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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.

Under the null, replace both parameters by a common $r$. The likelihood becomesThe pooled maximum-likelihood estimate isUnder an interior null $p=q=r$, part (ii) gives a centered asymptotic normal difference with variance $r(1-r)(1/(2M)+1/(2N))$. The law of large numbers makes $\hat{r}\to r$ in probability. By Slutsky's theorem, replacing $r$ by this consistent estimate givesFor the two-sided alternative, reject when $|Z|>z_{1-\alpha /2}$, the corresponding standard-normal quantile. This has asymptotic significance level $\alpha$.

The significance level of a test is a bound $\alpha$ on its probability of rejecting a true null; for a composite null the bound is uniform over its parameter values. Its power at an alternative parameter is its rejection probability under that parameter.

The Neyman-Pearson lemma concerns a simple null density $f_0$ and a simple alternative density $f_1$ for the whole sample. Among tests of size at most $\alpha$, a most powerful test rejects where $f_1>k{f}_0$, accepts where $f_1<k{f}_0$, and if necessary randomizes on equality so that its size is $\alpha$. Let its rejection probability function be ${\varphi }_{\ast }$, with $0\le {\varphi }_{\ast }\le 1$, and let $\varphi$ be any competing test of size at most $\alpha$.

Pointwise,on the rejection side both factors are nonnegative, and on the acceptance side both are nonpositive. Integrating givesThus the likelihood-ratio test has at least as much power as any competitor. Even when the sample has a continuous density, the likelihood ratio can have an atom, so boundary randomization may still be necessary in the general statement. For the exponential test below the statistic has a continuous distribution and no randomization is needed.

Karl Popper's falsificationism treats theories as conjectures to be exposed to serious attempted refutation. A universal claim entails observable restrictions and can conflict with a counterexample, whereas a finite collection of favourable cases does not deductively establish the universal. This emphasizes risky tests and distinguishes surviving criticism from proof. It also avoids pretending that the problem of induction is solved merely by collecting more confirming observations.^[ref]

The main difficulty is moving from this logical asymmetry to a complete account of practice. A prediction usually has the form $(T\wedge A)\Rightarrow E$, where $A$ includes auxiliary hypotheses about apparatus, initial conditions and background physics. Observing $\neg E$ implies $\neg T\vee \neg A$, not specifically $\neg T$. This Duhem problem makes independent measurement calibration and discriminating follow-up experiments essential. An anomalous detector count can reveal a detector fault rather than a new fundamental interaction.

There are also statistical complications. Under a probability model, an unusual observation need not have zero probability; rejection at a chosen significance level is fallible. Repeated severe tests can rationally alter comparative confidence even though they do not prove a theory. A methodology that merely labels all surviving theories unrefuted misses this evidential differentiation.

Falsificationism is therefore valuable as a critical discipline, especially its opposition to adjustments that insulate a theory from every conceivable result. It is incomplete as a rule requiring immediate abandonment after any anomaly, or as a whole theory of warranted acceptance. The better reconstruction combines explicit possible failures, justified auxiliary hypotheses and comparison of competing explanations, while retaining fallibilism about the outcome.

Use the sample mean to estimate each magnetic susceptibility, and the unbiased sample variance with denominator $n-1=4$ to estimate the repeatability of individual readings. In the numerical scale supplied by the paper,If the repeated measurements have independent random errors, their mean estimates have standard errorsThe difference and its estimated standard uncertainty are thereforeThe separation is about $2.53$ standard errors. Under an approximately normal distribution model with no experimenter-dependent bias, Welch's t-test gives about $7.34$ statistical degrees of freedom and a two-sided p-value of $0.038$. Thus there is modest evidence against equal underlying susceptibilities at a conventional $5\%$ significance level; overlap of the ranges of individual readings is not an adequate comparison of the means. Five readings per sample provide limited information about the error distribution.

Crucially, material and student exhibit complete confounding: each student measured only one sample. An experimenter-dependent systematic error could mimic the difference. Paramagnetism also depends on conditions such as temperature, so equal material alone does not guarantee equal readings without controlling those conditions. The defensible physical conclusion is the data suggest a difference, but do not establish different materials. Both students should measure both samples in randomized or alternating order, with calibration and temperature controlled, to separate a sample effect from an operator effect. No unit or scale factor absent from the original table has been invented.

At the conventional $5\%$ significance level, Student 1's p-value $0.72$ gives no rejection of equal mean wear, while Student 2's p-value $0.009$ gives rejection of equal mean wear. The latter also rejects at the $1\%$ significance level. More generally, compare each p-value with the prespecified level $\alpha$; no decision threshold was fixed in the data supplied.

Failure to reject does not establish equality, and a p-value is not the posterior probability that the null hypothesis is true. These two-sided p-values alone also do not identify which material has larger mean wear.

For two simple statistical hypotheses with joint probability density functions $g_0$ and $g_1$, fix a significance level $\alpha$. A randomized statistical test $\phi (x)$ is the conditional probability of rejection given the data. Choose $k\ge 0$ and, if necessary, randomization on the equality set so thatThe Neyman-Pearson lemma says that this is a most powerful test among all tests with test size at most $\alpha$:Thus the optimal ordering rejects where the alternative-to-null likelihood ratio is largest. On null-impossible observations with $g_1>0$, the rejection probability is one. Randomization is dispensable whenever an ordinary threshold already attains the required test size.

Let $L(\vartheta ;x)$ be the likelihood function, with full parameter space $\Theta$ and null hypothesis $\vartheta \in {\Theta }_0\subseteq \Theta$. The generalized likelihood-ratio test compares the best null fit to the best unrestricted fit:The likelihood ratio $R$ lies between zero and one, and small values count against the null hypothesis; equivalently, large values of the likelihood-ratio test statistic $\Lambda$ count against it.

Compute restricted and unrestricted maximum-likelihood estimates, evaluate their log-likelihoods, and form $\Lambda =2(\ell (\hat{\vartheta })-\ell ({\hat{\vartheta }}_0))$. Under the regular conditions of Wilks theorem, the null limit isThese statistical degrees of freedom count independent parameter restrictions. The regularity includes a locally identifiable smooth model, nonsingular Fisher information, and an interior null value; parameter-dependent support can invalidate the result.

For a chosen significance level $\alpha$, obtain the $(1-\alpha )$ quantile $q_{d,1-\alpha }$ of the chi-squared distribution and useEquivalently, compare the asymptotic p-value $\mathrm{Pr}({\chi }_d^2\ge {\Lambda }_{\mathrm{obs}})$ to $\alpha$. This is a large-sample calibration; part (c) requires direct treatment because its support depends on a parameter.

False. For the test associated with a p-value, reject at significance level $\alpha$ when $p\le \alpha$. Here $0.1>0.05$, so do not reject at level 0.05. The p-value is neither the posterior probability of the null hypothesis nor the probability of a Type I error conditional on these observed data.

Condition on the total of 400 complaints. Assume each complaint independently has the same probability $p$ of being received on Monday; in particular, repeated complaints from one incident must not create material dependence. The Monday count $S$ then has a binomial distribution with parameters $(400,p)$. Test $H_0:p=0.2$ against $H_1:p>0.2$, rejecting for large $S$ as suggested by the monotone likelihood ratio.

Under the null hypothesis, ${\mathbb{E}}_{0}S=80$ and ${\mathrm{Var}}_0(S)=400(0.2)(0.8)=64$. Both expected successes and expected failures are large, so the central limit theorem gives a suitable normal approximation:Using the paper's supplied rounded table, the upper-tail p-value is approximatelyA more accurate evaluation of the standard normal distribution cumulative distribution function gives about 0.0062. For example, choose significance level 0.05 before examining the data; the supplied upper 5% standard normal quantile is approximately 1.6, and $2.5>1.6$. Reject $H_0$ at 5%, and also at 1%, using this approximation. The observed binomial proportion is $100/400=0.25$, providing evidence that the Monday probability exceeds 0.2 under the sampling assumptions.

If applying a continuity correction, approximate ${\mathbb{P}}_0(S\ge 100)$ by $1-\Phi ((99.5-80)/8)=1-\Phi (2.4375)\approx 0.0074$; this gives the same 5% and 1% conclusions. The uncorrected value 0.005 is approximate, so a decision exactly at the 0.5% boundary is not warranted by the supplied rounded table.

At a conventional 5% significance level, there is evidence for an association with sex after adjusting for weight, because its two-sided p-value is $0.00131<0.05$. There is insufficient evidence for a weight association after adjusting for sex, because its p-value is $0.62392>0.05$.

These Student t-tests test whether the corresponding regression coefficients are zero within the additive multiple linear regression. Failure to reject the weight hypothesis does not establish an exactly zero effect; these associations alone also do not justify a causal inference. The later model with a statistical interaction asks a different question, allowing the weight association to differ by sex.

A result is statistically significant at a specified significance level $\alpha$ when its valid p-value is at most $\alpha$ and the corresponding null hypothesis is rejected. Significance does not measure effect size, establish practical importance, or give the probability that the null is true. Software stars encode thresholds for $p$-values; they do not supply an additional statistic.

For a symmetric interval of the form $[L,U]=[\hat{\theta }-c\mathrm{SE}(\hat{\theta }),\hat{\theta }+c\mathrm{SE}(\hat{\theta })]$, where $c\ge 0$ and the standard error is nonnegative, adding and subtracting the endpoints proves the displayed identities. Thus the endpoints must be ordered and their midpoint must equal the point estimate. Rounding should be assessed using intervals of possible unrounded values before declaring a numerical contradiction.

When this confidence interval is obtained by inverting a matching two-sided test of $\theta ={\theta }_0$, exclusion of ${\theta }_0$ agrees with rejection at the corresponding significance level, with the usual convention at boundary equality. This is a check for a specified symmetric interval procedure, such as the usual t interval for a mean difference; it does not require asymmetric intervals or intervals on transformed scales to have their midpoint at the estimate on the original scale.