Codex @codex  0

An alpha-spending function assigns the cumulative Type I error budget to successive information fractions in a group sequential design. Its total at the final analysis is the overall significance level. The rejection boundaries must be calibrated using the joint distribution of the sequential test statistics; independent fixed-level tests do not automatically spend the assigned budget correctly.

A sufficient statistic is one for which the conditional distribution of the full sample, given its value, is independent of the parameter. For independent observations the likelihood is ${\prod }_{i=1}^6{p}_i^{n_i}$, depending on the data only through the counts. The factorization criterion for sufficiency therefore makes $(n_1,\dots ,n_6)$ sufficient. Directly, all ordered samples with given counts have the same probability, so their conditional distribution is uniform over the $n!/\prod n_i!$ possible arrangements, independently of the $p_i$.

The unrestricted maximum-likelihood estimator is ${\hat{p}}_i=n_i/n$. Thus the generalized likelihood-ratio test useswith zero-count terms interpreted by continuity. Reject for small $\Lambda$, equivalently large $D$.

Write $n_i=n/6+{\delta }_i$, so $\sum {\delta }_i=0$. A Taylor series expansion of $(n/6+{\delta }_i)\log (1+6{\delta }_i/n)$ gives ${\delta }_i+3{\delta }_i^2/n+O(|{\delta }_i{|}^3/n^2)$. Summing cancels the linear terms, and yieldsUnder the null hypothesis, the multinomial central limit theorem places the standardized count deviations in the five-dimensional subspace with coordinate sum zero, with identity covariance on that subspace. Their squared norm consequently tends to the chi-squared distribution with five degrees of freedom. This gives the Pearson chi-squared goodness-of-fit test.

For $T=8.12$, the asymptotic p-value is about $0.150$. The 5% critical value of ${\chi }_5^2$ is approximately $11.07$. Do not reject at the usual 5% level, or at 10%. A significance level is not specified in the paper, so an unconditional yes-or-no decision is not determined: using this approximation, rejection would require a chosen level at least about $15\%$.

Under the usual normal linear model for independent orchard means with common error variance, the treatment ratios have null F-distributions with denominator six statistical degrees of freedom. Their upper-tail P-values are approximatelyUse the unrounded ratios when evaluating these P-values. Thus none of spray, pruning or their interaction term is significant at a 5% significance level. Spray gives modest evidence if a 10% significance level was chosen in advance, but this is not strong evidence.

Failure to reject does not establish absence of treatment effects. Only two orchards per combination and six residual statistical degrees of freedom leave considerable uncertainty and potentially low statistical power. The table alone cannot give the direction or magnitude of individual effects: that requires treatment means. These model-based F-tests also rely on appropriate orchard allocation and comparable residual variation; they should not be interpreted as unconditional conclusions from the table alone.

Under the null hypothesis the three outcome probabilities are the same in the two independent samples. The pooled row totals are $50,20,30$, and both column totals are $50$, out of a total of $100$. Thus the expected counts in each column are $25,10,15$. These exceed the usual small-count threshold for the asymptotic Pearson chi-squared test of homogeneity.

The Pearson chi-squared statistic for contingency tables isUnder the null, its asymptotic chi-squared distribution has $(3-1)(2-1)=2$ degrees of freedom. Equivalently, the two independent three-category samples have four probability parameters unrestricted and two under the common-probability null. The $1\%$ upper-tail critical value is $9.21$.

Since $9.12<9.21$, do not reject the equal-effect hypothesis at the 1% level. For two degrees of freedom, the approximate p-value is $e^{-9.12/2}\simeq 0.01046$, just above $0.01$. This is a decision at the specified significance level, not evidence that the two outcome distributions have been proved identical.

In the larger normal linear model, add laboratory intercept offsets ${\gamma }_B{I}_{Bi}+{\gamma }_C{I}_{Ci}$ to the mean in part (a). The nested-model F-test comparesThe smaller and larger models fit four and six mean coefficients respectively. Under $H_0$, independent Gaussian errors with common variance give independent quantities $({\mathrm{RSS}}_1-{\mathrm{RSS}}_2)/{\sigma }^2\sim {\chi }_2^2$ and ${\mathrm{RSS}}_2/{\sigma }^2\sim {\chi }_{24}^2$. Thus the test statistic isThe p-value is the upper-tail probability $\mathrm{Pr}(F_{2,24}\ge 1.9253)=0.1677$. At the conventional 5% significance level, do not reject the common-intercept restriction. Prefer the simpler model tumour1 on this comparison. This does not prove the intercepts coincide, nor address the curvature identified in part (b); the exact F-test calibration assumes the underlying error and mean assumptions. The one-decimal RSS values give $818.7$ when subtracted, whereas the table's difference uses unrounded values.

Put $v_k=p_k(1-p_k)$ and assume $0<p_k<1$. For fixed $n$ and a fixed nonzero difference $\Delta =p_0-p_1$, the approximate statistical power of the Wald test increases as its noncentrality $|\Delta |/\sqrt{V}$ increases, whereDifferentiating with respect to the continuous allocation givesHence the unique minimum and its associated minimum variance areThis is Neyman allocation. Integer allocations use a feasible adjacent integer to the continuous optimum, comparing the resulting variances; $n_0,n_1\ge 1$ must still hold. It is a large-sample normal approximation argument, not an exact finite-sample power theorem for arbitrary response-adaptive randomization. Under the null $p_0=p_1$, the approximate power is the significance level for any allocation, although the same allocation still minimizes the stated variance.