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For independent standard normal random variables ${\eta }_1,\dots ,{\eta }_n$, define the displayed errors for $i\ge 2$. Each has mean zero and variance one. Nonadjacent errors depend on disjoint sets of independent variables. Adjacent covariances vanish because their product contains an independent centered factor. Thus their covariance matrix is $I_n$.

They are nevertheless dependent: the Gaussian fourth moment gives $\mathbb{E}[{ϵ}_i^2{ϵ}_{i+1}^2]=3$, and hence $\mathrm{Cov}({ϵ}_i^2,{ϵ}_{i+1}^2)=2$. Constant variance and zero pairwise covariance can conceal dependence in magnitudes. This is useful when interpreting a residual plot in linear regression.

The fitted simple linear regression for the numerical vaccination intention scale isHere age is measured in years. The fitted mean increases by about 0.01649 score points per year, or 0.16489 points per decade. The regression coefficient is an association across respondents, not an estimate of how an individual would change on ageing or of a causal effect of age.

The standard error of the age regression coefficient is $0.006806$. Dividing gives $t\approx 2.423$, with 2,133 residual statistical degrees of freedom, corresponding to $2135-2$ fitted residual degrees. The reported two-sided p-value is $0.0155$, so the zero-slope null hypothesis is rejected at a prespecified 5% level, but not at 1%. An approximate 95% confidence interval isIts positive endpoints are consistent with the test decision; they do not make the slope large.

The coefficient of determination is $R^2=0.002744$, so the fitted age trend accounts for only 0.2744% of the observed score variation, compared with an intercept-only prediction. Its adjusted value is $0.002276$. In an intercept-containing simple linear regression, the positive slope implies a positive Pearson correlation coefficient $r=\sqrt{R^2}\approx 0.05238$. The reported overall $F$ statistic, $5.869$ on one and 2,133 degrees of freedom, agrees with the squared slope test statistic up to printed rounding.

The residual standard deviation estimate is $5.618$ score points, much larger than the fitted decade contrast. At age 50 the fitted value is about $13.4697$, and at 60 it is about $13.6346$. Across the observed ages 19 to 91, the fitted mean changes by only $1.1872$ points. Statistical detection with 2,135 observations therefore coexists with weak individual prediction.

The intercept predicts $12.645291$ at age zero, outside the observed age range; it should not be given a substantive infant interpretation. Centring age at its observed sample mean, $50.174$, giveswhich agrees with the reported score sample mean of $13.473$ after rounding. The bounded scale and upper quartile at its maximum motivate attention to a possible ceiling effect; they do not establish homoskedasticity, a normal error distribution or a satisfactory linear conditional mean. Five residual quantiles cannot replace a residual plot or checks of the sampling assumptions.

There is also a source consistency problem. The output reports that all 2,135 records have age and score present, that the attained minimum score is one, and that ages lie between 19 and 91. The positive fitted slope therefore places every fitted value between $12.958582$ and $14.145790$. By residual extrema when response extrema are attained, the minimum regression residual must satisfyAt the observation with score one, its residual is already at most $1-12.958582=-11.958582$. The printed minimum residual, $-10.014$, lies outside this necessary interval. Those summaries cannot all describe the same complete sample and fitted model. The discrepancy is too large to be explained by the displayed rounding. It is present in the original PDF; the individual records would be needed to establish which printed item is wrong.

Compute fitted responses ${\hat{y}}_i=\hat{\alpha }+\hat{\beta }{x}_i$ and regression residuals $e_i=y_i-{\hat{y}}_i$. A conventional residual plot shows $e_i$ against ${\hat{y}}_i$ or $x_i$; plotting against observation order additionally looks for runs. One checks for curvature, changing spread and other structure rather than random scatter around zero.

An ordinary residual plot need not reveal this failure of independence. The errors have constant mean zero, constant variance one and zero pairwise covariances, so the usual trend and changing-variance clues need not appear. A visually patternless residual plot cannot validate independence; fitted regression residuals themselves also share the fitted coefficients even when the true errors are independent.

There is detectable dependence in the magnitudes. The Gaussian fourth moment is $\mathbb{E}{\eta }_i^4=3$, giving for every adjacent pairFor $i=1$, the product is ${\eta }_1^4{\eta }_2^2$; for $i\ge 2$, it is ${\eta }_{i-1}^2{\eta }_i^4{\eta }_{i+1}^2$, with the same expected value 3. Large magnitudes tend to occur together. A lagged plot of squared regression residuals or a test directed at dependence in their squares is more informative. Thus recognition is possible with suitable diagnostics, but is not guaranteed by the standard residual plot requested here.