TC3 → Stan Brown → TI-83/84/89 → Critical t
revised Jan 27, 2008

# Critical t on the TI-83/84/89

Summary:

In class we compute confidence intervals directly, using the Stat Tests menu of the TI-83 or Ints menu of the TI-89. Therefore we never need to compute inverse t or critical t on its own. However, you would need to find critical t in the classical approach to hypothesis test, for determining sample size, or for other applications. This note explains how to do it on the TI-83, TI-84, or TI-89.

For all three models, let’s assume that you want a 95% confidence interval with a sample size of 28. df = n−1 = 27, and the area in one tail is (100%−5%)÷2 = 2.5% or 0.025, so you need t(27,0.025).

## Notation

t(df, rtail) is concerned with the t distribution for df degrees of freedom (df = n−1). It is the t value such that the area in the right-hand tail is rtail. This t value is also known as critical t.

Caution: Some textbooks write the t function the other way, t(rtail,df). Since df is a whole number and rtail is a decimal between 0 and 1, you will be able to adapt.

## TI-89 Procedure

### Method 1: Fill in the Blanks

The flash application, Stats/List Editor, can compute critical t for any α/2 and df directly:

1. Press [`F5`] [`2`] [`2`] for inverse t.
2. In the “area” box, enter the area of the left-hand tail. For example, if you’re finding t(27,0.025), the right-hand tail is 0.025 and the left-hand tail is 1−0.025.
3. Enter the degrees of freedom (27), and press the [`ENTER`] key.

Try this method with the examples below.

### Method 2: Function

You can also find critical t right on the main screen by using the `invT` function. This is particularly handy when you intend to use the number in further calculations. The function format is `TIStat.inv_t(`ltail,df`)`. Here’s the procedure:

1. Press [`CATALOG`] [`F3`] [plain `9` makes `I`] [`ENTER`] for `TIStat.inv_t(`.
2. Enter the area of the left-hand tail. Since 0.025 is the right-hand tail, 1−0.025 is the left-hand tail.
3. Press the comma and enter the degrees of freedom, 27.
4. Press [`)`] [`ENTER`].

Answer: t(27,0.025) = `TIStat.inv_t(1−.025,27)` = 2.0518

Try this method with the examples below.

## TI-84 Procedure

The TI-84 offers direct access to inverse t — that’s one of its few differences from the TI-83. The format is `invT(`ltail,df`)`. Suppose you want to find critical t for 27 degrees of freedom and a 0.025 one-tailed test. Here’s the procedure:

1. Press [`2nd` `VARS` makes `DISTR`] [`4`] for `invT(`.
2. Enter the area of the left-hand tail. Since 0.025 is the right-hand tail, 1−0.025 is the left-hand tail.
3. Press the comma and enter the degrees of freedom, 27.
4. Press [`)`] [`ENTER`].

Answer: t(27,0.025) = `invT(1−.025,27)` = 2.0518

## TI-83 Procedure

The TI-83 doesn’t have a specific function for t(df, α/2) but you can get at it by a “back door”. Here’s how you would find critical t for 27 degrees of freedom in a 0.025 one-tailed test:

1. Select TInterval using the keystrokes you already know, [`STAT`] [`◄`] [`8`].
2. Pick “Stats”.
3. Set =0.
4. For Sx, you want the square root of the sample size. Press [`2nd` `x²` makes `√`] and then enter the sample size (which is df+1).
5. For n, again enter the sample size df+1.
6. For C-Level, enter the area in the middle (the nonrejection region). You can enter it as a calculation, 1−2× (right-hand tail). For instance, in computing t(27,0.025), you can enter 1−2×0.025 or just enter .95 directly.
7. Select `CALCULATE` and read the upper bound of the interval. That is the inverse t you’re looking for. In this case the answer is 2.0518.

Example: What is t(40,0.005)?

Solution: df=40 means n=41. If the area in one tail is 0.005, the area in two tails is 0.01 and the area in the nonrejection region is 1−0.01 = 0.99. Use the TInterval screen as shown below left. Sx is √41, which will change to 6.403... as soon as you press [`ENTER`]. The output screen is shown below right, and the upper end of the interval is the desired value.

## Example: Inverse t

What is t(3,0.01)?

TI-83 solution: df=3 ⇒ n=4; one tail=0.01 ⇒ nonrejection region=0.98. In TInterval, set Sx= √4, n=4, and C-Level=0.98. Answer: t(3,0.01) = 4.5407

TI-84 solution: t(3,0.01) = `invT(1−0.01,3)` = 4.5407

TI-89 solution: [`F5`] [`2`] [`2`]. Area = 1−0.01 and df = 3. Answer: t(3,0.01) = 4.5407

## Example: Critical t

Your sample size is 14. For a significance level of 0.05, what is the critical value of t?

Solution: for sample size 14, there are 13 degrees of freedom, so critical t is t(27,0.05). Use one of the methods from the previous example to find t(27,0.05) = 1.7709.

## Example: Margin of Error

What is the maximum error of estimate for a confidence interval when n=11, =45, s=6, with 95% confidence? Use the formula shown at right.

TI-83 solution: first calculate t(df,α/2), then use it in the formula for E.

(a) n=11 ⇒ df=10; 1−α=0.95 ⇒ α/2=0.025; therefore t(10,0.025) is needed. Compute it using TInterval with : 0, Sx:  √11 not the standard deviation of the sample, n: 11, C-Level: .95. Result: the critical t is t(10,0.025) = 2.2281.

(b) Use that in the formula: press the [`×`] key to chain calculations, then enter 6/√11. Answer: 4.03.

Check: [`STAT`] [`◄`] [`8`] for TInterval. Select `Stats`, : 45, Sx: 6, n: 11, C-Level: .95. The interval is (40.969 to 49.031). Half the width of that interval is 4.03, which agrees with the result calculated the long way.

TI-84 solution:

(a) n=11 ⇒ df=10; 1−α=0.95 ⇒ α/2=0.025; therefore t(10,0.025) is needed. Compute it using `invT(1−0.025,10)` = 2.2281.

(b) Use that in the formula: press the [`×`] key to chain calculations, then enter 6/√11. Answer: 4.03.

Check: same as for TI-83.

TI-89 solution: [`2nd` `F2` makes `F7`] [`2`] and select Stats. Enter the numbers, and read off the margin of error (ME) as 4.03.

This page is used in instruction at Tompkins Cortland Community College in Dryden, New York; it’s not an official statement of the College. Please visit www.tc3.edu/instruct/sbrown/ to report errors or ask to copy it.

For updates and new info, go to http://www.tc3.edu/instruct/sbrown/ti83/