Choosing Between Chi-Square and t for Normal-Sample Questions

Start With the Quantity Being Studied

A normal-sample problem may mention a sample mean, a sample variance, or both. The correct distribution depends on the statistic in the question, not merely on the fact that the observations are normal. A variance scaled by its population variance leads to a chi-square distribution. A sample mean standardized using the sample standard deviation leads to Student t. Using the wrong distribution usually gives the wrong degrees of freedom as well.

The classification is easiest when the statistic is written explicitly. Ask: Is the question about squared deviations, or is it about a mean measured in estimated standard-error units? In a normal sample, those correspond to different sampling distributions.

Model and notation: Let \(X_1,\ldots,X_n\) be independent observations from the same normal distribution, \(X_i\sim N(\mu,\sigma^2)\), where \(n\geq 2\). Define the sample mean and unbiased sample variance by

\bar{X}=\frac{1}{n}\sum_{i=1}^n X_i, \qquad S^2=\frac{1}{n-1}\sum_{i=1}^n(X_i-\bar{X})^2.

Here \(S\) is the positive square root of \(S^2\). The exact results below require independent, identically distributed normal observations.

The Two Main Classification Rules

The centered sum of squared deviations from the sample mean has \(n-1\) degrees of freedom. After scaling by the population variance, it has a chi-square distribution:

\frac{(n-1)S^2}{\sigma^2}\sim\chi^2_{n-1}.

The sample mean standardized using the estimated standard error has a Student t distribution with \(n-1\) degrees of freedom:

T=\frac{\bar{X}-\mu}{S/\sqrt{n}}\sim t_{n-1}.

The denominator \(S/\sqrt n\) estimates the standard deviation of \(\bar X\). If \(\sigma\) is known and the standardization instead uses \(\sigma/\sqrt n\), the statistic is standard normal, not Student t:

\frac{\bar{X}-\mu}{\sigma/\sqrt n}\sim N(0,1).

Thus, “a normal sample” alone does not tell you whether to use chi-square, t, or standard normal. You must inspect how the statistic is constructed and whether the population standard deviation is known.

Known-mean distinction: If \(\mu\) is specified and the statistic is the sum of squared deviations from that specified mean, then

\frac{\sum_{i=1}^n(X_i-\mu)^2}{\sigma^2}\sim\chi^2_n.

This has \(n\), not \(n-1\), degrees of freedom. The sample-variance result centers deviations at \(\bar X\), which uses one degree of freedom to estimate the mean.

Why the Studentized Mean Has a t Distribution

The distinction between chi-square and t follows from how the sample variance enters the mean statistic. For a normal sample, the standardized mean \(Z=(\bar X-\mu)/(\sigma/\sqrt n)\) has a standard normal distribution. Also, the scaled sample variance \(Q=(n-1)S^2/\sigma^2\) has a chi-square distribution with \(n-1\) degrees of freedom. A special property of a normal sample is that \(Z\) and \(Q\) are independent.

Substitute these definitions into the statistic that uses \(S\). Since \(S^2/\sigma^2=Q/(n-1)\), its estimated standard error is the known standard error multiplied by \(\sqrt{Q/(n-1)}\):

\frac{\bar{X}-\mu}{S/\sqrt n} = \frac{(\bar{X}-\mu)/(\sigma/\sqrt n)} {\sqrt{S^2/\sigma^2}} = \frac{Z}{\sqrt{Q/(n-1)}}.

By definition, the ratio of an independent standard normal variable to the square root of an independent chi-square variable divided by its degrees of freedom has a Student t distribution. Here the chi-square degrees of freedom are \(n-1\), so the ratio has \(t_{n-1}\) distribution. This explains both why estimating \(\sigma\) changes the distribution and why the degrees of freedom are \(n-1\).

A Practical Selection Process

Write the statistic.
Identify whether it contains squared deviations, a sample mean, or both. Do not choose a distribution from the wording alone.

Check the centering.
Squared deviations centered at \(\bar X\) lead to the sample-variance result with \(n-1\) degrees of freedom. Deviations centered at a specified \(\mu\) lead to a chi-square variable with \(n\) degrees of freedom.

Check the standard error.
A mean divided by \(S/\sqrt n\) uses an estimated standard deviation and has a t distribution. If it is divided by \(\sigma/\sqrt n\), with \(\sigma\) known, it has a standard normal distribution.

Set the degrees of freedom and evaluate.
For the sample variance and studentized mean, use \(n-1\). For a sum of squared deviations from a specified known mean, use \(n\). Then use the relevant distribution table or probability calculation.

Worked Examples

Worked Example: A Tail Probability for the Sample Variance

A hypothetical process produces independent measurements from \(N(\mu,3^2)\). A sample of \(n=5\) measurements is taken. Find the probability that the sample variance exceeds \(27\). Use the exact normal-sample distribution.

Method: identify the variance pivot. The question concerns \(S^2\), and the population variance is \(\sigma^2=9\). For \(n=5\), the scaled statistic is

\frac{(n-1)S^2}{\sigma^2} = \frac{4S^2}{9} \sim\chi^2_4.

The event \(S^2>27\) is equivalent to

\frac{4S^2}{9}>\frac{4(27)}{9}=12.

Therefore the required probability is \(P(\chi^2_4>12)\). For four degrees of freedom, the chi-square survival probability is \(e^{-x/2}(1+x/2)\). Substituting \(x=12\) gives

P(\chi^2_4>12) = e^{-12/2}\left(1+\frac{12}{2}\right) = 7e^{-6} \approx 0.0173513.

To four decimal places, the answer is 0.0174. The method avoids treating \(S^2\) itself as chi-square: it is the scaled quantity \(4S^2/9\) that has the chi-square distribution.

Worked Example: A Mean Standardized by the Sample Standard Deviation

Suppose a hypothetical normal population has null mean \(\mu_0=100\) and unknown standard deviation. A sample of \(n=6\) has observed \(\bar X=104\) and \(S=6\). First calculate the observed studentized statistic. Then, under the null model, find the probability that the statistic exceeds \(2.015\).

Method: use Student t because the standard error uses \(S\). The observed statistic is

t_{\mathrm{obs}} = \frac{\bar X-\mu_0}{S/\sqrt n} = \frac{104-100}{6/\sqrt 6} = \frac{4\sqrt 6}{6} = \frac{2\sqrt 6}{3} \approx 1.6330.

Under the null model, the repeated-sample statistic \((\bar X-\mu_0)/(S/\sqrt 6)\) has a \(t_{6-1}=t_5\) distribution. The value \(2.015\) is approximately the 95th percentile of \(t_5\), so

P\left(\frac{\bar X-\mu_0}{S/\sqrt 6}>2.015\right)\approx 0.0500.

The observed value \(1.6330\) is below \(2.015\); the probability calculation describes the upper-tail behavior of the statistic under repeated sampling, not the probability that the fixed observed statistic will change. The method avoids substituting a standard normal distribution when the population standard deviation is unknown and estimated by \(S\).

Worked Example: Known-Mean Sum of Squares Versus Sample Variance

A hypothetical sample of \(n=4\) observations comes from \(N(5,2^2)\). The observed values are \(1,4,7,10\). Calculate the statistic formed from deviations about the specified population mean, identify its distribution and degrees of freedom, and compare it with the sample-variance statistic.

Method: keep the two centers separate. The squared deviations from the specified mean \(\mu=5\) are

(1-5)^2=16,\quad(4-5)^2=1,\quad(7-5)^2=4,\quad(10-5)^2=25.

Their sum is \(16+1+4+25=46\). Since \(\sigma^2=4\), the known-mean statistic is

\frac{\sum_{i=1}^4(X_i-\mu)^2}{\sigma^2} = \frac{46}{4} = 11.5.

Its sampling distribution is \(\chi^2_4\), because the deviations are from the specified known mean and there are four independent normal observations.

For comparison, the observed sample mean is \(\bar X=(1+4+7+10)/4=5.5\). The sum of squared deviations from \(\bar X\) is \(20.25+2.25+2.25+20.25=45\), so \(S^2=45/3=15\). The sample-variance statistic is therefore

\frac{(n-1)S^2}{\sigma^2} = \frac{3(15)}{4} = 11.25 \sim\chi^2_3.

The observed values \(11.5\) and \(11.25\) are different, and the distributions have different degrees of freedom. The method avoids replacing the specified mean by the sample mean, or using \(n-1\) degrees of freedom for a known-mean sum of squares.

Common Mistakes and Why the Choice Matters

These distributions answer different questions. The chi-square pivot describes variation in squared deviations. Student t describes a standardized mean when its standard error includes an estimated standard deviation. Choosing correctly is essential for probabilities and for later inference, because a wrong distribution or degree of freedom changes the tail area or critical value.

Using chi-square for a studentized mean. The statistic \((\bar X-\mu)/(S/\sqrt n)\) is t, not chi-square. Its numerator can be positive or negative, while a chi-square statistic is nonnegative.
Using t for a sample variance. The scaled variance \((n-1)S^2/\sigma^2\) is chi-square, not t.
Using \(n\) degrees of freedom for \(S^2\). The sample mean is estimated from the observations, so the centered sample-variance pivot has \(n-1\) degrees of freedom. A sum of squares centered at a specified known \(\mu\) has \(n\).
Using t when \(\sigma\) is known. If the mean is standardized by \(\sigma/\sqrt n\), the result is standard normal. Student t applies when \(S\) estimates \(\sigma\).
Ignoring the normality condition. The stated chi-square and t laws are exact for an independent normal sample. For a nonnormal population they do not generally hold exactly.

For example, a common arithmetic trap is to stop at \(7e^{-6}\) and report the coefficient \(7\) as the probability. The coefficient must still be multiplied by the exponential:

7e^{-6}=7(0.00247875\ldots)=0.0173513\ldots,

so the probability to four decimal places is \(0.0174\), not \(7\). A probability must lie between zero and one, which is also a useful check.

Key takeaway: For a normal sample, a scaled sample variance has a chi-square distribution with \(n-1\) degrees of freedom; a mean divided by its estimated standard error has a Student t distribution with \(n-1\) degrees of freedom. Always write the statistic before choosing the distribution.

Check Your Understanding

For each problem, identify the distribution and degrees of freedom before using a probability or evaluating the statistic.

A normal sample has \(n=9\) and unknown \(\sigma\). Under a specified mean \(\mu_0\), find \(P((\bar X-\mu_0)/(S/3)<-2.306)\). (Answer: \(t_8\); approximately \(0.0250\).)
A normal sample has \(n=7\), with known population variance \(\sigma^2=16\). Find \(P(S^2>32)\). (Answer: \(\chi^2_6\); \(P(\chi^2_6>12)=25e^{-6}\approx0.0620\).)
For \(n=5\) independent normal observations with specified \(\mu\) and known \(\sigma^2\), classify \(\sum_{i=1}^5(X_i-\mu)^2/\sigma^2\). (Answer: \(\chi^2_5\), with 5 degrees of freedom.)
A normal sample has \(n=4\) and unknown population standard deviation. Under the null mean, what is the upper-tail probability for the studentized mean exceeding \(3.182\)? (Answer: \(t_3\); approximately \(0.0250\).)
A normal sample has \(n=10\) and unknown population variance. A problem asks separately about \(S^2\) and \((\bar X-\mu)/(S/\sqrt{10})\). Classify both statistics and give their degrees of freedom. (Answer: \((n-1)S^2/\sigma^2\) is \(\chi^2_9\); the studentized mean is \(t_9\).)

What You'll Learn