What a t Critical Value Represents
In “The t Distribution and Degrees of Freedom,” you saw that a one-sample t procedure uses \(df=n-1\), and that the t distribution’s tails depend on the degrees of freedom. This tutorial focuses on finding a cutoff, called the t critical value, for a specified confidence level and \(df\).
A t critical value is a point on the horizontal axis of a t distribution. For a two-sided confidence level, the positive critical value \(t^*\) leaves the stated confidence level in the middle of the curve, with equal areas in the two tails. The corresponding negative cutoff is \(-t^*\). For example, a central 95% area leaves 5% outside the central region: 2.5% in each tail.
The significance level associated with a central confidence level is \(\alpha=1-C\). For a two-sided critical value, each tail has area \(\alpha/2\). Because calculator commands such as invT use the cumulative area to the left of a cutoff, the left-tail area at the positive cutoff is \(1-\alpha/2\), or equivalently \((1+C)/2\).
Keep the central confidence area, the area in one tail, and the cumulative area entered into invT distinct. For a central 90% area, \(C=0.90\), so \(\alpha=1-0.90=0.10\). Each tail has area \(0.10/2=0.05\), and the positive cutoff has \(1-0.05=0.95\) to its left. Thus the central area is 0.90, while the invT input for the positive cutoff is 0.95.
Finding t* in a t Table
A t table organizes critical values by degrees of freedom and tail area. Tables do not all use the same column headings: a table may label columns by one-tail area, two-tail area, or confidence level. Read the heading before selecting a column. For a central confidence level \(C\), the two-tail area is \(1-C\), and the one-tail area is \((1-C)/2\).
For example, a central 95% critical value corresponds to two-tail area \(0.05\), or one-tail area \(0.025\). A central 90% critical value corresponds to two-tail area \(0.10\), or one-tail area \(0.05\). These are two ways of describing the same pair of tails.
| Central confidence level | Total area in both tails | Area in each tail | Left area at positive t* |
|---|---|---|---|
| 90% | 0.10 | 0.05 | 0.95 |
| 95% | 0.05 | 0.025 | 0.975 |
| 99% | 0.01 | 0.005 | 0.995 |
To look up \(t^*\), first find the row for \(df\), then find the column that represents the correct tail area or confidence level. The entry is the positive cutoff. For a one-sample t procedure with sample size \(n\), use \(df=n-1\), as established in the earlier tutorial on degrees of freedom.
Some printed tables do not include every possible \(df\). If a table gives only selected rows, a common conservative approach for a confidence interval is to use the next smaller listed \(df\). Because smaller \(df\) gives a larger \(t^*\) at the same confidence level, this choice does not understate the critical value. If the table provides the exact row, use it rather than substituting another row.
Finding t* with invT
A calculator’s invT function returns a t cutoff from a cumulative area to the left and a degrees-of-freedom value. For a two-sided confidence level, use the cumulative area at the positive cutoff—not the central confidence level and not the area in one tail.
For a one-sample t procedure, calculate \(df=n-1\).
For confidence level \(C\), calculate \(\alpha=1-C\), then divide \(\alpha\) by 2.
Calculate \(1-\alpha/2\), which is also \((1+C)/2\).
Enter the left area and \(df\). The result is the positive \(t^*\) for the central confidence level.
On a TI-84, the command is commonly written as invT(area, df). For instance, invT(0.975, 14) finds the t cutoff with area 0.975 to its left and 14 degrees of freedom. If a task asks for the negative cutoff instead, the distribution’s symmetry means it is \(-t^*\). A confidence interval uses the positive critical value as a multiplier.
Worked Examples: Finding t Critical Values
Worked Example: Using a Table for a 95% Critical Value
A random sample of 15 values will be used for a one-sample t procedure. Find the 95% t critical value using a table that includes a row for \(df=14\) and columns for central confidence levels.
Find the degrees of freedom. The sample size is \(n=15\), so
Match the confidence level to the table. The central confidence level is 95%, so the total tail area is
Each tail has area \(0.05/2=0.025\). In a table with a two-tail-area heading, use the 0.05 column; in a table with a one-tail-area heading, use the 0.025 column. In the \(df=14\) row, the critical value is approximately \(2.145\).
Check with invT. The area to the left of the positive cutoff is \(1-0.025=0.975\). Using invT gives
Answer. The 95% t critical value with 14 degrees of freedom is \(t^*\approx2.145\). The table and calculator agree when they use the same confidence level and \(df\).
Worked Example: Finding a 90% Critical Value with invT
A sample of 10 water filters is used to estimate a population mean. Find the positive critical value for a central 90% t area.
Find the degrees of freedom. With \(n=10\),
Find the correct invT area. The central area is \(C=0.90\), so the total area outside it is
This 0.10 is split equally between the tails, giving \(0.10/2=0.05\) in each tail. Therefore, the positive cutoff has \(1-0.05=0.95\) to its left. The central area is 0.90; the cumulative area entered in invT is 0.95.
Use invT and check the result.
A t table gives approximately 1.833 at \(df=9\) for a one-tail area of 0.05, or equivalently a two-tail area of 0.10. A common error is to enter 0.90 into invT for this two-sided critical value. That would find a different cutoff: it is the 90th percentile, not the positive cutoff that leaves 90% in the center.
Answer. For a central 90% t area with 9 degrees of freedom, \(t^*\approx1.833\).
Worked Example: Finding a 99% Critical Value from Either Tool
A one-sample t procedure has \(df=24\). Find its positive 99% critical value, and explain why it is larger than the 95% critical value with the same degrees of freedom.
Plan. The table and invT both require the degrees of freedom and the correct tail area. For a 99% central area, the remaining 1% is split into two tails.
Do: calculate the area. The total tail area is
The cumulative area to the left of the positive cutoff is \(1-0.005=0.995\). Thus,
A table lookup uses \(df=24\) and a two-tail area of 0.01, or a one-tail area of 0.005, and gives approximately 2.797. For comparison, the 95% critical value at \(df=24\) is approximately 2.064. The 99% critical value is larger because capturing more area in the center requires the cutoffs to be farther from zero.
Conclude. With 24 degrees of freedom, the positive t critical value for a central 99% area is approximately 2.797. The higher confidence level requires a larger critical value than the 95% level with the same \(df\).
Common Mistakes and AP Exam Tips
- Entering the confidence level directly into invT. For a central 95% critical value, enter 0.975, not 0.95. For a central 90% critical value, enter 0.95, not 0.90. First split the area outside the center between the two tails.
- Confusing central area with total tail area. For a central 90% area, the central area is 0.90 and the total tail area is 0.10. Each tail contains 0.05. State which area you mean instead of calling all of them “the tail.”
- Using the wrong table column. Check whether the heading gives one-tail area, two-tail area, or confidence level. For a central 95% critical value, the matching one-tail area is 0.025 and the matching two-tail area is 0.05.
- Using \(n\) for the degrees of freedom. A one-sample t procedure uses \(df=n-1\). A sample of 15 has 14 degrees of freedom, not 15.
- Forgetting that invT uses area to the left. The positive cutoff has most of the distribution to its left. For a central 99% area, the correct input is 0.995, not the 0.005 area in the upper tail.
- Mixing up a central confidence level and a one-sided percentile. invT(0.90, \(df\)) gives the 90th percentile, with 90% to the left. It is not the positive critical value for a central 90% area; that requires invT(0.95, \(df\)).
- Reporting only a calculator result. Show \(df\), the tail-area reasoning, and the invT inputs or the table row and column. For example: “With \(n=15\), \(df=14\). A central 95% level leaves 0.025 in each tail, so the left area at \(t^*\) is 0.975; therefore \(t^*\approx2.145\).”
A quick reasonableness check can catch many lookup errors. At a fixed confidence level, smaller \(df\) gives a larger \(t^*\), because the t distribution has heavier tails. At fixed \(df\), a higher confidence level gives a larger \(t^*\), because the cutoff must move farther from zero to capture more central area. A result that goes in the opposite direction is a reason to recheck the row, column, and tail area.
Check Your Understanding
For each question, distinguish the central confidence level from the area entered into invT.
- A one-sample t procedure has a sample size of 18. What are its degrees of freedom?
- For a central 95% t area, what is the total tail area, what is the area in each tail, and what left-tail area should be entered into invT for the positive cutoff?
- For a central 90% area with \(df=9\), explain why the invT input is 0.95 rather than 0.90.
- A t table labels columns by two-tail area. Which column should you use for a central 99% critical value?
- With the degrees of freedom fixed, should the 99% critical value be larger or smaller than the 90% critical value? Explain using the area under the curve.