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One-sample t confidence intervals · Tutorial 624 of 1000

Finding t* Critical Values With a Table or invT

Learn to match a confidence level and degrees of freedom to the correct t critical value using a table or invT.

Intermediate 9 min read

What You'll Learn

  • Connect a central confidence level to the total area in the tails of a t distribution
  • Find a positive t critical value using the correct degrees-of-freedom row and tail column
  • Use invT with the cumulative area to the left of the positive cutoff
  • Distinguish central confidence levels from one-sided cumulative areas
  • Check whether a table value or calculator result matches the intended confidence level and degrees of freedom

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.

Definition: For a central confidence level \(C\), the t critical value \(t^*\) with \(df\) degrees of freedom is the positive cutoff that leaves area \(C\) between \(-t^*\) and \(t^*\). The remaining area, \(1-C\), is split equally between the two tails.

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 \((1+C)/2\).

$$ \text{Left-tail area for }t^* = 1-\frac{\alpha}{2} =\frac{1+C}{2}, \qquad \alpha=1-C $$

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.

Formula: For a two-sided t critical value at confidence level \(C\), use \(t^*=\text{invT}\big((1+C)/2,\ df\big)\). The first input is the cumulative area to the left of the positive cutoff, not the central confidence level itself.

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 levelTotal area in both tailsArea in each tailLeft area at positive t*
90%0.100.050.95
95%0.050.0250.975
99%0.010.0050.995

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.

1
Find the degrees of freedom.
For a one-sample t procedure, calculate \(df=n-1\).
2
Find the area in each tail.
For confidence level \(C\), calculate \(\alpha=1-C\), then divide \(\alpha\) by 2.
3
Find the left area at the positive cutoff.
Calculate \(1-\alpha/2\), which is also \((1+C)/2\).
4
Use invT.
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.

Key idea: For two-sided confidence levels, split the area outside the central region between both tails. Then use the cumulative area to the left of the positive cutoff in invT. A central 90% level uses invT area 0.95; a central 95% level uses 0.975.

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

$$ df=n-1=15-1=14. $$

Match the confidence level to the table. The central confidence level is 95%, so the total tail area is

$$ \alpha=1-0.95=0.05. $$

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

$$ \text{invT}(0.975,14)\approx 2.1448\approx 2.145. $$

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 \(n=10\),

$$ df=n-1=10-1=9. $$

Find the correct invT area. The central area is \(C=0.90\), so the total area outside it is

$$ \alpha=1-C=1-0.90=0.10. $$

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.

$$ t^*=\text{invT}(0.95,9)\approx1.8331\approx1.833. $$

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

$$ \alpha=1-0.99=0.01, \qquad \frac{\alpha}{2}=0.005. $$

The cumulative area to the left of the positive cutoff is \(1-0.005=0.995\). Thus,

$$ t^*=\text{invT}(0.995,24)\approx2.7969\approx2.797. $$

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.

Key takeaway: Find \(df=n-1\), split the area outside a central confidence level equally between the tails, and use the left cumulative area at the positive cutoff. A table and invT should give matching critical values when the degrees of freedom and tail areas agree.

Check Your Understanding

For each question, distinguish the central confidence level from the area entered into invT.

  1. A one-sample t procedure has a sample size of 18. What are its degrees of freedom?
  2. 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?
  3. For a central 90% area with \(df=9\), explain why the invT input is 0.95 rather than 0.90.
  4. A t table labels columns by two-tail area. Which column should you use for a central 99% critical value?
  5. 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.