Critical t-Scores
This document concerns finding critical t-scores for constructing confidence
intervals and for conducting hypothesis tests at a given level of significance
using the TI-83/84 calculators. We proceed by example. You should review
the calculator document about the t-distribution.
Example 1: Find the critical t-score for a 93% confidence interval with 15
degrees-of-freedom.
Solution: The critical t-score, t0 , places 93% of the area in the t distribution with 15 df between it and its additive inverse, −t0 . You should draw
a picture of the distribution and place the score in the distribution. The
picture shows that 3.5% of the total area is left of −t0 . Therefore the total
area left of t0 itself is 96.5%. So the value of t0 is given by the command
invT(0.965, 15) on most TI-84 calculators.
If you do not have the command invT on your calculator (TI-83, some
versions of the TI-84), you can use the
√ command TInterval (STAT/TESTS)
as follows: Set x̄ = 0, set Sx = 16, set n = 16 (df = n − 1) and set
confidence level to 0.93. The right-hand endpoint of the interval given is
t0 .
Example 2: Find the critical t-score with 9 df, t0 , for a right-tailed test at
the α = 0.04 level of significance.
Solution: If you have the command invT available, use invT(0.96, 9) since
the area left of t0 is 0.96.
If not, note that between −t0 and t0 the area is 0.92 since the area left
of −t0√is also 0.04 (a picture is helpful). Now use TInterval with x̄ = 0,
Sx = 10, n = 10, confidence level 0.92. The right-hand endpoint of the
resulting interval is t0 .
Example 3: Find critical t-score with 9 df, t0 , for a left-tailed test at the
α = 0.04 level of significance.
Solution: If you have the command invT available, use invT(0.04, 9) since
the area left of t0 is 0.04.
If not, note that t0 is negative and the area between t0 and −t0 the area
is 0.92√as above (a picture is helpful). Now use TInterval with x̄ = 0,
Sx = 10, n = 10. The left-hand endpoint of the resulting interval is t0 .
Example 4: Find the critical t-scores with 9 df, t−0 and t0 , for a two-tailed
test at the α = 0.04 level of significance.
Solution: The between −t0 and t0 is 0.96 hence the area left of t0 is 0.98.
If you have the command invT available, use invT(0.98, 9) since the area
left of t0 is 0.98.
If not, note that the area between −t0 and t0 the area is 0.96
√ as above(a
picture is helpful). Now use TInterval with x̄ = 0, Sx = 10, n = 10,
confidence level is 0.96. The two endpoints of the resulting interval are −t0
and t0 .