Past exam of the options II course of the University of Oxford / 2005 / O

Minimize the residual sum of squares $\Vert Y-X\theta {\Vert }^2$. Differentiating gives the normal equations $X^{T}X\hat{\theta }=X^{T}Y$. Full column rank makes $X^{T}X$ positive definite and invertible, so the unique minimizer isSince $\mathbb{E}Y=X\theta$, it is unbiased. The independent common-variance errors give $\mathrm{Cov}(Y)={\sigma }^{2}I$, henceIndependent Gaussian errors with common variance are a sufficient condition for this estimator also to be the maximum likelihood estimator: their log-likelihood differs from $-\Vert Y-X\theta {\Vert }^2/(2{\sigma }^2)$ by terms independent of $\theta$. Profiling an unknown common variance gives the same minimizer. Mean and variance assumptions alone do not specify a likelihood.

For the seven design points, $\bar{x}=0$ and $\overline{x^2}=4$. In increasing time order the design matrix isIts columns are mutually orthogonal: the odd sums vanish and $\sum (x_i^2-4)=0$. Their squared lengths are $7,28,84$, respectively. ThereforeIn particular $\hat{\gamma }={\sum }_i(x_i^2-4)Y_i/84$. Testing for a quadratic contribution means testing $H_0:\gamma =0$ against $\gamma \ne 0$.

The cooling measurements are stated to have zero-mean errors and known common variance, but normality is not explicitly assumed for them. Under the independent-error linear regression model as stated, Chebyshev's inequality gives a valid conservative level-$\delta$ test: reject whenIts null rejection probability is at most $\delta$ because the estimator has mean zero and the calculated variance. If Gaussian errors are additionally assumed, the sharper exact test uses $Z=\sqrt{84}\hat{\gamma }/\sigma \sim N(0,1)$ under the null and rejects for $|Z|>z_{1-\delta /2}$. With only seven observations, exact normal calibration cannot be deduced from the printed moment assumptions alone.

For a sample of size $n\ge 2$, defineFor the zero-mean Gaussian sample, the standardized vector $(X_1/\sigma ,\dots ,X_n/\sigma )$ has independent standard normal coordinates. Apply an orthogonal transformation whose first row is $(1,\dots ,1)/\sqrt{n}$. The resulting coordinates $Z_1,\dots ,Z_n$ remain independent standard normals: their joint density depends only on squared Euclidean length and the transformation has unit absolute Jacobian. Its first coordinate is $Z_1=\sqrt{n}\overline{X}/\sigma$, while orthogonal decomposition givesThusThe ratio of the first independent standard normal and the square root of the normalized chi-square variable has a Student probability distribution, giving

For the paired height observations, form one cross-minus-self difference per pot. The sampling units for this test are the fifteen differences, not thirty separate plant heights. Under the usual independent normal model for differences, a null mean of zero gives a $t_{14}$ statistic. The observed statistic isFor the two-sided alternative, reject at significance level $\delta$ if $|t_{\mathrm{obs}}|>t_{14,1-\delta /2}$. Its two-sided $p$-value is $2[1-F_{t_{14}}(|t_{\mathrm{obs}}|)]$. This uses the recorded values without performing the numerical calculation forbidden by the question. The normal difference model is what makes the small-sample Student calibration exact.

For the positive-direction alternative, use the same paired statistic $t_{\mathrm{obs}}=2.607/(4.713/\sqrt{15})$, but reject at level $\delta$ only when $t_{\mathrm{obs}}>t_{14,1-\delta }$. Its one-sided $p$-value isBecause the observed mean difference is positive and the null Student probability distribution is symmetric, the two-sided $p$-value is twice this upper-tail probability, consistent with the supplied $0.05$ and $0.025$.

The two-sided value is the null-model probability of a test statistic at least as far from zero in either direction as the observation. The one-sided value uses only the direction of larger cross-fertilized heights. They are probabilities under the assumed null sampling distribution, not posterior probabilities that the null hypothesis is true.

At the reported precision, the two-sided result is on the conventional 5 percent boundary; rounding can matter for a strict numerical threshold. The one-sided result is significant at5 percent and supports a positive mean difference if that directional alternative was chosen in advance. Choosing the direction after observing the sign would not justify using the smaller one-sided probability as a prespecified test. The paired design controls shared pot conditions, while the inference still relies on the independence and probability distribution assumptions for the differences.