When the Population Spread Is Unknown
In “Point Estimates and Margin of Error for a Mean,” you saw that the sample mean \(\bar{x}\) estimates a population mean \(\mu\), and that \(s/\sqrt{n}\) estimates the standard error of \(\bar{x}\). The next question is which critical multiplier should go with that estimated standard error when we build a confidence interval.
If the population standard deviation \(\sigma\) is known, the standard error of \(\bar{x}\) is \(\sigma/\sqrt{n}\). Under the appropriate Normal model, a standard Normal critical value, often called a \(z^*\) value, can be used. In most real investigations, however, \(\sigma\) is unknown. We estimate it with the sample standard deviation \(s\), so the standard error in the interval is estimated as \(s/\sqrt{n}\).
That substitution matters. The sample standard deviation varies from sample to sample, so \(s/\sqrt{n}\) is itself an estimate and not a fixed, known standard error. A \(t\) critical value accounts for the added uncertainty. For a one-sample \(t\) confidence interval, the basic form is:
The \(t\) distribution has a shape similar to the standard Normal distribution, but its tails are heavier: it assigns more probability to values far from its center. This accounts for the extra variability from estimating \(\sigma\) with \(s\). For the same confidence level and a finite sample, the \(t^*\) critical value is larger than the corresponding \(z^*\) value. Holding the observed \(s\) and \(n\) fixed, that makes the \(t\) interval wider than a z-based interval.
The choice is not about whether the data are quantitative: both procedures concern a population mean. It is about whether the population standard deviation is known. If it is unknown, substituting \(s\) does not make the uncertainty disappear. The \(t\) distribution incorporates that uncertainty through its heavier tails.
What the Heavier Tails Accomplish
Imagine repeatedly taking samples from a Normal population. If \(\sigma\) were known, you could standardize each sample mean using \(\sigma/\sqrt{n}\). If \(\sigma\) is not known, you instead standardize using the sample-based estimate \(s/\sqrt{n}\). Because \(s\) changes from sample to sample, the resulting standardized values vary more than they would with a known \(\sigma\). Under the Normal population model, this standardized quantity follows a \(t\) distribution rather than the standard Normal distribution.
A \(t\) distribution is centered at zero and is symmetric, like the standard Normal distribution. But it has more probability in its tails. To capture a specified central confidence level, its critical value must reach farther from zero. That greater critical value produces a larger margin of error and a wider interval, allowing for the added uncertainty in estimating the population spread.
There is a family of \(t\) distributions. The particular member used depends on the sample size, through its degrees of freedom. For a one-sample \(t\) interval, the degrees of freedom are \(n-1\). The details of degrees of freedom and how to select a critical value are developed in the next tutorial. The key point here is that with more degrees of freedom, the \(t\) distribution gets closer to the standard Normal distribution. A larger sample generally provides a more stable estimate of spread, so the extra adjustment becomes smaller.
Worked Examples: Why the Multiplier Changes
Worked Example: A Small Sample of Sensor Readings
A technician takes a random sample of 9 sensors from a large production population. Their average reading is 42 units, and the sample standard deviation is 6 units. Assume the population distribution is approximately Normal. Compare the estimated standard error and 95% margins of error using \(t^*=2.306\) and, just for comparison, \(z^*=1.960\).
State. The parameter is \(\mu\), the population mean sensor reading. The sample mean is \(\bar{x}=42\) units, and the population standard deviation is not known, so the t-based interval is the appropriate choice.
Plan. Use \(s/\sqrt{n}\) to estimate the standard error. The sample is random, and it is less than 10% of the large production population, so the 10% condition is met and independence is reasonable. The approximately Normal population model supports using a one-sample \(t\) procedure with this small sample. Compare the supplied critical values while keeping the standard error the same.
Do. The estimated standard error is
Using the t critical value, the margin of error is
The t interval is \(42\pm4.612\), or \((37.388,\ 46.612)\) units. Using the z critical value only for comparison gives a margin of error of
and a z-based interval of \(42\pm3.920\), or \((38.080,\ 45.920)\) units.
Conclude. Because the population standard deviation is unknown, the t-based interval is appropriate. Its margin of error is larger than the comparison margin of error because \(2.306\) is larger than \(1.960\). The t distribution’s heavier tails account for the extra uncertainty from estimating spread using this small sample.
Worked Example: A Random Sample of Water-Filter Lifetimes
An engineer samples 25 water filters at random from a large production line to estimate mean lifetime. The sample mean is 18.4 months, and the sample standard deviation is 3.5 months. The population standard deviation is unknown. The observations show no strong skewness or outliers. Use \(t^*=2.064\) for a 95% interval and compare its margin of error with one calculated using \(z^*=1.960\).
State. The parameter is \(\mu\), the population mean filter lifetime. The point estimate is \(\bar{x}=18.4\) months. Since \(\sigma\) is unknown, use the one-sample \(t\) interval.
Plan. The data come from a random sample. The sample is less than 10% of the large population, so the 10% condition supports treating observations as independent. With 25 observations and no strong skewness or outliers, the sample is suitable for a one-sample \(t\) interval. Calculate \(s/\sqrt{n}\), then multiply by \(t^*\) for the t margin of error. Use \(z^*\) only to illustrate the difference in critical values.
Do. The estimated standard error is
The t margin of error is
The t interval is
or approximately \((16.96,\ 19.84)\) months. For comparison, the z-based margin of error is \(1.960(0.70)=1.372\) months, giving an interval of \((17.028,\ 19.772)\) months.
Conclude. The t-based interval gives plausible values of about 16.96 to 19.84 months for the population mean filter lifetime. It is a little wider than the comparison z interval because the t critical value is larger. The difference is modest here, but the t procedure is still the appropriate choice because the population standard deviation is unknown.
Worked Example: A Larger Sample and a Smaller Adjustment
A random sample of 100 seedlings from a large nursery population has a mean height of 5.2 centimeters and a sample standard deviation of 1.5 centimeters. Assume the sample’s distribution has no extreme outliers. For a 95% interval, compare \(t^*=1.984\) with \(z^*=1.960\).
State. The population mean seedling height is \(\mu\), and \(\bar{x}=5.2\) centimeters is its point estimate. Since \(\sigma\) is unknown, the t procedure is appropriate.
Plan. The sample is random and less than 10% of the large nursery population, meeting the 10% condition. With 100 observations and no extreme outliers, a t procedure is reasonable. Calculate the estimated standard error and compare the margins of error using the supplied critical values.
Do. The estimated standard error is
The t margin of error is \(1.984(0.15)=0.2976\) centimeters. The t interval is \(5.2\pm0.2976\), or \((4.9024,\ 5.4976)\) centimeters. Using the z value for comparison gives a margin of error of \(1.960(0.15)=0.294\) centimeters and an interval of \((4.906,\ 5.494)\) centimeters.
Conclude. The t interval is slightly wider, but the difference between the critical values is small in this example. As sample size increases, the t distribution approaches the standard Normal distribution, so the t adjustment typically becomes less noticeable. The t procedure remains the right choice because \(\sigma\) is unknown.
Common Mistakes and AP Exam Tips
- Using z just because the sample is large. A large sample can make the t and z critical values very close, but it does not make an unknown \(\sigma\) known. For a one-sample confidence interval for a mean, use \(t\) when \(\sigma\) is unknown.
- Confusing \(s\) with \(\sigma\). The sample standard deviation \(s\) describes spread in the sample and estimates the population standard deviation \(\sigma\). It is not the known population value.
- Saying the t interval is wider because \(s\) is always larger than \(\sigma\). That is not true. A sample standard deviation can be larger or smaller than the population standard deviation. The t adjustment comes from the uncertainty in estimating \(\sigma\), represented by the heavier tails and larger critical value at a fixed confidence level.
- Claiming t is always substantially wider. At a fixed confidence level, the t critical value exceeds the corresponding z critical value for finite degrees of freedom, but the difference can be very small for a large sample.
- Forgetting to check conditions. State that the sample is random, address independence using the 10% condition for sampling without replacement, and assess whether the population is approximately Normal or the sample is large enough without severe skewness or extreme outliers.
- Leaving out the reason for choosing t. A full-credit explanation connects the method to the parameter and the information available: “Because the population standard deviation is unknown and is estimated by the sample standard deviation, use a one-sample \(t\) interval for the population mean.”
The t procedure does not fix bias from a poor sampling method, and a heavier tail is not a claim that the population itself must have heavy tails. It describes the distribution used for the standardized sample mean when spread is estimated from the sample, under the model and conditions for the procedure.
Check Your Understanding
Focus on why the procedure uses t and what the t adjustment represents.
- A random sample of 16 garden plots has a sample standard deviation of 4 kilograms. The population standard deviation is unknown. Should a confidence interval for the population mean use t or z, and why?
- In your own words, explain why using \(s\) instead of \(\sigma\) adds uncertainty to a confidence interval for a mean.
- For a fixed confidence level and sample size, why is a t-based margin of error generally larger than a z-based margin of error when both use the same estimated standard error?
- How does the difference between the t and standard Normal distributions generally change as the sample size grows?
- A student says, “The t interval is wider because the sample standard deviation must be larger than the population standard deviation.” Explain what is wrong with this claim.