Why Small Calculation Errors Matter
Chi-square and Student t calculations often require only a few steps, but a small error in one step can change the result substantially. Dividing by the wrong variance, using \(n\) instead of \(n-1\) degrees of freedom, or reversing a tail can produce a plausible-looking answer that is nevertheless wrong. A reliable calculation starts by identifying the statistic, its assumptions, and the direction of the requested probability.
The exact sample-variance and Student t results below assume a random sample \(X_1,\ldots,X_n\) from a normal population with mean \(\mu\) and variance \(\sigma^2\). Define the sample mean and unbiased sample variance by
The denominator \(n-1\) is part of the definition of \(S^2\). It is not interchangeable with \(n\). Under the stated normal-sampling assumptions, \((n-1)S^2/\sigma^2\) has a chi-square distribution with \(n-1\) degrees of freedom, and \((\bar{X}-\mu)/(S/\sqrt{n})\) has a Student t distribution with \(n-1\) degrees of freedom. These exact distribution results need not hold for a non-normal population.
A Short Justification for the Sample-Variance Statistic
The factor \((n-1)/\sigma^2\) in the chi-square statistic follows from the normal-sample decomposition. Standardize each observation as \(Z_i=(X_i-\mu)/\sigma\). The sum of their squares has a chi-square distribution with \(n\) degrees of freedom. It splits into a component for the sample mean and a component for deviations around that mean:
For a normal sample, these two components are independent: they are squared lengths of projections onto perpendicular directions in the standardized observation space. The first component has a chi-square distribution with 1 degree of freedom. The full sum has \(n\) degrees of freedom, so the remaining component has \(n-1\) degrees of freedom. Since \(\sum (X_i-\bar{X})^2=(n-1)S^2\), this gives \((n-1)S^2/\sigma^2\sim\chi^2_{n-1}\).
Check the Denominator and the Degrees of Freedom
A dependable first check is to write the sample variance definition before substituting. The numerator \(\sum (X_i-\bar{X})^2\) is divided by \(n-1\) to obtain \(S^2\); multiplying \(S^2\) by \(n-1\) recovers that sum. That is why the chi-square statistic uses \(n-1\), rather than \(n\), in its numerator and degrees of freedom.
Worked Example: Correcting a Sample-Variance Denominator
A hypothetical laboratory measures a quantity on a random sample of \(n=3\) items from a normal population with known variance \(\sigma^2=4\). The reported sample variance is \(S^2\), calculated with denominator \(n-1\). Find \(P(S^2>12)\). A proposed solution uses \(3S^2/4\) as the chi-square statistic. Diagnose and correct the error.
Method and assumption. Use the exact normal-sample result. With \(n=3\), the degrees of freedom are \(n-1=2\), and the correctly scaled statistic is
Translate the event. Because the scale factor \(2/4\) is positive, the inequality direction stays the same:
For 2 degrees of freedom, \(P(\chi^2_2>x)=e^{-x/2}\) for \(x\geq 0\). Therefore,
The proposed calculation used \(3S^2/4\), which incorrectly substitutes \(n\) for \(n-1\). It would transform the threshold to \(3(12)/4=9\), but that is not the statistic with the stated sample-variance definition. The corrected probability is approximately \(0.0498\), to four decimal places. Common mistake avoided: replacing the sample-variance denominator \(n-1\) with \(n\).
Track the Scale and Its Square
When a statistic is rescaled, every factor must be applied to the quantity being calculated. In particular, variance has squared units. Under normal sampling, the sample variance has
The fourth power in \(\sigma^4\) is essential. It arises because \(S^2\) is proportional to \(\sigma^2\), and the variance of a quantity scaled by \(\sigma^2\) is scaled by \((\sigma^2)^2=\sigma^4\). The degrees of freedom in the denominator are \(n-1\), matching the chi-square variable that generates \(S^2\).
Worked Example: Correcting the Variance of a Sample Variance
A hypothetical instrument is used on \(n=5\) independent observations from a normal population with variance \(\sigma^2=4\). Find the mean and variance of the sample variance \(S^2\). A proposed calculation gives \(\operatorname{Var}(S^2)=2\sigma^2/(n-1)\).
Method. Apply the normal-sample formulas, with \(n-1=4\) degrees of freedom. First, the mean is
For the variance, square the population variance before substituting. Since \(\sigma^2=4\), \(\sigma^4=(\sigma^2)^2=4^2=16\). Thus,
The proposed expression uses \(\sigma^2\) instead of \(\sigma^4\), giving \(2(4)/4=2\), which is incorrect. The variance of \(S^2\) has units equal to the square of the units of \(S^2\), so it must involve \(\sigma^4\). The checked results are \(E[S^2]=4\) and \(\operatorname{Var}(S^2)=8\). Common mistake avoided: treating the variance parameter as though it were not being squared.
Translate the Tail Direction Before Looking Up a Probability
A chi-square variable is nonnegative and generally asymmetric. An event of the form \(S^2<c\) becomes a lower-tail chi-square event; an event of the form \(S^2>c\) becomes an upper-tail event. Multiplication or division by a positive constant preserves the inequality direction. After translating the event, check whether the table or calculator reports a lower-tail cumulative probability or an upper-tail probability.
Worked Example: Avoiding a Reversed Chi-Square Tail
Suppose a random sample of \(n=3\) observations comes from a normal population with variance \(\sigma^2=2\). Find \(P(S^2<1)\), where \(S^2\) uses denominator \(n-1\). A proposed solution uses the upper-tail probability above 1.
Method and assumption. The sample is normal, so \(W=2S^2/2=S^2\sim\chi^2_2\). The event is a lower-tail event because the scaling factor is positive:
For 2 degrees of freedom, the lower-tail probability at 1 is
The upper-tail probability is \(e^{-1/2}\approx0.6065\), which answers \(P(S^2>1)\), not the requested probability. The two probabilities sum to 1, as they should for complementary events. Common mistake avoided: selecting the upper tail merely because a chi-square table is arranged by upper-tail areas, without first translating the inequality.
Use Student t When the Population Variance Is Unknown
For a normal sample, if \(\sigma\) is known, standardizing the sample mean uses the normal statistic \(Z=(\bar{X}-\mu)/(\sigma/\sqrt{n})\). If \(\sigma\) is unknown and replaced by the sample standard deviation \(S=\sqrt{S^2}\), the statistic is instead Student t:
The denominator \(S/\sqrt{n}\) is the estimated standard error of the sample mean. The degrees of freedom are \(n-1\), because \(S^2\) is calculated from deviations around \(\bar{X}\). Using a standard normal probability after substituting \(S\) for \(\sigma\) ignores the additional uncertainty from estimating the population variance.
Worked Example: Correcting Normal Standardization to Student t
A hypothetical study observes a random sample of \(n=5\) measurements from a normal population. The sample mean is \(\bar{X}=12\) and the sample standard deviation is \(S=2.5\). To assess \(H_0:\mu=10\), calculate the observed test statistic and its two-sided null probability. A proposed solution uses a standard normal distribution after substituting \(S\) for \(\sigma\).
Method and assumption. The population is normal, the population variance is unknown, and the sample standard deviation is used. Therefore, use Student t with \(n-1=4\) degrees of freedom. The standard error is
The observed statistic is
For \(t_4\), this value has \(t_{\text{obs}}^2=3.2\), and the two-sided probability is \(4/27\approx0.1481\). Thus the null probability is approximately \(0.1481\), to four decimal places. A standard normal calculation would give a smaller two-sided probability because the normal tails are lighter than the \(t_4\) tails. Common mistake avoided: using a normal distribution after estimating \(\sigma\) with \(S\). The normal standardization applies when \(\sigma\) is known; this calculation uses \(S\), so its reference distribution is \(t_4\).
A Practical Error-Checking Sequence
Before accepting a chi-square or Student t result, check each link between the data and the reference distribution. The following sequence is useful when solving a multi-step problem or reviewing a proposed calculation.
For the exact sample-variance chi-square and Student t results, verify that observations are an independent random sample from a normal population.
Use \((n-1)S^2/\sigma^2\) for the sample-variance chi-square statistic and \((\bar{X}-\mu)/(S/\sqrt{n})\) for Student t.
For these one-sample normal results, both reference distributions have \(n-1\) degrees of freedom.
Apply positive scale factors to both sides of an inequality, then identify whether the result is a lower-tail or upper-tail probability.
Known \(\sigma\) gives normal standardization of the sample mean; replacing it with \(S\) gives a Student t statistic under normal sampling.
Common Exam Traps
| Error | Why it fails | Correction |
|---|---|---|
| Using \(nS^2/\sigma^2\) | The defined sample variance uses denominator \(n-1\). | Use \((n-1)S^2/\sigma^2\), with \(n-1\) degrees of freedom. |
| Using \(2\sigma^2/(n-1)\) for \(\operatorname{Var}(S^2)\) | The variance of \(S^2\) scales with \((\sigma^2)^2\). | Use \(2\sigma^4/(n-1)\). |
| Reversing a lower-tail event | Upper-tail tables can make the wrong tail easy to select. | Translate the inequality first, then use the matching tail. |
| Using normal probabilities with \(S\) in the denominator | Estimating \(\sigma\) adds uncertainty to the standardized statistic. | For a normal sample, use Student t with \(n-1\) degrees of freedom. |
These checks are aligned with the SOA VEE Mathematical Statistics syllabus because they focus on applying sampling-distribution results correctly, not simply recalling a distribution name. A useful final check is dimensional: a variance has squared units, and a sample-variance statistic divided by \(\sigma^2\) is dimensionless.
Check Your Understanding
For each problem, identify the appropriate scaling or reference distribution before calculating.
- A normal sample has \(n=4\), population variance \(\sigma^2=2\), and sample variance \(S^2\) defined with denominator \(n-1\). Express \(P(S^2>5)\) as a chi-square tail probability. (Answer: \(P(\chi^2_3>7.5)\).)
- A normal sample has \(n=8\) and population variance \(\sigma^2=9\). Find \(\operatorname{Var}(S^2)\). (Answer: \(2(9^2)/7=162/7\approx23.1429\).)
- A normal sample has \(n=6\), \(\bar{X}=18\), and \(S=3\). For \(H_0:\mu=15\), find the observed Student t statistic and degrees of freedom. (Answer: \(t_{\text{obs}}=3/(3/\sqrt{6})=\sqrt{6}\approx2.4495\), with 5 degrees of freedom.)
- A normal sample has \(n=3\), population variance \(\sigma^2=2\), and sample variance \(S^2\) with denominator \(n-1\). Find \(P(S^2<1)\). (Answer: \(P(\chi^2_2<1)=1-e^{-1/2}\approx0.3935\).)
- A normal sample has \(n=9\), known population standard deviation \(\sigma=6\), and sample mean \(\bar{X}=52\). Under \(H_0:\mu=50\), find the one-sided upper-tail probability using the correct standardization. (Answer: \(Z=(52-50)/(6/\sqrt{9})=1\), so \(P(Z>1)\approx0.1587\).)