# Critical t on the TI-83/84/89

Copyright © 2001–2008 by Stan Brown, Oak Road Systems

Copyright © 2001–2008 by Stan Brown, Oak Road Systems

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).**

**Contents:**

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.

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

- Press [
`F5`

] [`2`

] [`2`

] for inverse t. - 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.
- Enter the degrees of freedom (27), and press the [
`ENTER`

] key.

Answer: t(27,0.025) = 2.0518

Try this method with the examples below.

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:

- Press [
`CATALOG`

] [`F3`

] [*plain*`9`

*makes*`I`

] [`ENTER`

] for`TIStat.inv_t(`

. - Enter the area of the
*left-hand tail*. Since 0.025 is the right-hand tail, 1−0.025 is the left-hand tail. - Press the comma and enter the degrees of freedom, 27.
- Press [
`)`

] [`ENTER`

].

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

= 2.0518

Try this method with the examples below.

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:

- Press [
`2nd`

`VARS`

*makes*`DISTR`

] [`4`

] for`invT(`

. - Enter the area of the
*left-hand tail*. Since 0.025 is the right-hand tail, 1−0.025 is the left-hand tail. - Press the comma and enter the degrees of freedom, 27.
- Press [
`)`

] [`ENTER`

].

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

= 2.0518

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:

- Select TInterval using the keystrokes you already know,
[
`STAT`

] [`◄`

] [`8`

]. - Pick “Stats”.
- Set x̄=0.
- 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). - For n, again enter the sample size
*df*+1. - 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.
- 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.

Answer: t(40,0.005) = 2.7045

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

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.

What is the maximum error of estimate for a confidence interval when n=11, x̄=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 x̄: 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`

, x̄: 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.

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