Replacing Sigma Without Changing The Critical Value

Why the Critical Value Must Match the Standard Error

A confidence interval for a population mean uses a critical value to determine how far its endpoints extend from the sample mean. That value is tied to the distribution of the standardized statistic. When the population standard deviation is known, a Z critical value can be appropriate. When the standard deviation is unknown and replaced by a sample estimate, the statistic has additional variability. Keeping the Z critical value while changing the standard error does not keep the same exact confidence level.

This distinction matters in actuarial work whenever a mean is estimated from a limited sample, such as hypothetical claim-processing times or hypothetical costs. A plug-in Z interval can be a useful large-sample approximation, but it is not the exact normal-theory interval. Under a normal-sample model, the exact interval uses a Student t critical value with \(n-1\) degrees of freedom.

Definition: For a sample \(X_1,\ldots,X_n\), the sample mean is \(\bar X=\frac{1}{n}\sum_{i=1}^n X_i\), and the sample standard deviation is \(S=\sqrt{\frac{1}{n-1}\sum_{i=1}^n(X_i-\bar X)^2}\). The observed values are denoted \(\bar x\) and \(s\). The population standard deviation is \(\sigma\); it is a parameter, not the observed sample standard deviation.

Known and Estimated Standard Deviations

Suppose observations are independent and identically distributed from a normal population with mean \(\mu\) and variance \(\sigma^2\). If \(\sigma\) is known, then \((\bar X-\mu)/(\sigma/\sqrt n)\) has a standard normal distribution. A two-sided \(100(1-\alpha)\%\) confidence interval is therefore:

\bar x \mathbin{\pm} z_{1-\alpha/2}\frac{\sigma}{\sqrt n}.

Here \(z_{1-\alpha/2}\) is the standard normal quantile with probability \(1-\alpha/2\) to its left. For example, \(z_{0.975}=1.96\) for a 95% interval. This interval is exact under the normal model with known \(\sigma\), for any sample size.

If \(\sigma\) is unknown, replacing it with \(s\) changes the statistic. Under the independent normal-sample model, the exact two-sided interval is:

\bar x \mathbin{\pm} t_{1-\alpha/2,\ n-1}\frac{s}{\sqrt n},

where \(t_{1-\alpha/2,\ n-1}\) is the corresponding Student t quantile. The degrees of freedom are \(n-1\), because estimating the sample mean imposes the constraint \(\sum_{i=1}^n(X_i-\bar X)=0\) on the \(n\) deviations.

Key distinction: Substituting \(s\) for \(\sigma\) but retaining \(z_{1-\alpha/2}\) gives a plug-in Z interval. It is not the exact t interval. For an exact normal-theory interval with unknown \(\sigma\), use the t critical value with \(n-1\) degrees of freedom.

Why Estimating the Standard Deviation Changes the Critical Value

The reason can be seen from the standardized statistic. With known \(\sigma\), the sample mean is standardized by the fixed quantity \(\sigma/\sqrt n\). With unknown \(\sigma\), it is standardized by the random quantity \(S/\sqrt n\). In a normal sample, the ratio

T=\frac{\bar X-\mu}{S/\sqrt n}

has a Student t distribution with \(n-1\) degrees of freedom, not a standard normal distribution. The random denominator adds uncertainty, especially when the sample is small. The t distribution has heavier tails than the standard normal distribution, so its critical value is larger at a given confidence level.

A short justification shows how the exact interval follows. Under the normal-sample assumptions, \(T\) has the stated t distribution. Thus, by the definition of the t quantiles:

P\left(-t_{1-\alpha/2,\ n-1}\leq \frac{\bar X-\mu}{S/\sqrt n} \leq t_{1-\alpha/2,\ n-1}\right)=1-\alpha.

Since \(S/\sqrt n\) is positive, multiplying all parts of the inequality by it and rearranging to isolate \(\mu\) gives:

P\left(\bar X-t_{1-\alpha/2,\ n-1}\frac{S}{\sqrt n} \leq \mu \leq \bar X+t_{1-\alpha/2,\ n-1}\frac{S}{\sqrt n}\right)=1-\alpha.

This is the exact t interval. If the t quantile in this argument is replaced by a Z quantile, the probability statement no longer follows: the statistic in the middle has a t distribution, not a standard normal distribution.

Worked Examples

Worked Example: Comparing Z and t for a Small Normal Sample

A hypothetical insurer models the processing time, in days, for a particular type of claim as normally distributed. A random sample of \(n=25\) independent claims has \(\bar x=72.4\) days and \(s=10.0\) days. The population standard deviation is unknown. Calculate and compare the plug-in 95% Z interval and the exact 95% t interval.

Method. The estimated standard error is:

\frac{s}{\sqrt n}=\frac{10.0}{\sqrt{25}}=\frac{10.0}{5}=2.0\text{ days}.

For the plug-in Z interval, use \(z_{0.975}=1.96\). Its margin of error is \(1.96(2.0)=3.92\) days, so the interval is:

72.4\mathbin{\pm}3.92=[68.48,\ 76.32]\text{ days}.

For the exact t interval, the degrees of freedom are \(25-1=24\), and \(t_{0.975,24}=2.064\). Its margin of error is \(2.064(2.0)=4.128\) days, giving:

72.4\mathbin{\pm}4.128=[68.272,\ 76.528]\text{ days}.

Rounded to two decimals, the exact t interval is \([68.27,\ 76.53]\) days. It is wider because the t critical value accounts for estimating \(\sigma\). Common mistake avoided: using \(s\) in the standard error but treating the resulting statistic as standard normal. The plug-in Z interval is a different procedure, not the exact normal-theory interval.

Worked Example: A 90% Interval for a Hypothetical Cost Measure

In a hypothetical study, nine independent observations are drawn from a normal population of annual service costs, measured in thousands of dollars. The sample mean is \(\bar x=14.2\), and the sample standard deviation is \(s=3.0\). The population standard deviation is unknown. Find both the plug-in 90% Z interval and the exact 90% t interval.

The estimated standard error is \(3.0/\sqrt 9=3.0/3=1.0\) thousand dollars. For a two-sided 90% interval, \(z_{0.95}=1.645\). The plug-in Z interval is therefore:

14.2\mathbin{\pm}1.645(1.0)=[12.555,\ 15.845].

For the exact t interval, \(n-1=8\) degrees of freedom and \(t_{0.95,8}=1.860\). Its margin of error is \(1.860(1.0)=1.860\) thousand dollars, so:

14.2\mathbin{\pm}1.860=[12.340,\ 16.060].

Rounded to two decimals, the intervals are \([12.56,\ 15.85]\) and \([12.34,\ 16.06]\) thousand dollars, respectively. The t interval is wider. Common mistake avoided: using the 90% Z quantile merely because the confidence level is 90%. The choice of critical-value distribution depends on the statistic, not just the confidence level.

Worked Example: How the Difference Shrinks in a Larger Sample

A hypothetical sample of \(n=100\) independent observations comes from a normal population of measurement values. The observed mean is \(\bar x=508\), and the observed standard deviation is \(s=40\). The population standard deviation is unknown. Compare the 99% plug-in Z interval with the exact t interval.

The estimated standard error is \(40/\sqrt{100}=40/10=4\). For 99% confidence, \(z_{0.995}=2.576\), giving a Z margin of \(2.576(4)=10.304\). The plug-in interval is:

508\mathbin{\pm}10.304=[497.696,\ 518.304].

The exact t interval has \(100-1=99\) degrees of freedom. Using \(t_{0.995,99}=2.626\), its margin is \(2.626(4)=10.504\), so:

508\mathbin{\pm}10.504=[497.496,\ 518.504].

Rounded to two decimals, the intervals are \([497.70,\ 518.30]\) and \([497.50,\ 518.50]\). They are close, but not identical. Common mistake avoided: concluding that a large sample makes the Z and t procedures exactly the same. The t critical value approaches the Z critical value as the degrees of freedom grow, but for finite \(n\) they can differ.

When the Plug-In Z Interval Is an Approximation

Under suitable large-sample conditions, replacing \(\sigma\) with a consistent estimate \(S\) can produce an interval with coverage close to its nominal level. In particular, for independent, identically distributed observations with finite variance, the sample mean is approximately normal for large \(n\), and \(S\) estimates \(\sigma\). This motivates a plug-in Z interval as an approximation. It does not make that interval exact for a finite sample.

For a normal population with unknown \(\sigma\), the exact t interval is available for every sample size. If the data are not from a normal population, the t interval is not generally exact; its performance depends on the sample size and the population distribution. A large sample may support an approximation, but a small sample with substantial skewness or outliers needs particular care. The normality assumptions should not be silently replaced by a claim that using \(s\) automatically justifies a Z interval.

For a normal sample, the coverage of a plug-in Z interval using the usual two-sided critical value is \(P(-z_{1-\alpha/2}\leq T\leq z_{1-\alpha/2})\), where \(T\) has a t distribution with \(n-1\) degrees of freedom. Because the t distribution has heavier tails than the standard normal, this central probability is less than \(1-\alpha\) at the same finite degrees of freedom. This explains the undercoverage risk of retaining the Z critical value.

Exam trap: Do not choose a critical value solely by recognizing the expression “sample mean plus or minus critical value times standard error.” First identify whether the population standard deviation is known. If it is unknown, determine whether the question calls for an exact t procedure under a normal-sample model or an explicitly approximate large-sample procedure.

A Practical Decision Sequence

Check whether \(\sigma\) is known.
If it is known and the normal model applies, use the Z interval with \(\sigma/\sqrt n\). Do not substitute \(s\) unless an approximation is intended.

If \(\sigma\) is unknown, check the model and sample size.
For an independent normal sample, use the exact t interval. For a large-sample approximation, state that it is approximate and assess whether the conditions are reasonable.

Match the critical value to the statistic.
Use the t quantile with \(n-1\) degrees of freedom when the exact statistic uses \(S/\sqrt n\). Then calculate the standard error and interval endpoints.

Confidence level describes the long-run coverage of a procedure over repeated samples, not the probability that a fixed parameter changes after an interval has been calculated. If an exact 95% t procedure is used under its assumptions, 95% of intervals constructed this way cover \(\mu\) in repeated sampling. A plug-in Z interval may be close to 95% in a large sample, but the substitution alone does not establish exact 95% coverage.

Check Your Understanding

For each problem, distinguish the plug-in Z calculation from the exact t procedure where requested. Use the stated normal-sample assumptions for exact t intervals.

A normal population has unknown standard deviation. A sample of 36 has \(\bar x=41.8\) and \(s=6.0\). Calculate the 95% plug-in Z interval and the exact t interval using \(t_{0.975,35}=2.030\). Round endpoints to two decimals. (Answer: Z \([39.84,\ 43.76]\); t \([39.77,\ 43.83]\).)
A hypothetical normal model for independent service durations has unknown population standard deviation. For \(n=10\), \(\bar x=18.5\) hours and \(s=4.0\) hours. Compare 90% intervals using \(z_{0.95}=1.645\) and \(t_{0.95,9}=1.833\). Round endpoints to two decimals. (Answer: Z \([16.42,\ 20.58]\) hours; t \([16.18,\ 20.82]\) hours.)
For a normal population with unknown standard deviation, a sample of 64 has \(\bar x=102\), \(s=16\), and the requested confidence level is 99%. Use \(z_{0.995}=2.576\) and \(t_{0.995,63}=2.656\). Calculate both intervals to two decimals and state which is wider. (Answer: Z \([96.85,\ 107.15]\); t \([96.69,\ 107.31]\); the t interval is wider.)
A report gives \(n=15\), \(\bar x=6.4\), and \(s=1.5\) from an independent normal sample with unknown \(\sigma\). The analyst reports a 95% interval using \(z_{0.975}=1.96\). Calculate the exact interval using \(t_{0.975,14}=2.145\), and identify the issue with the analyst’s procedure. Round to two decimals. (Answer: exact t interval \([5.57,\ 7.23]\); the analyst used the plug-in Z critical value instead of the exact t critical value.)
A sample of eight independent observations is strongly right-skewed, and the population standard deviation is unknown. Is the exact normal-theory t interval automatically justified, and does substituting the sample standard deviation make a Z interval exact? Give the reason. (Answer: No to both; the stated small-sample data do not support the normal-sample assumption for exact t coverage, and substituting \(s\) does not make a Z interval exact.)

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