Critical t on the TI-83/84/89

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.

Answer: t(27,0.025) = 2.0518

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 x̄=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.
Answer: t(40,0.005) = 2.7045

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, 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.