq130, calculate the posterior odds \mathbb{P}(H_0\mid x)/\mathbb{P}(H_1\mid x) and state which hypothesis they favour.

  3. Calculate \operatorname{BF}_{01} and reconcile its direction with that of the posterior odds by examining the prior odds.

Question. Model population IQ by \mu\sim\operatorname{N}(100,15^2). Conditional on a person’s true IQ \mu, a standardised test score satisfies X\mid\mu\sim\operatorname{N}(\mu,10^2). A candidate obtains x=135, and the classification threshold for “gifted” is 130.

  1. Find an equal-tailed 95\% posterior credible interval for \mu.

  2. For H_0:\mu<130 and H_1:\mu\geq130, calculate the posterior odds \mathbb{P}(H_0\mid x)/\mathbb{P}(H_1\mid x) and state which hypothesis they favour.

  3. Calculate \operatorname{BF}_{01} and reconcile its direction with that of the posterior odds by examining the prior odds.

Solution. The normal prior and likelihood give posterior precision

\frac1{v_1}=\frac1{15^2}+\frac1{10^2} =\frac1{225}+\frac1{100} =\frac{325}{22500}.

Hence

v_1=\frac{22500}{325}=69.2308.

The posterior mean is \begin{align*} m_1 &=v_1\left(\frac{100}{15^2}+\frac{135}{10^2}\right)\\ &=\frac{15^2(135)+10^2(100)}{15^2+10^2}\\ &=\frac{30375+10000}{325}\\ &=124.2308. \end{align*} Therefore

\boxed{\mu\mid x=135\sim\operatorname{N}(124.2308,69.2308)}.

  1. The posterior standard deviation is \sqrt{69.2308}=8.3205. Thus the equal-tailed interval is \begin{align*} m_1\pm z_{0.975}\sqrt{v_1} &=124.2308\pm1.959964(8.3205)\\ &=\boxed{(107.923,140.539)}. \end{align*}

  2. Under the posterior, \begin{align*} \mathbb{P}(H_0\mid x) &=\mathbb{P}(\mu<130\mid x)\\ &=\Phi\!\left(\frac{130-124.2308}{8.3205}\right)\\ &=0.7560, \end{align*} so \mathbb{P}(H_1\mid x)=1-0.7560=0.2440. The posterior odds are

\boxed{ \frac{\mathbb{P}(H_0\mid x)}{\mathbb{P}(H_1\mid x)} =\frac{0.7560}{0.2440}=3.0977}.

On posterior odds alone, H_0 is favoured by about 3.1:1.

  1. Under the prior, \begin{align*} \mathbb{P}(H_0) &=\mathbb{P}(\mu<130)\\ &=\Phi\!\left(\frac{130-100}{15}\right)\\ &=\Phi(2)=0.97725. \end{align*} The prior odds are therefore

\frac{\mathbb{P}(H_0)}{\mathbb{P}(H_1)} =\frac{0.97725}{1-0.97725} =42.9558.

Consequently \begin{align*} \operatorname{BF}_{01} &=\frac{\text{posterior odds for $H_0$}} {\text{prior odds for $H_0$}}\\ &=\frac{3.0977}{42.9558}\\ &=\boxed{0.0721}. \end{align*} Equivalently, \operatorname{BF}_{10}=1/0.0721\approx13.87: the score is about fourteen times more probable under H_1 than under H_0 after averaging within each hypothesis. The posterior nevertheless favours H_0 because the prior odds favoured it by about 43:1. The Bayes factor records the movement supplied by the data; the posterior odds also contain the starting odds.