Why Check Normality Before Inferring a Variance?
A confidence interval for a population variance can look precise while relying on an unsuitable model. The exact chi-square interval for a variance is derived for an independent random sample from a normal population. If the population is strongly skewed, heavy-tailed, or a mixture of different groups, that interval may not have its advertised coverage. A small sample can make those departures difficult to detect, so a diagnostic is a check on the model—not a certificate that the model is true.
For observations \(X_1,\ldots,X_n\), the sample variance is \(S^2=\sum_{i=1}^n(X_i-\bar X)^2/(n-1)\). If the observations are independent and identically distributed as \(N(\mu,\sigma^2)\), then \((n-1)S^2/\sigma^2\) has a chi-square distribution with \(n-1\) degrees of freedom. The resulting two-sided interval is
where \(\chi^2_{p,\nu}\) denotes the \(p\)-quantile of a chi-square distribution with \(\nu\) degrees of freedom. The order of the quantiles matters: the larger quantile is in the lower endpoint. Diagnostics are used to assess whether the normal-sample model underlying this result is plausible. They do not change the formula or prove its assumptions.
Three Complementary Checks
Use the checks together rather than treating any one graph or rule as decisive. Before interpreting them, verify that the data are observations of the same quantity in the same units, and consider whether the sample was collected in a way that makes independence plausible. A histogram and a normal probability plot cannot repair dependence caused by repeated measurements, clustering, or a time trend.
| Diagnostic | What to inspect | Important limitation |
|---|---|---|
| Histogram | Symmetry, one main cluster, and tail behavior | Appearance depends on bin widths; small samples look irregular |
| Normal probability plot | Whether ordered data follow an approximately straight pattern | A few points can dominate, and small samples provide little resolution |
| Outlier check | Observations unusually far from the rest, using context and a stated rule | A flagged value is not automatically an error or a reason for deletion |
A histogram is a useful first view. A roughly symmetric, single-peaked shape is compatible with normality; a long right tail, a second cluster, or an isolated extreme value raises a question. But a histogram can conceal features when there are few observations, and changing the bin width may change its appearance. It is evidence to weigh, not a pass/fail test.
A normal probability plot, also called a normal Q-Q plot, compares ordered sample values with theoretical normal quantiles. If the sample is approximately normal, the plotted points tend to lie near a straight line. A curved pattern can indicate skewness; systematic departures at both ends can indicate tails that are heavier or lighter than normal. A single distant point may indicate an outlier, but the context and data quality still need review.
One common plotting convention assigns observation \(j\), in increasing order, the plotting probability \(p_j=(j-0.375)/(n+0.25)\), for \(j=1,\ldots,n\). Its corresponding standard normal quantile is \(z_j=\Phi^{-1}(p_j)\), where \(\Phi\) is the standard normal cumulative distribution function. Plot \(X_{(j)}\) against \(z_j\). Other plotting conventions are also used; they generally give similar diagnostic guidance. The goal is to assess the pattern, not to demand exact collinearity.
Why the Chi-Square Result Depends on Normality
The normality assumption is not a cosmetic detail in the variance interval. Under the model \(X_i\sim N(\mu,\sigma^2)\) independently, standardize the observations as \(Z_i=(X_i-\mu)/\sigma\). These are independent standard normal variables. The normal-sample result decomposes their total squared variation into a component for the sample mean and a component for deviations around that mean:
For a normal sample, the second component has a chi-square distribution with \(n-1\) degrees of freedom; it equals \((n-1)S^2/\sigma^2\). This result depends on normality as well as independence. If the observations instead have substantial skewness or unusually heavy tails, the standardized sum of squared deviations need not follow that chi-square distribution. Thus the interval's exact coverage claim may no longer apply.
This justification also explains why a variance interval can be sensitive to the shape of the population: squaring deviations gives extreme observations substantial influence. A diagnostic should therefore pay attention to the tails, not only to whether the center looks symmetric.
Worked Diagnostic Examples
Worked Example: A Small Sample With a Fairly Straight Normal Plot
A hypothetical laboratory records seven measurement deviations, in millimetres, from a calibration target: \(-4,-2.2,-1.1,0,1.1,2.2,4\). Assess whether the normal-sample assumption for a chi-square variance interval appears plausible. Use the plotting probability convention above and Tukey fences, taking each quartile as the median of its half of the data.
Step 1: Inspect the shape and center. The ordered values are already listed. They are symmetric around zero: each negative value has a matching positive value, and the middle value is zero. A histogram would have one central region and balanced sides, although seven measurements are too few for a detailed shape assessment.
Step 2: Check the normal probability pattern. For \(n=7\), the plotting positions are \((j-0.375)/7.25\). The probabilities for ranks 1 through 7 are approximately \(0.0862,0.2241,0.3621,0.5000,0.6379,0.7759,0.9138\). The corresponding standard normal quantiles are approximately \(-1.366,-0.758,-0.353,0,0.353,0.758,1.366\). The ordered observations rise symmetrically as these quantiles rise. For example, the absolute data-to-quantile ratios are \(4/1.366\approx2.93\), \(2.2/0.758\approx2.90\), and \(1.1/0.353\approx3.12\). These similar ratios support an approximately linear pattern.
Step 3: Apply the outlier check. The lower half is \(-4,-2.2,-1.1\), so \(Q_1=-2.2\); the upper half is \(1.1,2.2,4\), so \(Q_3=2.2\). Thus \(\mathrm{IQR}=2.2-(-2.2)=4.4\). The fences are \(-2.2-1.5(4.4)=-8.8\) and \(2.2+1.5(4.4)=8.8\). No observation lies outside them.
Conclusion. The symmetry, roughly linear normal plot, and absence of a Tukey-flagged value provide no clear warning against the normal model. With only seven observations, however, the diagnostics have limited power; they support, but do not prove, the assumption. The method used is a combined graphical and fence check, avoiding the mistake of treating a small, tidy sample as conclusive evidence of normality.
Worked Example: A Right-Tailed Sample
A hypothetical claims team records the following eight processing times, in hours, for a single type of uncomplicated claim: \(1,1,2,2,3,4,7,16\). Assess whether a normal-sample chi-square interval for the variance is credible. Use the same quartile convention as in the preceding example.
Step 1: Sort and inspect. The ordered values are \(1,1,2,2,3,4,7,16\). Most times are between 1 and 4 hours, with two larger times, especially 16. The pattern has a long right tail rather than a balanced shape, so a histogram would be expected to show right-skewness.
Step 2: Check the normal plot positions. For \(n=8\), the plotting probabilities are approximately \(0.0758,0.1970,0.3182,0.4394,0.5606,0.6818,0.8030,0.9242\), with normal quantiles approximately \(-1.43,-0.85,-0.47,-0.15,0.15,0.47,0.85,1.43\). The ordered data have repeated low values, followed by a much larger final gap: the last increase is \(16-7=9\), while the preceding increase is \(7-4=3\). The upper end therefore departs sharply from a steady, roughly linear rise.
Step 3: Apply Tukey fences. The lower half is \(1,1,2,2\), giving \(Q_1=(1+2)/2=1.5\). The upper half is \(3,4,7,16\), giving \(Q_3=(4+7)/2=5.5\). Therefore \(\mathrm{IQR}=5.5-1.5=4\), and the upper fence is \(5.5+1.5(4)=11.5\). Since \(16>11.5\), the value 16 is flagged; the lower fence is \(1.5-6=-4.5\), and no observation is below it.
Conclusion. The right tail, the sharp upper-end departure in the normal plot, and the flagged observation all raise concerns about normality. The method avoids the trap of assuming that a large sample variance itself proves a normal model is appropriate. Before using an exact chi-square interval, investigate whether 16 is a recording error, a different claim type, or a genuine long processing time. Do not remove it merely because a rule flags it.
Worked Example: A Flagged Value and Its Effect on the Sample Variance
A hypothetical instrument produces the following twelve readings, in a common unit: \(8,9,9,10,10,11,11,12,12,13,14,30\). Check for a possible outlier, calculate the sample variance with all observations, and explain what the diagnostic implies for a normal-sample variance interval.
Step 1: Calculate the quartiles and fences. The lower six readings are \(8,9,9,10,10,11\), so \(Q_1=(9+10)/2=9.5\). The upper six are \(11,12,12,13,14,30\), so \(Q_3=(12+13)/2=12.5\). Thus \(\mathrm{IQR}=12.5-9.5=3\). The fences are \(9.5-1.5(3)=5\) and \(12.5+1.5(3)=17\). The reading 30 exceeds 17 and is flagged.
Step 2: Calculate the variance using all readings. Their sum is \(149\), and their sum of squares is \(2221\). Hence \(\bar X=149/12\), and the corrected sum of squares is \(2221-149^2/12=2221-22201/12=4451/12\). The sample variance is \(S^2=(4451/12)/11=4451/132\approx33.7197\) square units. Rounded to two decimals, \(S^2=33.72\) square units.
Step 3: See why deleting the flagged value is not a routine fix. If 30 were removed, the remaining sum would be 119, the sum of squares would be 1321, and the sample size would be 11. The resulting sample variance would be \((1321-119^2/11)/10=(1321-14161/11)/10=370/110\approx3.36\) square units. That differs greatly from 33.72 because the extreme observation contributes heavily to squared deviations. This comparison is a warning about influence, not a justification for discarding the reading.
Conclusion. The value 30 merits investigation, and a normal probability plot would be expected to show a strong upper-tail departure. If 30 is a valid observation from the population of interest, deleting it changes the data and the quantity being estimated. The method combines an explicit outlier rule with the sample-variance calculation, avoiding the mistake of removing a genuine observation solely to make the normality assumption look more plausible. The chi-square interval should not be treated as automatically credible until the data source and model are reviewed.
Interpreting Diagnostics Without Overclaiming
The purpose of these checks is to identify reasons for caution. Their interpretation depends on sample size and context. With a small sample, a histogram has few bars and a probability plot has few points; either may look irregular even when the population is normal. Conversely, a sample may look nearly linear simply because there is too little information to reveal a departure. As the sample grows, plots show more detail, but even a modest departure can matter for a procedure whose exact distribution assumes normality.
An outlier rule is a flagging convention, not a test of whether an observation is valid. Check units, transcription, measurement procedures, eligibility, and whether all observations come from the same target population. If the value is an error, correct it using verified source information; if it is a genuine observation, report and analyze it appropriately. Do not remove data just because doing so produces a more convenient interval.
A defensible assessment records what each diagnostic showed and what action followed. If the plots and data context support a normal model, the exact chi-square interval may be reasonable, with the limitation that diagnostics cannot prove normality. If they show pronounced skewness, heavy tails, multiple groups, or a genuine influential observation, state that the exact normal-theory interval may not provide its nominal coverage. Consider a method justified for the actual model rather than silently applying a normal-theory formula.
Check Your Understanding
For each situation, decide what the diagnostic evidence implies and show the requested calculation. Answers are provided for checking.
- A sample of nine instrument errors is \(-6,-4,-3,-1,0,1,3,4,6\). Using median-of-halves quartiles, find the IQR and Tukey fences, then state whether any observation is flagged. (Answer: \(Q_1=-3.5\), \(Q_3=3.5\), IQR \(=7\), fences \(-14\) and \(14\); none is flagged.)
- A sorted sample has observations \(2,3,3,4,5,6,7,18\). Find the upper Tukey fence using median-of-halves quartiles, and explain whether the maximum is flagged. (Answer: \(Q_1=3\), \(Q_3=6.5\), IQR \(=3.5\), upper fence \(=11.75\); 18 is flagged.)
- For \(n=5\), compute the first and last Blom plotting probabilities, \(p_j=(j-0.375)/(n+0.25)\), to four decimals. What does this tell you about assessing tails from such a plot? (Answer: \(p_1=0.1190\), \(p_5=0.8810\); the plot has only five points and limited tail detail.)
- A normal probability plot is nearly straight in the center but bends upward sharply at the largest two observations, and the histogram has a long right tail. State whether an exact normal-sample chi-square interval is automatically justified and give one appropriate diagnostic follow-up. (Answer: No; investigate the upper observations and data context, since the patterns raise a normality concern.)
- One observation exceeds the upper Tukey fence but is confirmed as a valid measurement from the target population. Should it be deleted before calculating the variance interval solely because it was flagged? (Answer: No; a flag prompts investigation, not automatic deletion.)